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How to find the sum of this series : $1+frac12+ frac13+frac14+dots+frac1n$


Is there a partial sum formula for the Harmonic Series?Summation of 1/kFinding the sum of this series: $1+frac 12 + frac 13 + cdots + frac 150$Closed form of $sum_i=1^N frac1i$?what is the formula to calculate the irreducible fraction series as sum?Summation series formulaQ: Prove 1+1/2+1/3+…+1/(n-1) = lnn+c+o(n)Why does the series $sum_n=1^inftyfrac1n$ not converge?Is there any formula for the series $1 + frac12 + frac13 + cdots + frac 1 n = ?$What would Gauss do in this case: adding $1+frac12+frac13+frac14+ dots +frac1100$?How to find the sum of the sequence $frac11+1^2+1^4 +frac21+2^2+2^4 +frac31+3^2+3^4+…$If an A.P ( Arithmetic Progression) , a G.P( Geometric Progression) and a H.P ( Harmonic Progression) ..Find the sum $frac1sqrt1+sqrt2 + frac1sqrt2+sqrt3 + …+ frac1sqrt99+sqrt100$Find the sum of the following infinite seriesFinding the sum of this series: $1+frac 12 + frac 13 + cdots + frac 150$Find sum of first $n$ terms of the series : $1+frac1^3+2^31+2+frac1^3+2^3+3^31+2+3+dots$Find sum of the infinite seriesWhat is the proper way of approaching this question about related AP, GP and HP series?Alternative ways to find the sum of this seriesHow to find the total number of terms and the sum in a not explicit geometric series?













33












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Problem :



How to find the sum of this series : $1+frac12+ frac13+frac14+dots+frac1n$



This is a Harmonic progression : So is this formula correct to sum the series :



$frac(number ~of ~terms)^2sum~ of~ all ~the~ denominators$



$Rightarrow $ if $frac1A + frac1B +frac1C$ are in H.P.



Therefore the sum of the series can be written as :



$Rightarrow frac(3)^3(A+B+C)$



Is this correct please suggest.










share|cite|improve this question











$endgroup$







  • 4




    $begingroup$
    possible duplicate of Why does the series $frac 1 1 + frac 12 + frac 13 + cdots$ not converge?
    $endgroup$
    – Zev Chonoles
    Jul 25 '13 at 2:53






  • 6




    $begingroup$
    Or are you trying to come up with a closed form for the $n$th harmonic number?
    $endgroup$
    – Zev Chonoles
    Jul 25 '13 at 2:54






  • 1




    $begingroup$
    How did you arrive at that formula?? Plus are you aware of convergence tests for series? The simple ones should suffice.
    $endgroup$
    – Vishesh
    Jul 25 '13 at 3:02






  • 1




    $begingroup$
    duplicate of math.stackexchange.com/questions/163921/harmonic-series ?
    $endgroup$
    – Arjang
    Jul 25 '13 at 6:51







  • 5




    $begingroup$
    Actually $frac1A+frac1B+frac1C=fracAB+BC+CAABC neq frac(3)^3(A+B+C)$.
    $endgroup$
    – Pentapolis
    Oct 12 '16 at 21:48















33












$begingroup$


Problem :



How to find the sum of this series : $1+frac12+ frac13+frac14+dots+frac1n$



This is a Harmonic progression : So is this formula correct to sum the series :



$frac(number ~of ~terms)^2sum~ of~ all ~the~ denominators$



$Rightarrow $ if $frac1A + frac1B +frac1C$ are in H.P.



Therefore the sum of the series can be written as :



$Rightarrow frac(3)^3(A+B+C)$



Is this correct please suggest.










share|cite|improve this question











$endgroup$







  • 4




    $begingroup$
    possible duplicate of Why does the series $frac 1 1 + frac 12 + frac 13 + cdots$ not converge?
    $endgroup$
    – Zev Chonoles
    Jul 25 '13 at 2:53






  • 6




    $begingroup$
    Or are you trying to come up with a closed form for the $n$th harmonic number?
    $endgroup$
    – Zev Chonoles
    Jul 25 '13 at 2:54






  • 1




    $begingroup$
    How did you arrive at that formula?? Plus are you aware of convergence tests for series? The simple ones should suffice.
    $endgroup$
    – Vishesh
    Jul 25 '13 at 3:02






  • 1




    $begingroup$
    duplicate of math.stackexchange.com/questions/163921/harmonic-series ?
    $endgroup$
    – Arjang
    Jul 25 '13 at 6:51







  • 5




    $begingroup$
    Actually $frac1A+frac1B+frac1C=fracAB+BC+CAABC neq frac(3)^3(A+B+C)$.
    $endgroup$
    – Pentapolis
    Oct 12 '16 at 21:48













33












33








33


26



$begingroup$


Problem :



How to find the sum of this series : $1+frac12+ frac13+frac14+dots+frac1n$



This is a Harmonic progression : So is this formula correct to sum the series :



$frac(number ~of ~terms)^2sum~ of~ all ~the~ denominators$



$Rightarrow $ if $frac1A + frac1B +frac1C$ are in H.P.



Therefore the sum of the series can be written as :



$Rightarrow frac(3)^3(A+B+C)$



Is this correct please suggest.










share|cite|improve this question











$endgroup$




Problem :



How to find the sum of this series : $1+frac12+ frac13+frac14+dots+frac1n$



This is a Harmonic progression : So is this formula correct to sum the series :



$frac(number ~of ~terms)^2sum~ of~ all ~the~ denominators$



$Rightarrow $ if $frac1A + frac1B +frac1C$ are in H.P.



Therefore the sum of the series can be written as :



$Rightarrow frac(3)^3(A+B+C)$



Is this correct please suggest.







sequences-and-series algebra-precalculus summation harmonic-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 3 '18 at 9:09









Abcd

3,05531337




3,05531337










asked Jul 25 '13 at 2:51









sachinsachin

4,323554119




4,323554119







  • 4




    $begingroup$
    possible duplicate of Why does the series $frac 1 1 + frac 12 + frac 13 + cdots$ not converge?
    $endgroup$
    – Zev Chonoles
    Jul 25 '13 at 2:53






  • 6




    $begingroup$
    Or are you trying to come up with a closed form for the $n$th harmonic number?
    $endgroup$
    – Zev Chonoles
    Jul 25 '13 at 2:54






  • 1




    $begingroup$
    How did you arrive at that formula?? Plus are you aware of convergence tests for series? The simple ones should suffice.
    $endgroup$
    – Vishesh
    Jul 25 '13 at 3:02






  • 1




    $begingroup$
    duplicate of math.stackexchange.com/questions/163921/harmonic-series ?
    $endgroup$
    – Arjang
    Jul 25 '13 at 6:51







  • 5




    $begingroup$
    Actually $frac1A+frac1B+frac1C=fracAB+BC+CAABC neq frac(3)^3(A+B+C)$.
    $endgroup$
    – Pentapolis
    Oct 12 '16 at 21:48












  • 4




    $begingroup$
    possible duplicate of Why does the series $frac 1 1 + frac 12 + frac 13 + cdots$ not converge?
    $endgroup$
    – Zev Chonoles
    Jul 25 '13 at 2:53






  • 6




    $begingroup$
    Or are you trying to come up with a closed form for the $n$th harmonic number?
    $endgroup$
    – Zev Chonoles
    Jul 25 '13 at 2:54






  • 1




    $begingroup$
    How did you arrive at that formula?? Plus are you aware of convergence tests for series? The simple ones should suffice.
    $endgroup$
    – Vishesh
    Jul 25 '13 at 3:02






  • 1




    $begingroup$
    duplicate of math.stackexchange.com/questions/163921/harmonic-series ?
    $endgroup$
    – Arjang
    Jul 25 '13 at 6:51







  • 5




    $begingroup$
    Actually $frac1A+frac1B+frac1C=fracAB+BC+CAABC neq frac(3)^3(A+B+C)$.
    $endgroup$
    – Pentapolis
    Oct 12 '16 at 21:48







4




4




$begingroup$
possible duplicate of Why does the series $frac 1 1 + frac 12 + frac 13 + cdots$ not converge?
$endgroup$
– Zev Chonoles
Jul 25 '13 at 2:53




$begingroup$
possible duplicate of Why does the series $frac 1 1 + frac 12 + frac 13 + cdots$ not converge?
$endgroup$
– Zev Chonoles
Jul 25 '13 at 2:53




6




6




$begingroup$
Or are you trying to come up with a closed form for the $n$th harmonic number?
$endgroup$
– Zev Chonoles
Jul 25 '13 at 2:54




$begingroup$
Or are you trying to come up with a closed form for the $n$th harmonic number?
$endgroup$
– Zev Chonoles
Jul 25 '13 at 2:54




1




1




$begingroup$
How did you arrive at that formula?? Plus are you aware of convergence tests for series? The simple ones should suffice.
$endgroup$
– Vishesh
Jul 25 '13 at 3:02




$begingroup$
How did you arrive at that formula?? Plus are you aware of convergence tests for series? The simple ones should suffice.
$endgroup$
– Vishesh
Jul 25 '13 at 3:02




1




1




$begingroup$
duplicate of math.stackexchange.com/questions/163921/harmonic-series ?
$endgroup$
– Arjang
Jul 25 '13 at 6:51





$begingroup$
duplicate of math.stackexchange.com/questions/163921/harmonic-series ?
$endgroup$
– Arjang
Jul 25 '13 at 6:51





5




5




$begingroup$
Actually $frac1A+frac1B+frac1C=fracAB+BC+CAABC neq frac(3)^3(A+B+C)$.
$endgroup$
– Pentapolis
Oct 12 '16 at 21:48




$begingroup$
Actually $frac1A+frac1B+frac1C=fracAB+BC+CAABC neq frac(3)^3(A+B+C)$.
$endgroup$
– Pentapolis
Oct 12 '16 at 21:48










5 Answers
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The exact expression for $displaystyle H_n:=1+frac12+frac13+cdots +frac1n $ is not known, but you can estimate $H_n$ as below



Let us consider the area under the curve $displaystyle frac1x$ when $x$ varies from $1$ to $n$.



Now note that $displaystyle H_n-frac1n=1+frac12+frac13+cdots +frac1n-1$ is an overestimation of this area by rectangles. See below



Overestimation



And $displaystyle H_n-1=frac12+frac13+cdots +frac1n $ is an underestimation of the area. See below



Underestimation
(source: uark.edu)



Hence $$large H_n-1<int_1^nfrac1xdx<H_n-frac1n\
Rightarrow ln n+frac1n<H_n<ln n+1$$



Also, Euler discovered this beautiful property of harmonic number $H_n$ that $$large lim_nrightarrow inftyleft(H_n-ln nright)=gammaapprox 0.57721566490153286060651209008240243104215933593992…$$ $gamma$ is called the Euler-Mascheroni constant.






share|cite|improve this answer











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  • 9




    $begingroup$
    I wouldn't really call this a beautiful property of harmonic numbers, when this is in fact usually taken to be the definition of the Euler-Mascheroni constant.
    $endgroup$
    – Ethan
    Jul 25 '13 at 8:32







  • 14




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    But surely the fact that such a definition is allowed to EXIST is beautiful?
    $endgroup$
    – fretty
    Jul 25 '13 at 8:35






  • 1




    $begingroup$
    Yes @fretty , the fact that such wonderful relation exists, makes the Harmonic series more interesting and intriguing.
    $endgroup$
    – Samrat Mukhopadhyay
    Jul 25 '13 at 9:39






  • 1




    $begingroup$
    Yes, I know, I was trying to convince Ethan...
    $endgroup$
    – fretty
    Jul 26 '13 at 17:52






  • 1




    $begingroup$
    And I think you succeeded in that :-)
    $endgroup$
    – Samrat Mukhopadhyay
    Jul 26 '13 at 17:53


















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There are other ways to represent the harmonic series, for example, recall the geometric sum:



$$frac1-r^n1-r=1+r+r^2+dots+r^n-1$$



And integrate both sides with respect to $r$ from $0$ to $1$:



$$beginalignint_0^1frac1-r^n1-rdr&=int_0^11+r+r^2+dots+r^n-1dr\&=left.frac11r+frac12r^2+frac13r^3+dots+frac1nr^nright|_0^1\&=frac11+frac12+frac13+dots+frac1nendalign$$



So we can rewrite the harmonic series as



$$1+frac12+frac13+dots+frac1n=int_0^1frac1-r^n1-rdr$$



Which is most useful for deriving many formulas. I used in deriving a representation of the gamma function, for example.



Note that:



$$ln(Gamma(n+1))=ln(nGamma(n))=ln(Gamma(n))+ln(n)$$



Take the derivative of both sides to get



$$psi(n+1)=psi(n)+frac1n$$



Repeated application of this formula gives:



$$beginalignpsi(n+1)&=psi(n)+frac1n\&=psi(n-1)+frac1n-1+frac1n\&=psi(1)+1+frac12+frac13+dots+frac1n\psi(n+1)+gamma&=1+frac12+frac13+dots+frac1nendalign$$






share|cite|improve this answer











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  • $begingroup$
    simply beautiful :-)
    $endgroup$
    – gt6989b
    Oct 12 '18 at 15:00


















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That formula is not correct to sum the first few terms of the harmonic series. Trying it with even the first three would mean that $$frac3^21+2+3 = frac96 = 1.5 neq frac11 + frac12 + frac13$$.



The harmonic series actually diverges, so the sum of the series as we let $n$ get large doesn't exist... You can, however, get partial sums as the harmonic numbers, however this is somewhat outside the scope of the algebra/precalculus topic you have it listed under. You can find more information here.






share|cite|improve this answer









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    3












    $begingroup$

    While I won't take any credit for this approximation as I don't know if someone else discovered it before myself, I did stumble upon this while playing around with the bound $ln(n)+frac1n < H_n < 1+ln(n)$ for $n > 1$ which was mentioned above and the Euler-Mascheroni Constant.



    Kudos to the person who first discovered this approximation (assuming there was one before myself) if they happen to see this post.



    Approximation: $H_n approx ln(n) + frac1n + gammaleft(1+lnleft(fracnn+1right)right)$



    where $gamma = 0.577215664901532860606512...$ is the Euler-Mascheroni Constant.



    Some calculations to back up this approximation



    For $n = 100$



    Actual Value: $H_100 = 5.1873775...$



    Approximation: $H_100 approx 5.1866423...$



    Error: $-0.0007351...$



    Percent Error: $0.0141719...$%



    For $n = 1000$



    Actual Value: $H_1000 = 7.4854708...$



    Approximation: $H_1000 approx 7.4853940...$



    Error: $-0.0000768...$



    Perent Error: $0.0010265...$%



    I've also been able to stumble upon the fact that using the bound $ln(n+1) < H_n < 1+ln(n), n > 1$ (source: http://www.math.drexel.edu/~tolya/123_harmonic.pdf) allows you to get an approximation that is just a slight bit more inaccurate than the first one I presented at the top.



    Approximation: $H_n approx ln(n+1) + gammaleft(1+lnleft(fracnn+1right)right)$



    This approximation is simpler, and a slight bit easier to calculate. However, it is slightly less accurate, so unless you are desiring extreme precision, this one may be more appealing.



    The above two approximations are is relatively easy to calculate compared to some others you may find out there (like Ramanujan).



    A very simplified form of Ramanujan's approximation is



    $H_n approx ln(n) + gamma$



    The above two approximations I presented are both quite a bit more accurate than this truly oversimplified version of Ramanujan's approximation. As @Winther pointed out, there is an error of approximately $frac0.077n$ which implies that (just like pretty much all other approximations) the approximations are much closer to the actual value as $n$ gets larger. However, Ramanujan's approximation in its complete form is extremely accurate, though it is extremely complex.



    If anyone wishes to know the motivation behind the derivation of these approximations (at least from how I did it), then I am happy to answer in the comments.



    EDIT: I stumbled upon one that is more accurate, again kudos to whoever may have discovered it before me. Also, this is more accurate than the approximation $ln(n)+gamma+frac12n$ up to some point.



    $H_n = ln(n) + gammaleft(1+frac5051n+lnleft(fracn-fracgamma10n+fracgamma10right)right)$






    share|cite|improve this answer











    $endgroup$








    • 1




      $begingroup$
      Series approximations for the harmonic number (like this one) has been known since the 1700s (see e.g. en.wikipedia.org/wiki/Harmonic_number#Calculation). By series expanding the logarithm we can show that the error of your approximation is approximately $frac2(1-gamma)-12n approx - frac0.077n$ which agrees well with what you find. You can do a bit better by simply using the approximation $H_n approx log(n) + gamma + frac12n$. This also avoids the additional logarithm
      $endgroup$
      – Winther
      Nov 15 '16 at 3:45











    • $begingroup$
      @Winther That's pretty cool! Yeah I just stumbled upon this while playing around with the bounds so as I said before I am not claiming that I am the first to discover either of these two approximations. Though, they do seem to differ from the ones you suggest.
      $endgroup$
      – Thomas Kim
      Nov 15 '16 at 4:15










    • $begingroup$
      @Winther also, I know that these aren't the best approximations, but they are considerably good compared to some others that are out there.
      $endgroup$
      – Thomas Kim
      Nov 15 '16 at 4:29










    • $begingroup$
      @Winther Could you actually show the calculation for error? I don't really follow on how that works w.r.t. what I initially found.
      $endgroup$
      – Thomas Kim
      Nov 23 '16 at 3:31










    • $begingroup$
      It's based on using the approximation $log((n+1)/n) approx frac1n$ valid for large $n$ and comparing this with the full series for $H_n$. Using this we see that your first approximation is $approx log(n) + gamma + frac1n - fracgamman$ for large $n$ while the known series says that it's $approx log(n) + gamma + frac12n$ so the difference (error) is $approx frac2(1-gamma) - 12n$
      $endgroup$
      – Winther
      Nov 24 '16 at 13:41


















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    There is a new formula, that I've created. It can be written in a few different ways, below are two of them:



    $sum_k=1^nfrac1k=frac12n+piint_0^1 (1-u)cotpi uleft(1-cos2pi n uright),du$



    $sum_k=1^nfrac1k=frac12n+fracpi2int_0^1 (1-u)cotfracpi u2left(1-cospi n uright),du$



    There is also a generalization that goes beyond the harmonic numbers.
    For a harmonic progression where $a$ and $b$ are integers:



    $sum _k=1^n frac1a k+b=-frac12b+frac12(a n+b)+2piint_0^1 (1-u)sin[pi a n u]sin[(a n+2b)pi u]cot[pi a u],du$



    (Notice the formula at the top is a particular case where $a=1$, $b=0$.) For odd powers, the general formula is:
    beginmultline
    sum_j=1^nfrac1(a j+b)^2k+1=-frac12b^2k+1+frac12(a n+b)^2k+1+(-1)^k(2pi)^2k+1\ int_0^1sum_j=0^kfracB_2k-2jleft(2-2^2k-2jright)(2k-2j)!(2j+1)!(1-u)^2j+1sin[pi a nu]sin[(a n+2b)pi u]cot[pi au],du
    endmultline



    And for a harmonic progression where $a$ and $b$ are complex:
    beginmultlinenonumber
    sum_j=1^nfrac1(ai j+b)^k =-frac12b^k+frac12(ai n+b)^k+ e^-2pi b/aleft(frac2piaright)^k\ int_0^1sum_j=1^kfracphileft(e^-2pi b/a,-j+1,0right)(1-u)^k-j(j-1)!(k-j)!e^2pi bu/aleft(fracsin2pi n u2+i frac1-cos2pi n u2right)cotpi u,du
    endmultline



    The proofs are on papers that I've posted to the arXiv.






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      5 Answers
      5






      active

      oldest

      votes








      5 Answers
      5






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

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      37












      $begingroup$

      The exact expression for $displaystyle H_n:=1+frac12+frac13+cdots +frac1n $ is not known, but you can estimate $H_n$ as below



      Let us consider the area under the curve $displaystyle frac1x$ when $x$ varies from $1$ to $n$.



      Now note that $displaystyle H_n-frac1n=1+frac12+frac13+cdots +frac1n-1$ is an overestimation of this area by rectangles. See below



      Overestimation



      And $displaystyle H_n-1=frac12+frac13+cdots +frac1n $ is an underestimation of the area. See below



      Underestimation
      (source: uark.edu)



      Hence $$large H_n-1<int_1^nfrac1xdx<H_n-frac1n\
      Rightarrow ln n+frac1n<H_n<ln n+1$$



      Also, Euler discovered this beautiful property of harmonic number $H_n$ that $$large lim_nrightarrow inftyleft(H_n-ln nright)=gammaapprox 0.57721566490153286060651209008240243104215933593992…$$ $gamma$ is called the Euler-Mascheroni constant.






      share|cite|improve this answer











      $endgroup$








      • 9




        $begingroup$
        I wouldn't really call this a beautiful property of harmonic numbers, when this is in fact usually taken to be the definition of the Euler-Mascheroni constant.
        $endgroup$
        – Ethan
        Jul 25 '13 at 8:32







      • 14




        $begingroup$
        But surely the fact that such a definition is allowed to EXIST is beautiful?
        $endgroup$
        – fretty
        Jul 25 '13 at 8:35






      • 1




        $begingroup$
        Yes @fretty , the fact that such wonderful relation exists, makes the Harmonic series more interesting and intriguing.
        $endgroup$
        – Samrat Mukhopadhyay
        Jul 25 '13 at 9:39






      • 1




        $begingroup$
        Yes, I know, I was trying to convince Ethan...
        $endgroup$
        – fretty
        Jul 26 '13 at 17:52






      • 1




        $begingroup$
        And I think you succeeded in that :-)
        $endgroup$
        – Samrat Mukhopadhyay
        Jul 26 '13 at 17:53















      37












      $begingroup$

      The exact expression for $displaystyle H_n:=1+frac12+frac13+cdots +frac1n $ is not known, but you can estimate $H_n$ as below



      Let us consider the area under the curve $displaystyle frac1x$ when $x$ varies from $1$ to $n$.



      Now note that $displaystyle H_n-frac1n=1+frac12+frac13+cdots +frac1n-1$ is an overestimation of this area by rectangles. See below



      Overestimation



      And $displaystyle H_n-1=frac12+frac13+cdots +frac1n $ is an underestimation of the area. See below



      Underestimation
      (source: uark.edu)



      Hence $$large H_n-1<int_1^nfrac1xdx<H_n-frac1n\
      Rightarrow ln n+frac1n<H_n<ln n+1$$



      Also, Euler discovered this beautiful property of harmonic number $H_n$ that $$large lim_nrightarrow inftyleft(H_n-ln nright)=gammaapprox 0.57721566490153286060651209008240243104215933593992…$$ $gamma$ is called the Euler-Mascheroni constant.






      share|cite|improve this answer











      $endgroup$








      • 9




        $begingroup$
        I wouldn't really call this a beautiful property of harmonic numbers, when this is in fact usually taken to be the definition of the Euler-Mascheroni constant.
        $endgroup$
        – Ethan
        Jul 25 '13 at 8:32







      • 14




        $begingroup$
        But surely the fact that such a definition is allowed to EXIST is beautiful?
        $endgroup$
        – fretty
        Jul 25 '13 at 8:35






      • 1




        $begingroup$
        Yes @fretty , the fact that such wonderful relation exists, makes the Harmonic series more interesting and intriguing.
        $endgroup$
        – Samrat Mukhopadhyay
        Jul 25 '13 at 9:39






      • 1




        $begingroup$
        Yes, I know, I was trying to convince Ethan...
        $endgroup$
        – fretty
        Jul 26 '13 at 17:52






      • 1




        $begingroup$
        And I think you succeeded in that :-)
        $endgroup$
        – Samrat Mukhopadhyay
        Jul 26 '13 at 17:53













      37












      37








      37





      $begingroup$

      The exact expression for $displaystyle H_n:=1+frac12+frac13+cdots +frac1n $ is not known, but you can estimate $H_n$ as below



      Let us consider the area under the curve $displaystyle frac1x$ when $x$ varies from $1$ to $n$.



      Now note that $displaystyle H_n-frac1n=1+frac12+frac13+cdots +frac1n-1$ is an overestimation of this area by rectangles. See below



      Overestimation



      And $displaystyle H_n-1=frac12+frac13+cdots +frac1n $ is an underestimation of the area. See below



      Underestimation
      (source: uark.edu)



      Hence $$large H_n-1<int_1^nfrac1xdx<H_n-frac1n\
      Rightarrow ln n+frac1n<H_n<ln n+1$$



      Also, Euler discovered this beautiful property of harmonic number $H_n$ that $$large lim_nrightarrow inftyleft(H_n-ln nright)=gammaapprox 0.57721566490153286060651209008240243104215933593992…$$ $gamma$ is called the Euler-Mascheroni constant.






      share|cite|improve this answer











      $endgroup$



      The exact expression for $displaystyle H_n:=1+frac12+frac13+cdots +frac1n $ is not known, but you can estimate $H_n$ as below



      Let us consider the area under the curve $displaystyle frac1x$ when $x$ varies from $1$ to $n$.



      Now note that $displaystyle H_n-frac1n=1+frac12+frac13+cdots +frac1n-1$ is an overestimation of this area by rectangles. See below



      Overestimation



      And $displaystyle H_n-1=frac12+frac13+cdots +frac1n $ is an underestimation of the area. See below



      Underestimation
      (source: uark.edu)



      Hence $$large H_n-1<int_1^nfrac1xdx<H_n-frac1n\
      Rightarrow ln n+frac1n<H_n<ln n+1$$



      Also, Euler discovered this beautiful property of harmonic number $H_n$ that $$large lim_nrightarrow inftyleft(H_n-ln nright)=gammaapprox 0.57721566490153286060651209008240243104215933593992…$$ $gamma$ is called the Euler-Mascheroni constant.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited 2 days ago









      Glorfindel

      3,42981830




      3,42981830










      answered Jul 25 '13 at 6:18









      Samrat MukhopadhyaySamrat Mukhopadhyay

      13.8k2047




      13.8k2047







      • 9




        $begingroup$
        I wouldn't really call this a beautiful property of harmonic numbers, when this is in fact usually taken to be the definition of the Euler-Mascheroni constant.
        $endgroup$
        – Ethan
        Jul 25 '13 at 8:32







      • 14




        $begingroup$
        But surely the fact that such a definition is allowed to EXIST is beautiful?
        $endgroup$
        – fretty
        Jul 25 '13 at 8:35






      • 1




        $begingroup$
        Yes @fretty , the fact that such wonderful relation exists, makes the Harmonic series more interesting and intriguing.
        $endgroup$
        – Samrat Mukhopadhyay
        Jul 25 '13 at 9:39






      • 1




        $begingroup$
        Yes, I know, I was trying to convince Ethan...
        $endgroup$
        – fretty
        Jul 26 '13 at 17:52






      • 1




        $begingroup$
        And I think you succeeded in that :-)
        $endgroup$
        – Samrat Mukhopadhyay
        Jul 26 '13 at 17:53












      • 9




        $begingroup$
        I wouldn't really call this a beautiful property of harmonic numbers, when this is in fact usually taken to be the definition of the Euler-Mascheroni constant.
        $endgroup$
        – Ethan
        Jul 25 '13 at 8:32







      • 14




        $begingroup$
        But surely the fact that such a definition is allowed to EXIST is beautiful?
        $endgroup$
        – fretty
        Jul 25 '13 at 8:35






      • 1




        $begingroup$
        Yes @fretty , the fact that such wonderful relation exists, makes the Harmonic series more interesting and intriguing.
        $endgroup$
        – Samrat Mukhopadhyay
        Jul 25 '13 at 9:39






      • 1




        $begingroup$
        Yes, I know, I was trying to convince Ethan...
        $endgroup$
        – fretty
        Jul 26 '13 at 17:52






      • 1




        $begingroup$
        And I think you succeeded in that :-)
        $endgroup$
        – Samrat Mukhopadhyay
        Jul 26 '13 at 17:53







      9




      9




      $begingroup$
      I wouldn't really call this a beautiful property of harmonic numbers, when this is in fact usually taken to be the definition of the Euler-Mascheroni constant.
      $endgroup$
      – Ethan
      Jul 25 '13 at 8:32





      $begingroup$
      I wouldn't really call this a beautiful property of harmonic numbers, when this is in fact usually taken to be the definition of the Euler-Mascheroni constant.
      $endgroup$
      – Ethan
      Jul 25 '13 at 8:32





      14




      14




      $begingroup$
      But surely the fact that such a definition is allowed to EXIST is beautiful?
      $endgroup$
      – fretty
      Jul 25 '13 at 8:35




      $begingroup$
      But surely the fact that such a definition is allowed to EXIST is beautiful?
      $endgroup$
      – fretty
      Jul 25 '13 at 8:35




      1




      1




      $begingroup$
      Yes @fretty , the fact that such wonderful relation exists, makes the Harmonic series more interesting and intriguing.
      $endgroup$
      – Samrat Mukhopadhyay
      Jul 25 '13 at 9:39




      $begingroup$
      Yes @fretty , the fact that such wonderful relation exists, makes the Harmonic series more interesting and intriguing.
      $endgroup$
      – Samrat Mukhopadhyay
      Jul 25 '13 at 9:39




      1




      1




      $begingroup$
      Yes, I know, I was trying to convince Ethan...
      $endgroup$
      – fretty
      Jul 26 '13 at 17:52




      $begingroup$
      Yes, I know, I was trying to convince Ethan...
      $endgroup$
      – fretty
      Jul 26 '13 at 17:52




      1




      1




      $begingroup$
      And I think you succeeded in that :-)
      $endgroup$
      – Samrat Mukhopadhyay
      Jul 26 '13 at 17:53




      $begingroup$
      And I think you succeeded in that :-)
      $endgroup$
      – Samrat Mukhopadhyay
      Jul 26 '13 at 17:53











      15












      $begingroup$

      There are other ways to represent the harmonic series, for example, recall the geometric sum:



      $$frac1-r^n1-r=1+r+r^2+dots+r^n-1$$



      And integrate both sides with respect to $r$ from $0$ to $1$:



      $$beginalignint_0^1frac1-r^n1-rdr&=int_0^11+r+r^2+dots+r^n-1dr\&=left.frac11r+frac12r^2+frac13r^3+dots+frac1nr^nright|_0^1\&=frac11+frac12+frac13+dots+frac1nendalign$$



      So we can rewrite the harmonic series as



      $$1+frac12+frac13+dots+frac1n=int_0^1frac1-r^n1-rdr$$



      Which is most useful for deriving many formulas. I used in deriving a representation of the gamma function, for example.



      Note that:



      $$ln(Gamma(n+1))=ln(nGamma(n))=ln(Gamma(n))+ln(n)$$



      Take the derivative of both sides to get



      $$psi(n+1)=psi(n)+frac1n$$



      Repeated application of this formula gives:



      $$beginalignpsi(n+1)&=psi(n)+frac1n\&=psi(n-1)+frac1n-1+frac1n\&=psi(1)+1+frac12+frac13+dots+frac1n\psi(n+1)+gamma&=1+frac12+frac13+dots+frac1nendalign$$






      share|cite|improve this answer











      $endgroup$












      • $begingroup$
        simply beautiful :-)
        $endgroup$
        – gt6989b
        Oct 12 '18 at 15:00















      15












      $begingroup$

      There are other ways to represent the harmonic series, for example, recall the geometric sum:



      $$frac1-r^n1-r=1+r+r^2+dots+r^n-1$$



      And integrate both sides with respect to $r$ from $0$ to $1$:



      $$beginalignint_0^1frac1-r^n1-rdr&=int_0^11+r+r^2+dots+r^n-1dr\&=left.frac11r+frac12r^2+frac13r^3+dots+frac1nr^nright|_0^1\&=frac11+frac12+frac13+dots+frac1nendalign$$



      So we can rewrite the harmonic series as



      $$1+frac12+frac13+dots+frac1n=int_0^1frac1-r^n1-rdr$$



      Which is most useful for deriving many formulas. I used in deriving a representation of the gamma function, for example.



      Note that:



      $$ln(Gamma(n+1))=ln(nGamma(n))=ln(Gamma(n))+ln(n)$$



      Take the derivative of both sides to get



      $$psi(n+1)=psi(n)+frac1n$$



      Repeated application of this formula gives:



      $$beginalignpsi(n+1)&=psi(n)+frac1n\&=psi(n-1)+frac1n-1+frac1n\&=psi(1)+1+frac12+frac13+dots+frac1n\psi(n+1)+gamma&=1+frac12+frac13+dots+frac1nendalign$$






      share|cite|improve this answer











      $endgroup$












      • $begingroup$
        simply beautiful :-)
        $endgroup$
        – gt6989b
        Oct 12 '18 at 15:00













      15












      15








      15





      $begingroup$

      There are other ways to represent the harmonic series, for example, recall the geometric sum:



      $$frac1-r^n1-r=1+r+r^2+dots+r^n-1$$



      And integrate both sides with respect to $r$ from $0$ to $1$:



      $$beginalignint_0^1frac1-r^n1-rdr&=int_0^11+r+r^2+dots+r^n-1dr\&=left.frac11r+frac12r^2+frac13r^3+dots+frac1nr^nright|_0^1\&=frac11+frac12+frac13+dots+frac1nendalign$$



      So we can rewrite the harmonic series as



      $$1+frac12+frac13+dots+frac1n=int_0^1frac1-r^n1-rdr$$



      Which is most useful for deriving many formulas. I used in deriving a representation of the gamma function, for example.



      Note that:



      $$ln(Gamma(n+1))=ln(nGamma(n))=ln(Gamma(n))+ln(n)$$



      Take the derivative of both sides to get



      $$psi(n+1)=psi(n)+frac1n$$



      Repeated application of this formula gives:



      $$beginalignpsi(n+1)&=psi(n)+frac1n\&=psi(n-1)+frac1n-1+frac1n\&=psi(1)+1+frac12+frac13+dots+frac1n\psi(n+1)+gamma&=1+frac12+frac13+dots+frac1nendalign$$






      share|cite|improve this answer











      $endgroup$



      There are other ways to represent the harmonic series, for example, recall the geometric sum:



      $$frac1-r^n1-r=1+r+r^2+dots+r^n-1$$



      And integrate both sides with respect to $r$ from $0$ to $1$:



      $$beginalignint_0^1frac1-r^n1-rdr&=int_0^11+r+r^2+dots+r^n-1dr\&=left.frac11r+frac12r^2+frac13r^3+dots+frac1nr^nright|_0^1\&=frac11+frac12+frac13+dots+frac1nendalign$$



      So we can rewrite the harmonic series as



      $$1+frac12+frac13+dots+frac1n=int_0^1frac1-r^n1-rdr$$



      Which is most useful for deriving many formulas. I used in deriving a representation of the gamma function, for example.



      Note that:



      $$ln(Gamma(n+1))=ln(nGamma(n))=ln(Gamma(n))+ln(n)$$



      Take the derivative of both sides to get



      $$psi(n+1)=psi(n)+frac1n$$



      Repeated application of this formula gives:



      $$beginalignpsi(n+1)&=psi(n)+frac1n\&=psi(n-1)+frac1n-1+frac1n\&=psi(1)+1+frac12+frac13+dots+frac1n\psi(n+1)+gamma&=1+frac12+frac13+dots+frac1nendalign$$







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Apr 13 '17 at 12:21









      Community

      1




      1










      answered Oct 12 '16 at 21:11









      Simply Beautiful ArtSimply Beautiful Art

      50.7k579183




      50.7k579183











      • $begingroup$
        simply beautiful :-)
        $endgroup$
        – gt6989b
        Oct 12 '18 at 15:00
















      • $begingroup$
        simply beautiful :-)
        $endgroup$
        – gt6989b
        Oct 12 '18 at 15:00















      $begingroup$
      simply beautiful :-)
      $endgroup$
      – gt6989b
      Oct 12 '18 at 15:00




      $begingroup$
      simply beautiful :-)
      $endgroup$
      – gt6989b
      Oct 12 '18 at 15:00











      6












      $begingroup$

      That formula is not correct to sum the first few terms of the harmonic series. Trying it with even the first three would mean that $$frac3^21+2+3 = frac96 = 1.5 neq frac11 + frac12 + frac13$$.



      The harmonic series actually diverges, so the sum of the series as we let $n$ get large doesn't exist... You can, however, get partial sums as the harmonic numbers, however this is somewhat outside the scope of the algebra/precalculus topic you have it listed under. You can find more information here.






      share|cite|improve this answer









      $endgroup$

















        6












        $begingroup$

        That formula is not correct to sum the first few terms of the harmonic series. Trying it with even the first three would mean that $$frac3^21+2+3 = frac96 = 1.5 neq frac11 + frac12 + frac13$$.



        The harmonic series actually diverges, so the sum of the series as we let $n$ get large doesn't exist... You can, however, get partial sums as the harmonic numbers, however this is somewhat outside the scope of the algebra/precalculus topic you have it listed under. You can find more information here.






        share|cite|improve this answer









        $endgroup$















          6












          6








          6





          $begingroup$

          That formula is not correct to sum the first few terms of the harmonic series. Trying it with even the first three would mean that $$frac3^21+2+3 = frac96 = 1.5 neq frac11 + frac12 + frac13$$.



          The harmonic series actually diverges, so the sum of the series as we let $n$ get large doesn't exist... You can, however, get partial sums as the harmonic numbers, however this is somewhat outside the scope of the algebra/precalculus topic you have it listed under. You can find more information here.






          share|cite|improve this answer









          $endgroup$



          That formula is not correct to sum the first few terms of the harmonic series. Trying it with even the first three would mean that $$frac3^21+2+3 = frac96 = 1.5 neq frac11 + frac12 + frac13$$.



          The harmonic series actually diverges, so the sum of the series as we let $n$ get large doesn't exist... You can, however, get partial sums as the harmonic numbers, however this is somewhat outside the scope of the algebra/precalculus topic you have it listed under. You can find more information here.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jul 25 '13 at 3:45









          NessNess

          1134




          1134





















              3












              $begingroup$

              While I won't take any credit for this approximation as I don't know if someone else discovered it before myself, I did stumble upon this while playing around with the bound $ln(n)+frac1n < H_n < 1+ln(n)$ for $n > 1$ which was mentioned above and the Euler-Mascheroni Constant.



              Kudos to the person who first discovered this approximation (assuming there was one before myself) if they happen to see this post.



              Approximation: $H_n approx ln(n) + frac1n + gammaleft(1+lnleft(fracnn+1right)right)$



              where $gamma = 0.577215664901532860606512...$ is the Euler-Mascheroni Constant.



              Some calculations to back up this approximation



              For $n = 100$



              Actual Value: $H_100 = 5.1873775...$



              Approximation: $H_100 approx 5.1866423...$



              Error: $-0.0007351...$



              Percent Error: $0.0141719...$%



              For $n = 1000$



              Actual Value: $H_1000 = 7.4854708...$



              Approximation: $H_1000 approx 7.4853940...$



              Error: $-0.0000768...$



              Perent Error: $0.0010265...$%



              I've also been able to stumble upon the fact that using the bound $ln(n+1) < H_n < 1+ln(n), n > 1$ (source: http://www.math.drexel.edu/~tolya/123_harmonic.pdf) allows you to get an approximation that is just a slight bit more inaccurate than the first one I presented at the top.



              Approximation: $H_n approx ln(n+1) + gammaleft(1+lnleft(fracnn+1right)right)$



              This approximation is simpler, and a slight bit easier to calculate. However, it is slightly less accurate, so unless you are desiring extreme precision, this one may be more appealing.



              The above two approximations are is relatively easy to calculate compared to some others you may find out there (like Ramanujan).



              A very simplified form of Ramanujan's approximation is



              $H_n approx ln(n) + gamma$



              The above two approximations I presented are both quite a bit more accurate than this truly oversimplified version of Ramanujan's approximation. As @Winther pointed out, there is an error of approximately $frac0.077n$ which implies that (just like pretty much all other approximations) the approximations are much closer to the actual value as $n$ gets larger. However, Ramanujan's approximation in its complete form is extremely accurate, though it is extremely complex.



              If anyone wishes to know the motivation behind the derivation of these approximations (at least from how I did it), then I am happy to answer in the comments.



              EDIT: I stumbled upon one that is more accurate, again kudos to whoever may have discovered it before me. Also, this is more accurate than the approximation $ln(n)+gamma+frac12n$ up to some point.



              $H_n = ln(n) + gammaleft(1+frac5051n+lnleft(fracn-fracgamma10n+fracgamma10right)right)$






              share|cite|improve this answer











              $endgroup$








              • 1




                $begingroup$
                Series approximations for the harmonic number (like this one) has been known since the 1700s (see e.g. en.wikipedia.org/wiki/Harmonic_number#Calculation). By series expanding the logarithm we can show that the error of your approximation is approximately $frac2(1-gamma)-12n approx - frac0.077n$ which agrees well with what you find. You can do a bit better by simply using the approximation $H_n approx log(n) + gamma + frac12n$. This also avoids the additional logarithm
                $endgroup$
                – Winther
                Nov 15 '16 at 3:45











              • $begingroup$
                @Winther That's pretty cool! Yeah I just stumbled upon this while playing around with the bounds so as I said before I am not claiming that I am the first to discover either of these two approximations. Though, they do seem to differ from the ones you suggest.
                $endgroup$
                – Thomas Kim
                Nov 15 '16 at 4:15










              • $begingroup$
                @Winther also, I know that these aren't the best approximations, but they are considerably good compared to some others that are out there.
                $endgroup$
                – Thomas Kim
                Nov 15 '16 at 4:29










              • $begingroup$
                @Winther Could you actually show the calculation for error? I don't really follow on how that works w.r.t. what I initially found.
                $endgroup$
                – Thomas Kim
                Nov 23 '16 at 3:31










              • $begingroup$
                It's based on using the approximation $log((n+1)/n) approx frac1n$ valid for large $n$ and comparing this with the full series for $H_n$. Using this we see that your first approximation is $approx log(n) + gamma + frac1n - fracgamman$ for large $n$ while the known series says that it's $approx log(n) + gamma + frac12n$ so the difference (error) is $approx frac2(1-gamma) - 12n$
                $endgroup$
                – Winther
                Nov 24 '16 at 13:41















              3












              $begingroup$

              While I won't take any credit for this approximation as I don't know if someone else discovered it before myself, I did stumble upon this while playing around with the bound $ln(n)+frac1n < H_n < 1+ln(n)$ for $n > 1$ which was mentioned above and the Euler-Mascheroni Constant.



              Kudos to the person who first discovered this approximation (assuming there was one before myself) if they happen to see this post.



              Approximation: $H_n approx ln(n) + frac1n + gammaleft(1+lnleft(fracnn+1right)right)$



              where $gamma = 0.577215664901532860606512...$ is the Euler-Mascheroni Constant.



              Some calculations to back up this approximation



              For $n = 100$



              Actual Value: $H_100 = 5.1873775...$



              Approximation: $H_100 approx 5.1866423...$



              Error: $-0.0007351...$



              Percent Error: $0.0141719...$%



              For $n = 1000$



              Actual Value: $H_1000 = 7.4854708...$



              Approximation: $H_1000 approx 7.4853940...$



              Error: $-0.0000768...$



              Perent Error: $0.0010265...$%



              I've also been able to stumble upon the fact that using the bound $ln(n+1) < H_n < 1+ln(n), n > 1$ (source: http://www.math.drexel.edu/~tolya/123_harmonic.pdf) allows you to get an approximation that is just a slight bit more inaccurate than the first one I presented at the top.



              Approximation: $H_n approx ln(n+1) + gammaleft(1+lnleft(fracnn+1right)right)$



              This approximation is simpler, and a slight bit easier to calculate. However, it is slightly less accurate, so unless you are desiring extreme precision, this one may be more appealing.



              The above two approximations are is relatively easy to calculate compared to some others you may find out there (like Ramanujan).



              A very simplified form of Ramanujan's approximation is



              $H_n approx ln(n) + gamma$



              The above two approximations I presented are both quite a bit more accurate than this truly oversimplified version of Ramanujan's approximation. As @Winther pointed out, there is an error of approximately $frac0.077n$ which implies that (just like pretty much all other approximations) the approximations are much closer to the actual value as $n$ gets larger. However, Ramanujan's approximation in its complete form is extremely accurate, though it is extremely complex.



              If anyone wishes to know the motivation behind the derivation of these approximations (at least from how I did it), then I am happy to answer in the comments.



              EDIT: I stumbled upon one that is more accurate, again kudos to whoever may have discovered it before me. Also, this is more accurate than the approximation $ln(n)+gamma+frac12n$ up to some point.



              $H_n = ln(n) + gammaleft(1+frac5051n+lnleft(fracn-fracgamma10n+fracgamma10right)right)$






              share|cite|improve this answer











              $endgroup$








              • 1




                $begingroup$
                Series approximations for the harmonic number (like this one) has been known since the 1700s (see e.g. en.wikipedia.org/wiki/Harmonic_number#Calculation). By series expanding the logarithm we can show that the error of your approximation is approximately $frac2(1-gamma)-12n approx - frac0.077n$ which agrees well with what you find. You can do a bit better by simply using the approximation $H_n approx log(n) + gamma + frac12n$. This also avoids the additional logarithm
                $endgroup$
                – Winther
                Nov 15 '16 at 3:45











              • $begingroup$
                @Winther That's pretty cool! Yeah I just stumbled upon this while playing around with the bounds so as I said before I am not claiming that I am the first to discover either of these two approximations. Though, they do seem to differ from the ones you suggest.
                $endgroup$
                – Thomas Kim
                Nov 15 '16 at 4:15










              • $begingroup$
                @Winther also, I know that these aren't the best approximations, but they are considerably good compared to some others that are out there.
                $endgroup$
                – Thomas Kim
                Nov 15 '16 at 4:29










              • $begingroup$
                @Winther Could you actually show the calculation for error? I don't really follow on how that works w.r.t. what I initially found.
                $endgroup$
                – Thomas Kim
                Nov 23 '16 at 3:31










              • $begingroup$
                It's based on using the approximation $log((n+1)/n) approx frac1n$ valid for large $n$ and comparing this with the full series for $H_n$. Using this we see that your first approximation is $approx log(n) + gamma + frac1n - fracgamman$ for large $n$ while the known series says that it's $approx log(n) + gamma + frac12n$ so the difference (error) is $approx frac2(1-gamma) - 12n$
                $endgroup$
                – Winther
                Nov 24 '16 at 13:41













              3












              3








              3





              $begingroup$

              While I won't take any credit for this approximation as I don't know if someone else discovered it before myself, I did stumble upon this while playing around with the bound $ln(n)+frac1n < H_n < 1+ln(n)$ for $n > 1$ which was mentioned above and the Euler-Mascheroni Constant.



              Kudos to the person who first discovered this approximation (assuming there was one before myself) if they happen to see this post.



              Approximation: $H_n approx ln(n) + frac1n + gammaleft(1+lnleft(fracnn+1right)right)$



              where $gamma = 0.577215664901532860606512...$ is the Euler-Mascheroni Constant.



              Some calculations to back up this approximation



              For $n = 100$



              Actual Value: $H_100 = 5.1873775...$



              Approximation: $H_100 approx 5.1866423...$



              Error: $-0.0007351...$



              Percent Error: $0.0141719...$%



              For $n = 1000$



              Actual Value: $H_1000 = 7.4854708...$



              Approximation: $H_1000 approx 7.4853940...$



              Error: $-0.0000768...$



              Perent Error: $0.0010265...$%



              I've also been able to stumble upon the fact that using the bound $ln(n+1) < H_n < 1+ln(n), n > 1$ (source: http://www.math.drexel.edu/~tolya/123_harmonic.pdf) allows you to get an approximation that is just a slight bit more inaccurate than the first one I presented at the top.



              Approximation: $H_n approx ln(n+1) + gammaleft(1+lnleft(fracnn+1right)right)$



              This approximation is simpler, and a slight bit easier to calculate. However, it is slightly less accurate, so unless you are desiring extreme precision, this one may be more appealing.



              The above two approximations are is relatively easy to calculate compared to some others you may find out there (like Ramanujan).



              A very simplified form of Ramanujan's approximation is



              $H_n approx ln(n) + gamma$



              The above two approximations I presented are both quite a bit more accurate than this truly oversimplified version of Ramanujan's approximation. As @Winther pointed out, there is an error of approximately $frac0.077n$ which implies that (just like pretty much all other approximations) the approximations are much closer to the actual value as $n$ gets larger. However, Ramanujan's approximation in its complete form is extremely accurate, though it is extremely complex.



              If anyone wishes to know the motivation behind the derivation of these approximations (at least from how I did it), then I am happy to answer in the comments.



              EDIT: I stumbled upon one that is more accurate, again kudos to whoever may have discovered it before me. Also, this is more accurate than the approximation $ln(n)+gamma+frac12n$ up to some point.



              $H_n = ln(n) + gammaleft(1+frac5051n+lnleft(fracn-fracgamma10n+fracgamma10right)right)$






              share|cite|improve this answer











              $endgroup$



              While I won't take any credit for this approximation as I don't know if someone else discovered it before myself, I did stumble upon this while playing around with the bound $ln(n)+frac1n < H_n < 1+ln(n)$ for $n > 1$ which was mentioned above and the Euler-Mascheroni Constant.



              Kudos to the person who first discovered this approximation (assuming there was one before myself) if they happen to see this post.



              Approximation: $H_n approx ln(n) + frac1n + gammaleft(1+lnleft(fracnn+1right)right)$



              where $gamma = 0.577215664901532860606512...$ is the Euler-Mascheroni Constant.



              Some calculations to back up this approximation



              For $n = 100$



              Actual Value: $H_100 = 5.1873775...$



              Approximation: $H_100 approx 5.1866423...$



              Error: $-0.0007351...$



              Percent Error: $0.0141719...$%



              For $n = 1000$



              Actual Value: $H_1000 = 7.4854708...$



              Approximation: $H_1000 approx 7.4853940...$



              Error: $-0.0000768...$



              Perent Error: $0.0010265...$%



              I've also been able to stumble upon the fact that using the bound $ln(n+1) < H_n < 1+ln(n), n > 1$ (source: http://www.math.drexel.edu/~tolya/123_harmonic.pdf) allows you to get an approximation that is just a slight bit more inaccurate than the first one I presented at the top.



              Approximation: $H_n approx ln(n+1) + gammaleft(1+lnleft(fracnn+1right)right)$



              This approximation is simpler, and a slight bit easier to calculate. However, it is slightly less accurate, so unless you are desiring extreme precision, this one may be more appealing.



              The above two approximations are is relatively easy to calculate compared to some others you may find out there (like Ramanujan).



              A very simplified form of Ramanujan's approximation is



              $H_n approx ln(n) + gamma$



              The above two approximations I presented are both quite a bit more accurate than this truly oversimplified version of Ramanujan's approximation. As @Winther pointed out, there is an error of approximately $frac0.077n$ which implies that (just like pretty much all other approximations) the approximations are much closer to the actual value as $n$ gets larger. However, Ramanujan's approximation in its complete form is extremely accurate, though it is extremely complex.



              If anyone wishes to know the motivation behind the derivation of these approximations (at least from how I did it), then I am happy to answer in the comments.



              EDIT: I stumbled upon one that is more accurate, again kudos to whoever may have discovered it before me. Also, this is more accurate than the approximation $ln(n)+gamma+frac12n$ up to some point.



              $H_n = ln(n) + gammaleft(1+frac5051n+lnleft(fracn-fracgamma10n+fracgamma10right)right)$







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Nov 20 '16 at 0:35

























              answered Nov 15 '16 at 1:05









              Thomas KimThomas Kim

              313




              313







              • 1




                $begingroup$
                Series approximations for the harmonic number (like this one) has been known since the 1700s (see e.g. en.wikipedia.org/wiki/Harmonic_number#Calculation). By series expanding the logarithm we can show that the error of your approximation is approximately $frac2(1-gamma)-12n approx - frac0.077n$ which agrees well with what you find. You can do a bit better by simply using the approximation $H_n approx log(n) + gamma + frac12n$. This also avoids the additional logarithm
                $endgroup$
                – Winther
                Nov 15 '16 at 3:45











              • $begingroup$
                @Winther That's pretty cool! Yeah I just stumbled upon this while playing around with the bounds so as I said before I am not claiming that I am the first to discover either of these two approximations. Though, they do seem to differ from the ones you suggest.
                $endgroup$
                – Thomas Kim
                Nov 15 '16 at 4:15










              • $begingroup$
                @Winther also, I know that these aren't the best approximations, but they are considerably good compared to some others that are out there.
                $endgroup$
                – Thomas Kim
                Nov 15 '16 at 4:29










              • $begingroup$
                @Winther Could you actually show the calculation for error? I don't really follow on how that works w.r.t. what I initially found.
                $endgroup$
                – Thomas Kim
                Nov 23 '16 at 3:31










              • $begingroup$
                It's based on using the approximation $log((n+1)/n) approx frac1n$ valid for large $n$ and comparing this with the full series for $H_n$. Using this we see that your first approximation is $approx log(n) + gamma + frac1n - fracgamman$ for large $n$ while the known series says that it's $approx log(n) + gamma + frac12n$ so the difference (error) is $approx frac2(1-gamma) - 12n$
                $endgroup$
                – Winther
                Nov 24 '16 at 13:41












              • 1




                $begingroup$
                Series approximations for the harmonic number (like this one) has been known since the 1700s (see e.g. en.wikipedia.org/wiki/Harmonic_number#Calculation). By series expanding the logarithm we can show that the error of your approximation is approximately $frac2(1-gamma)-12n approx - frac0.077n$ which agrees well with what you find. You can do a bit better by simply using the approximation $H_n approx log(n) + gamma + frac12n$. This also avoids the additional logarithm
                $endgroup$
                – Winther
                Nov 15 '16 at 3:45











              • $begingroup$
                @Winther That's pretty cool! Yeah I just stumbled upon this while playing around with the bounds so as I said before I am not claiming that I am the first to discover either of these two approximations. Though, they do seem to differ from the ones you suggest.
                $endgroup$
                – Thomas Kim
                Nov 15 '16 at 4:15










              • $begingroup$
                @Winther also, I know that these aren't the best approximations, but they are considerably good compared to some others that are out there.
                $endgroup$
                – Thomas Kim
                Nov 15 '16 at 4:29










              • $begingroup$
                @Winther Could you actually show the calculation for error? I don't really follow on how that works w.r.t. what I initially found.
                $endgroup$
                – Thomas Kim
                Nov 23 '16 at 3:31










              • $begingroup$
                It's based on using the approximation $log((n+1)/n) approx frac1n$ valid for large $n$ and comparing this with the full series for $H_n$. Using this we see that your first approximation is $approx log(n) + gamma + frac1n - fracgamman$ for large $n$ while the known series says that it's $approx log(n) + gamma + frac12n$ so the difference (error) is $approx frac2(1-gamma) - 12n$
                $endgroup$
                – Winther
                Nov 24 '16 at 13:41







              1




              1




              $begingroup$
              Series approximations for the harmonic number (like this one) has been known since the 1700s (see e.g. en.wikipedia.org/wiki/Harmonic_number#Calculation). By series expanding the logarithm we can show that the error of your approximation is approximately $frac2(1-gamma)-12n approx - frac0.077n$ which agrees well with what you find. You can do a bit better by simply using the approximation $H_n approx log(n) + gamma + frac12n$. This also avoids the additional logarithm
              $endgroup$
              – Winther
              Nov 15 '16 at 3:45





              $begingroup$
              Series approximations for the harmonic number (like this one) has been known since the 1700s (see e.g. en.wikipedia.org/wiki/Harmonic_number#Calculation). By series expanding the logarithm we can show that the error of your approximation is approximately $frac2(1-gamma)-12n approx - frac0.077n$ which agrees well with what you find. You can do a bit better by simply using the approximation $H_n approx log(n) + gamma + frac12n$. This also avoids the additional logarithm
              $endgroup$
              – Winther
              Nov 15 '16 at 3:45













              $begingroup$
              @Winther That's pretty cool! Yeah I just stumbled upon this while playing around with the bounds so as I said before I am not claiming that I am the first to discover either of these two approximations. Though, they do seem to differ from the ones you suggest.
              $endgroup$
              – Thomas Kim
              Nov 15 '16 at 4:15




              $begingroup$
              @Winther That's pretty cool! Yeah I just stumbled upon this while playing around with the bounds so as I said before I am not claiming that I am the first to discover either of these two approximations. Though, they do seem to differ from the ones you suggest.
              $endgroup$
              – Thomas Kim
              Nov 15 '16 at 4:15












              $begingroup$
              @Winther also, I know that these aren't the best approximations, but they are considerably good compared to some others that are out there.
              $endgroup$
              – Thomas Kim
              Nov 15 '16 at 4:29




              $begingroup$
              @Winther also, I know that these aren't the best approximations, but they are considerably good compared to some others that are out there.
              $endgroup$
              – Thomas Kim
              Nov 15 '16 at 4:29












              $begingroup$
              @Winther Could you actually show the calculation for error? I don't really follow on how that works w.r.t. what I initially found.
              $endgroup$
              – Thomas Kim
              Nov 23 '16 at 3:31




              $begingroup$
              @Winther Could you actually show the calculation for error? I don't really follow on how that works w.r.t. what I initially found.
              $endgroup$
              – Thomas Kim
              Nov 23 '16 at 3:31












              $begingroup$
              It's based on using the approximation $log((n+1)/n) approx frac1n$ valid for large $n$ and comparing this with the full series for $H_n$. Using this we see that your first approximation is $approx log(n) + gamma + frac1n - fracgamman$ for large $n$ while the known series says that it's $approx log(n) + gamma + frac12n$ so the difference (error) is $approx frac2(1-gamma) - 12n$
              $endgroup$
              – Winther
              Nov 24 '16 at 13:41




              $begingroup$
              It's based on using the approximation $log((n+1)/n) approx frac1n$ valid for large $n$ and comparing this with the full series for $H_n$. Using this we see that your first approximation is $approx log(n) + gamma + frac1n - fracgamman$ for large $n$ while the known series says that it's $approx log(n) + gamma + frac12n$ so the difference (error) is $approx frac2(1-gamma) - 12n$
              $endgroup$
              – Winther
              Nov 24 '16 at 13:41











              1












              $begingroup$

              There is a new formula, that I've created. It can be written in a few different ways, below are two of them:



              $sum_k=1^nfrac1k=frac12n+piint_0^1 (1-u)cotpi uleft(1-cos2pi n uright),du$



              $sum_k=1^nfrac1k=frac12n+fracpi2int_0^1 (1-u)cotfracpi u2left(1-cospi n uright),du$



              There is also a generalization that goes beyond the harmonic numbers.
              For a harmonic progression where $a$ and $b$ are integers:



              $sum _k=1^n frac1a k+b=-frac12b+frac12(a n+b)+2piint_0^1 (1-u)sin[pi a n u]sin[(a n+2b)pi u]cot[pi a u],du$



              (Notice the formula at the top is a particular case where $a=1$, $b=0$.) For odd powers, the general formula is:
              beginmultline
              sum_j=1^nfrac1(a j+b)^2k+1=-frac12b^2k+1+frac12(a n+b)^2k+1+(-1)^k(2pi)^2k+1\ int_0^1sum_j=0^kfracB_2k-2jleft(2-2^2k-2jright)(2k-2j)!(2j+1)!(1-u)^2j+1sin[pi a nu]sin[(a n+2b)pi u]cot[pi au],du
              endmultline



              And for a harmonic progression where $a$ and $b$ are complex:
              beginmultlinenonumber
              sum_j=1^nfrac1(ai j+b)^k =-frac12b^k+frac12(ai n+b)^k+ e^-2pi b/aleft(frac2piaright)^k\ int_0^1sum_j=1^kfracphileft(e^-2pi b/a,-j+1,0right)(1-u)^k-j(j-1)!(k-j)!e^2pi bu/aleft(fracsin2pi n u2+i frac1-cos2pi n u2right)cotpi u,du
              endmultline



              The proofs are on papers that I've posted to the arXiv.






              share|cite|improve this answer











              $endgroup$

















                1












                $begingroup$

                There is a new formula, that I've created. It can be written in a few different ways, below are two of them:



                $sum_k=1^nfrac1k=frac12n+piint_0^1 (1-u)cotpi uleft(1-cos2pi n uright),du$



                $sum_k=1^nfrac1k=frac12n+fracpi2int_0^1 (1-u)cotfracpi u2left(1-cospi n uright),du$



                There is also a generalization that goes beyond the harmonic numbers.
                For a harmonic progression where $a$ and $b$ are integers:



                $sum _k=1^n frac1a k+b=-frac12b+frac12(a n+b)+2piint_0^1 (1-u)sin[pi a n u]sin[(a n+2b)pi u]cot[pi a u],du$



                (Notice the formula at the top is a particular case where $a=1$, $b=0$.) For odd powers, the general formula is:
                beginmultline
                sum_j=1^nfrac1(a j+b)^2k+1=-frac12b^2k+1+frac12(a n+b)^2k+1+(-1)^k(2pi)^2k+1\ int_0^1sum_j=0^kfracB_2k-2jleft(2-2^2k-2jright)(2k-2j)!(2j+1)!(1-u)^2j+1sin[pi a nu]sin[(a n+2b)pi u]cot[pi au],du
                endmultline



                And for a harmonic progression where $a$ and $b$ are complex:
                beginmultlinenonumber
                sum_j=1^nfrac1(ai j+b)^k =-frac12b^k+frac12(ai n+b)^k+ e^-2pi b/aleft(frac2piaright)^k\ int_0^1sum_j=1^kfracphileft(e^-2pi b/a,-j+1,0right)(1-u)^k-j(j-1)!(k-j)!e^2pi bu/aleft(fracsin2pi n u2+i frac1-cos2pi n u2right)cotpi u,du
                endmultline



                The proofs are on papers that I've posted to the arXiv.






                share|cite|improve this answer











                $endgroup$















                  1












                  1








                  1





                  $begingroup$

                  There is a new formula, that I've created. It can be written in a few different ways, below are two of them:



                  $sum_k=1^nfrac1k=frac12n+piint_0^1 (1-u)cotpi uleft(1-cos2pi n uright),du$



                  $sum_k=1^nfrac1k=frac12n+fracpi2int_0^1 (1-u)cotfracpi u2left(1-cospi n uright),du$



                  There is also a generalization that goes beyond the harmonic numbers.
                  For a harmonic progression where $a$ and $b$ are integers:



                  $sum _k=1^n frac1a k+b=-frac12b+frac12(a n+b)+2piint_0^1 (1-u)sin[pi a n u]sin[(a n+2b)pi u]cot[pi a u],du$



                  (Notice the formula at the top is a particular case where $a=1$, $b=0$.) For odd powers, the general formula is:
                  beginmultline
                  sum_j=1^nfrac1(a j+b)^2k+1=-frac12b^2k+1+frac12(a n+b)^2k+1+(-1)^k(2pi)^2k+1\ int_0^1sum_j=0^kfracB_2k-2jleft(2-2^2k-2jright)(2k-2j)!(2j+1)!(1-u)^2j+1sin[pi a nu]sin[(a n+2b)pi u]cot[pi au],du
                  endmultline



                  And for a harmonic progression where $a$ and $b$ are complex:
                  beginmultlinenonumber
                  sum_j=1^nfrac1(ai j+b)^k =-frac12b^k+frac12(ai n+b)^k+ e^-2pi b/aleft(frac2piaright)^k\ int_0^1sum_j=1^kfracphileft(e^-2pi b/a,-j+1,0right)(1-u)^k-j(j-1)!(k-j)!e^2pi bu/aleft(fracsin2pi n u2+i frac1-cos2pi n u2right)cotpi u,du
                  endmultline



                  The proofs are on papers that I've posted to the arXiv.






                  share|cite|improve this answer











                  $endgroup$



                  There is a new formula, that I've created. It can be written in a few different ways, below are two of them:



                  $sum_k=1^nfrac1k=frac12n+piint_0^1 (1-u)cotpi uleft(1-cos2pi n uright),du$



                  $sum_k=1^nfrac1k=frac12n+fracpi2int_0^1 (1-u)cotfracpi u2left(1-cospi n uright),du$



                  There is also a generalization that goes beyond the harmonic numbers.
                  For a harmonic progression where $a$ and $b$ are integers:



                  $sum _k=1^n frac1a k+b=-frac12b+frac12(a n+b)+2piint_0^1 (1-u)sin[pi a n u]sin[(a n+2b)pi u]cot[pi a u],du$



                  (Notice the formula at the top is a particular case where $a=1$, $b=0$.) For odd powers, the general formula is:
                  beginmultline
                  sum_j=1^nfrac1(a j+b)^2k+1=-frac12b^2k+1+frac12(a n+b)^2k+1+(-1)^k(2pi)^2k+1\ int_0^1sum_j=0^kfracB_2k-2jleft(2-2^2k-2jright)(2k-2j)!(2j+1)!(1-u)^2j+1sin[pi a nu]sin[(a n+2b)pi u]cot[pi au],du
                  endmultline



                  And for a harmonic progression where $a$ and $b$ are complex:
                  beginmultlinenonumber
                  sum_j=1^nfrac1(ai j+b)^k =-frac12b^k+frac12(ai n+b)^k+ e^-2pi b/aleft(frac2piaright)^k\ int_0^1sum_j=1^kfracphileft(e^-2pi b/a,-j+1,0right)(1-u)^k-j(j-1)!(k-j)!e^2pi bu/aleft(fracsin2pi n u2+i frac1-cos2pi n u2right)cotpi u,du
                  endmultline



                  The proofs are on papers that I've posted to the arXiv.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Mar 2 at 4:25

























                  answered Mar 1 at 23:52









                  JR SousaJR Sousa

                  1008




                  1008



























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