Evaluation of Euler's Constant $gamma$What is the fastest/most efficient algorithm for estimating Euler's Constant $gamma$?Showing that $lim_ntoinftysum^n_k=1frac1k-ln(n)=0.5772ldots$How do we show that $ln2-gamma=sum_n=1^inftyzeta(2n+1)over 2^2n(2n+1)?$Riemann's thinking on symmetrizing the zeta functional equationZeta function values in terms of Bernoulli numbers.What is the half-derivative of zeta at $s=0$ (and how to compute it)?How could I get access to more than the first 2 mln non-trivial zeros of $zeta(s)$?Riemann Zeta Function, Stirling's Numbers, and Infinite SeriesCalculating values of the Riemann Zeta FunctionQuestion regarding an inequality in Titchmarsh p. 267A method for assigning values to divergent series that seems to coincide with the Riemann Zeta FunctionHow to locate zeros of the Riemann Zeta function?Evaluate $int_0^1fracsin (pi xs)sinleft(pi x (1-s) right)sin(pi s),ds$
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Evaluation of Euler's Constant $gamma$
What is the fastest/most efficient algorithm for estimating Euler's Constant $gamma$?Showing that $lim_ntoinftysum^n_k=1frac1k-ln(n)=0.5772ldots$How do we show that $ln2-gamma=sum_n=1^inftyzeta(2n+1)over 2^2n(2n+1)?$Riemann's thinking on symmetrizing the zeta functional equationZeta function values in terms of Bernoulli numbers.What is the half-derivative of zeta at $s=0$ (and how to compute it)?How could I get access to more than the first 2 mln non-trivial zeros of $zeta(s)$?Riemann Zeta Function, Stirling's Numbers, and Infinite SeriesCalculating values of the Riemann Zeta FunctionQuestion regarding an inequality in Titchmarsh p. 267A method for assigning values to divergent series that seems to coincide with the Riemann Zeta FunctionHow to locate zeros of the Riemann Zeta function?Evaluate $int_0^1fracsin (pi xs)sinleft(pi x (1-s) right)sin(pi s),ds$
$begingroup$
Long back I had seen (in some obscure book) a formula to calculate the value of Euler's constant $gamma$ based on a table of values of Riemann zeta function $zeta(s)$. I am not able to recall the formula, but it used the fact that $zeta(s) to 1$ as $s to infty$ very fast and used terms of the form $zeta(s) - 1$ for odd values of $s > 1$ (something like a series $sum(zeta(s) - 1)$). If anyone has access to this formula please let me know and it would be great to have a proof.
riemann-zeta
asked Nov 12 '13 at 8:31
Paramanand SinghParamanand Singh
50.7k557168
$endgroup$
add a comment |
$begingroup$
Long back I had seen (in some obscure book) a formula to calculate the value of Euler's constant $gamma$ based on a table of values of Riemann zeta function $zeta(s)$. I am not able to recall the formula, but it used the fact that $zeta(s) to 1$ as $s to infty$ very fast and used terms of the form $zeta(s) - 1$ for odd values of $s > 1$ (something like a series $sum(zeta(s) - 1)$). If anyone has access to this formula please let me know and it would be great to have a proof.
riemann-zeta
asked Nov 12 '13 at 8:31
Paramanand SinghParamanand Singh
50.7k557168
$endgroup$
4
$begingroup$
The following paper has appeared just a few weeks ago: Jeffrey C. Lagarias: Euler's constant – Euler's work and modern developments. Bull. Am. Math. Soc. 50 (2013), 527–628. In this paper Lagarias has done for Euler's constant what Melville did for the whale.
$endgroup$
– Christian Blatter
Nov 12 '13 at 14:27
$begingroup$
Only a note (off-topic). The series with value of $zeta(n)$ are not the best way to evaluate $gamma$. The best two easy ways to calculate Euler's constant numerically are:<br> (1) Euler-Maclaurin sumformula<br> (2) With an asymptotic expansion of li(x) [integrallogarithm]. This Method is going back to Heinrich Wilhelm Brandes (1777–1834) in 1824.
$endgroup$
– skraemer
Mar 11 at 19:17
add a comment |
$begingroup$
Long back I had seen (in some obscure book) a formula to calculate the value of Euler's constant $gamma$ based on a table of values of Riemann zeta function $zeta(s)$. I am not able to recall the formula, but it used the fact that $zeta(s) to 1$ as $s to infty$ very fast and used terms of the form $zeta(s) - 1$ for odd values of $s > 1$ (something like a series $sum(zeta(s) - 1)$). If anyone has access to this formula please let me know and it would be great to have a proof.
riemann-zeta
asked Nov 12 '13 at 8:31
Paramanand SinghParamanand Singh
50.7k557168
$endgroup$
Long back I had seen (in some obscure book) a formula to calculate the value of Euler's constant $gamma$ based on a table of values of Riemann zeta function $zeta(s)$. I am not able to recall the formula, but it used the fact that $zeta(s) to 1$ as $s to infty$ very fast and used terms of the form $zeta(s) - 1$ for odd values of $s > 1$ (something like a series $sum(zeta(s) - 1)$). If anyone has access to this formula please let me know and it would be great to have a proof.
riemann-zeta
riemann-zeta
asked Nov 12 '13 at 8:31
Paramanand SinghParamanand Singh
50.7k557168
asked Nov 12 '13 at 8:31
Paramanand SinghParamanand Singh
50.7k557168
asked Nov 12 '13 at 8:31
Paramanand SinghParamanand Singh
50.7k557168
asked Nov 12 '13 at 8:31
Paramanand SinghParamanand Singh
50.7k557168
asked Nov 12 '13 at 8:31
Paramanand SinghParamanand Singh
50.7k557168
50.7k557168
4
$begingroup$
The following paper has appeared just a few weeks ago: Jeffrey C. Lagarias: Euler's constant – Euler's work and modern developments. Bull. Am. Math. Soc. 50 (2013), 527–628. In this paper Lagarias has done for Euler's constant what Melville did for the whale.
$endgroup$
– Christian Blatter
Nov 12 '13 at 14:27
$begingroup$
Only a note (off-topic). The series with value of $zeta(n)$ are not the best way to evaluate $gamma$. The best two easy ways to calculate Euler's constant numerically are:<br> (1) Euler-Maclaurin sumformula<br> (2) With an asymptotic expansion of li(x) [integrallogarithm]. This Method is going back to Heinrich Wilhelm Brandes (1777–1834) in 1824.
$endgroup$
– skraemer
Mar 11 at 19:17
add a comment |
4
$begingroup$
The following paper has appeared just a few weeks ago: Jeffrey C. Lagarias: Euler's constant – Euler's work and modern developments. Bull. Am. Math. Soc. 50 (2013), 527–628. In this paper Lagarias has done for Euler's constant what Melville did for the whale.
$endgroup$
– Christian Blatter
Nov 12 '13 at 14:27
$begingroup$
Only a note (off-topic). The series with value of $zeta(n)$ are not the best way to evaluate $gamma$. The best two easy ways to calculate Euler's constant numerically are:<br> (1) Euler-Maclaurin sumformula<br> (2) With an asymptotic expansion of li(x) [integrallogarithm]. This Method is going back to Heinrich Wilhelm Brandes (1777–1834) in 1824.
$endgroup$
– skraemer
Mar 11 at 19:17
4
4
$begingroup$
The following paper has appeared just a few weeks ago: Jeffrey C. Lagarias: Euler's constant – Euler's work and modern developments. Bull. Am. Math. Soc. 50 (2013), 527–628. In this paper Lagarias has done for Euler's constant what Melville did for the whale.
$endgroup$
– Christian Blatter
Nov 12 '13 at 14:27
$begingroup$
The following paper has appeared just a few weeks ago: Jeffrey C. Lagarias: Euler's constant – Euler's work and modern developments. Bull. Am. Math. Soc. 50 (2013), 527–628. In this paper Lagarias has done for Euler's constant what Melville did for the whale.
$endgroup$
– Christian Blatter
Nov 12 '13 at 14:27
$begingroup$
Only a note (off-topic). The series with value of $zeta(n)$ are not the best way to evaluate $gamma$. The best two easy ways to calculate Euler's constant numerically are:<br> (1) Euler-Maclaurin sumformula<br> (2) With an asymptotic expansion of li(x) [integrallogarithm]. This Method is going back to Heinrich Wilhelm Brandes (1777–1834) in 1824.
$endgroup$
– skraemer
Mar 11 at 19:17
$begingroup$
Only a note (off-topic). The series with value of $zeta(n)$ are not the best way to evaluate $gamma$. The best two easy ways to calculate Euler's constant numerically are:<br> (1) Euler-Maclaurin sumformula<br> (2) With an asymptotic expansion of li(x) [integrallogarithm]. This Method is going back to Heinrich Wilhelm Brandes (1777–1834) in 1824.
$endgroup$
– skraemer
Mar 11 at 19:17
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Note that for Harmonic numbers, $H_n$,
$$
sum_k=1^nleft(frac1k-logleft(1+frac1kright)right)=H_n-log(n+1)tag1
$$
Taking $(1)$ to the limit gives
$$
gamma=sum_k=1^inftyleft(frac1k-logleft(1+frac1kright)right)tag2
$$
We have the power series
$$
logleft(frac1+x1-xright)=2left(x+fracx^33+fracx^55+fracx^77dotsright)tag3
$$
If we set $x=frac12k+1$, then $frac1+x1-x=1+frac1k$; that is,
$$
logleft(1+frac1kright)=sum_j=0^inftyfrac22j+1left(frac12k+1right)^2j+1tag4
$$
Furthermore,
$$
beginalign
sum_k=1^inftyleft(frac12k+1right)^2j+1
&=sum_k=1^inftyleft(frac12kright)^2j+1
+left(frac12k+1right)^2j+1-frac12^2j+1left(frac1kright)^2j+1\
&=zeta(2j+1)-1-frac12^2j+1zeta(2j+1)\
&=frac2^2j+1-12^2j+1(zeta(2j+1)-1)-frac12^2j+1tag5
endalign
$$
Then, using $(4)$ in $(2)$, and then applying $(5)$, we get
$$
beginalign
gamma
&=sum_k=1^inftyleft(frac1k-logleft(1+frac1kright)right)\
&=sum_k=1^inftyfrac22k-frac22k+1-sum_j=1^inftyfrac22j+1left(frac12k+1right)^2j+1\
&=2(1-log(2))-sum_j=1^inftyfrac22j+1left(frac2^2j+1-12^2j+1(zeta(2j+1)-1)-frac12^2j+1right)\
&=1-log(4)+log(3)-sum_j=1^inftyfrac2^2j+1-12^2j(2j+1)(zeta(2j+1)-1)tag6
endalign
$$
The last equality in $(6)$ follows from plugging $k=frac12$ into $(4)$ to get
$$
log(3)-1=sum_j=1^inftyfrac2(2j+1)2^2j+1tag7
$$
We can accelerate the convergence of $(6)$ once, using $(3)$, we compute
$$
beginalign
sum_j=1^inftyfraczeta(2j+1)-12j+1
&=sum_j=1^inftysum_k=2^inftyfrac1(2j+1)k^2j+1\
&=lim_ntoinftysum_k=2^nfrac12logleft(frack+1k-1right)-frac1k\
&=lim_ntoinftyfrac12log(n(n+1)/2)-(H_n-1)\
&=1-log(2)/2-gammatag8
endalign
$$
If we add twice the left side of $(8)$ to the right side of $(6)$ and vice versa, we get
$$
2-log(2)-gamma
=1-log(4)+log(3)+sum_j=1^inftyfraczeta(2j+1)-12^2j(2j+1)tag9
$$
From which we get the Euler-Stieltjes series:
$$
gamma=1-log(3/2)-sum_j=1^inftyfraczeta(2j+1)-14^j(2j+1)tag10
$$
Using the following application of $(3)$
$$
sum_j=1^inftyfracfrac1n^2j+14^j(2j+1)=logleft(frac2n+12n-1right)-frac1ntag11
$$
we can accelerate the convergence of $(10)$:
$$
gamma=H_n-log(n+1/2)-sum_j=1^inftyfraczeta(2j+1)-sumlimits_k=1^nfrac1k^2j+14^j(2j+1)tag12
$$
$(12)$ converges about $2log_10(2n+2)$ digits per term. Euler-Stieltjes is the case $n=1$ of $(12)$. Note that $zeta(2j+1)-sumlimits_k=1^nfrac1k^2j+1=zeta(2j+1,n+1)$, the Hurwitz Zeta function.
In this answer, I give a another method for computing $gamma$ that uses the an accelerated function for the sum of the tail of the alternating harmonic series.
edited Apr 13 '17 at 12:20
Community♦
1
answered Nov 12 '13 at 12:51
robjohn♦robjohn
269k27311638
$endgroup$
$begingroup$
To robjohn: Thanks for the nice formula which is almost similar to Euler-Stieltjes. And the best part is the very very elementary proof which is accessible to anyone who knows the definition of $gamma$ and the logarithmic series. I also checked your other answer and found it very informative. Thanks a lot (+1)! I have not yet accepted any answer to check for any more interesting answers.
$endgroup$
– Paramanand Singh
Nov 12 '13 at 14:33
$begingroup$
To Robjohn: Before I could connect the dots from your answer to Euler-Stieltjes, you gave the full derivation.. gr8!
$endgroup$
– Paramanand Singh
Nov 12 '13 at 14:43
$begingroup$
@ParamanandSingh: I have added a faster converging extension of Euler-Stieltjes.
$endgroup$
– robjohn♦
Nov 12 '13 at 16:11
$begingroup$
To Robjohn: I am speechless!
$endgroup$
– Paramanand Singh
Nov 12 '13 at 16:30
add a comment |
$begingroup$
There are a lot of formulas of this type. Some of them are in the Collection of formulae for Euler's constant $gamma;$ by Xavier Gourdon and Pascal Sebah:
$$gamma = frac32 - ln 2 - sum_nge 2frac1nleft(zeta(n)-1- frac12^nright)$$
$$gamma = frac116 - ln 3 - sum_nge 2frac1nleft(zeta(n)-1
-frac12^n -frac13^nright)$$
$$gamma = 1- lnleft(frac32right) -sum_nge1fraczeta(2n+1)-14^n(2n+1) qquadtext(Euler-Stieltjes)
$$
The first two are derived from the Hurwitz zeta function as special cases. The Euler-Stieltjes formula seems near to your remembrance but is listed without proof.
Edit: You can find a proof in the Expansion of Euler's constant in terms of zeta numbers by M. Prévost.
answered Nov 12 '13 at 9:47
gammatestergammatester
16.8k21733
$endgroup$
$begingroup$
Yes I also think that the third one is what I saw long back. I will try to go through the linked PDF references
$endgroup$
– Paramanand Singh
Nov 12 '13 at 10:05
$begingroup$
The first two series are neat in that they are the series given by lhf with the first two or three terms of $zeta(n)$ taken out. They give better convergence at about $0.477$ and $.602$ places per term. The Euler-Stieltjes Series gives approximately $1.2$ places per term. (+1)
$endgroup$
– robjohn♦
Nov 12 '13 at 15:26
$begingroup$
In the spirit of your first two series (which extend the series given by lhf), I have similarly accelerated the Euler-Stieltjes series in my answer.
$endgroup$
– robjohn♦
Nov 12 '13 at 16:39
add a comment |
$begingroup$
Do you mean this?
$$sum_k=2^infty zeta(k)-1over k= 1-gamma $$
This formula can be found in MathWorld (eq 123).
(Quoted in What is the fastest/most efficient algorithm for estimating Euler's Constant γ?.)
edited Apr 13 '17 at 12:20
Community♦
1
answered Nov 12 '13 at 10:22
lhflhf
166k10171400
$endgroup$
$begingroup$
No! the sum I had seen is mentioned in answer by gammatester namely $$gamma = 1- lnleft(frac32right) -sum_nge1fraczeta(2n+1)-14^n(2n+1) qquadtext(Euler-Stieltjes)$$ The formula which you mention has an easy proof.
$endgroup$
– Paramanand Singh
Nov 12 '13 at 10:44
$begingroup$
Although it may not be the series that Paramanand was seeking, it is still a valid series, giving about $0.3$ places each term. (+1)
$endgroup$
– robjohn♦
Nov 12 '13 at 15:20
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that for Harmonic numbers, $H_n$,
$$
sum_k=1^nleft(frac1k-logleft(1+frac1kright)right)=H_n-log(n+1)tag1
$$
Taking $(1)$ to the limit gives
$$
gamma=sum_k=1^inftyleft(frac1k-logleft(1+frac1kright)right)tag2
$$
We have the power series
$$
logleft(frac1+x1-xright)=2left(x+fracx^33+fracx^55+fracx^77dotsright)tag3
$$
If we set $x=frac12k+1$, then $frac1+x1-x=1+frac1k$; that is,
$$
logleft(1+frac1kright)=sum_j=0^inftyfrac22j+1left(frac12k+1right)^2j+1tag4
$$
Furthermore,
$$
beginalign
sum_k=1^inftyleft(frac12k+1right)^2j+1
&=sum_k=1^inftyleft(frac12kright)^2j+1
+left(frac12k+1right)^2j+1-frac12^2j+1left(frac1kright)^2j+1\
&=zeta(2j+1)-1-frac12^2j+1zeta(2j+1)\
&=frac2^2j+1-12^2j+1(zeta(2j+1)-1)-frac12^2j+1tag5
endalign
$$
Then, using $(4)$ in $(2)$, and then applying $(5)$, we get
$$
beginalign
gamma
&=sum_k=1^inftyleft(frac1k-logleft(1+frac1kright)right)\
&=sum_k=1^inftyfrac22k-frac22k+1-sum_j=1^inftyfrac22j+1left(frac12k+1right)^2j+1\
&=2(1-log(2))-sum_j=1^inftyfrac22j+1left(frac2^2j+1-12^2j+1(zeta(2j+1)-1)-frac12^2j+1right)\
&=1-log(4)+log(3)-sum_j=1^inftyfrac2^2j+1-12^2j(2j+1)(zeta(2j+1)-1)tag6
endalign
$$
The last equality in $(6)$ follows from plugging $k=frac12$ into $(4)$ to get
$$
log(3)-1=sum_j=1^inftyfrac2(2j+1)2^2j+1tag7
$$
We can accelerate the convergence of $(6)$ once, using $(3)$, we compute
$$
beginalign
sum_j=1^inftyfraczeta(2j+1)-12j+1
&=sum_j=1^inftysum_k=2^inftyfrac1(2j+1)k^2j+1\
&=lim_ntoinftysum_k=2^nfrac12logleft(frack+1k-1right)-frac1k\
&=lim_ntoinftyfrac12log(n(n+1)/2)-(H_n-1)\
&=1-log(2)/2-gammatag8
endalign
$$
If we add twice the left side of $(8)$ to the right side of $(6)$ and vice versa, we get
$$
2-log(2)-gamma
=1-log(4)+log(3)+sum_j=1^inftyfraczeta(2j+1)-12^2j(2j+1)tag9
$$
From which we get the Euler-Stieltjes series:
$$
gamma=1-log(3/2)-sum_j=1^inftyfraczeta(2j+1)-14^j(2j+1)tag10
$$
Using the following application of $(3)$
$$
sum_j=1^inftyfracfrac1n^2j+14^j(2j+1)=logleft(frac2n+12n-1right)-frac1ntag11
$$
we can accelerate the convergence of $(10)$:
$$
gamma=H_n-log(n+1/2)-sum_j=1^inftyfraczeta(2j+1)-sumlimits_k=1^nfrac1k^2j+14^j(2j+1)tag12
$$
$(12)$ converges about $2log_10(2n+2)$ digits per term. Euler-Stieltjes is the case $n=1$ of $(12)$. Note that $zeta(2j+1)-sumlimits_k=1^nfrac1k^2j+1=zeta(2j+1,n+1)$, the Hurwitz Zeta function.
In this answer, I give a another method for computing $gamma$ that uses the an accelerated function for the sum of the tail of the alternating harmonic series.
edited Apr 13 '17 at 12:20
Community♦
1
answered Nov 12 '13 at 12:51
robjohn♦robjohn
269k27311638
$endgroup$
$begingroup$
To robjohn: Thanks for the nice formula which is almost similar to Euler-Stieltjes. And the best part is the very very elementary proof which is accessible to anyone who knows the definition of $gamma$ and the logarithmic series. I also checked your other answer and found it very informative. Thanks a lot (+1)! I have not yet accepted any answer to check for any more interesting answers.
$endgroup$
– Paramanand Singh
Nov 12 '13 at 14:33
$begingroup$
To Robjohn: Before I could connect the dots from your answer to Euler-Stieltjes, you gave the full derivation.. gr8!
$endgroup$
– Paramanand Singh
Nov 12 '13 at 14:43
$begingroup$
@ParamanandSingh: I have added a faster converging extension of Euler-Stieltjes.
$endgroup$
– robjohn♦
Nov 12 '13 at 16:11
$begingroup$
To Robjohn: I am speechless!
$endgroup$
– Paramanand Singh
Nov 12 '13 at 16:30
add a comment |
$begingroup$
Note that for Harmonic numbers, $H_n$,
$$
sum_k=1^nleft(frac1k-logleft(1+frac1kright)right)=H_n-log(n+1)tag1
$$
Taking $(1)$ to the limit gives
$$
gamma=sum_k=1^inftyleft(frac1k-logleft(1+frac1kright)right)tag2
$$
We have the power series
$$
logleft(frac1+x1-xright)=2left(x+fracx^33+fracx^55+fracx^77dotsright)tag3
$$
If we set $x=frac12k+1$, then $frac1+x1-x=1+frac1k$; that is,
$$
logleft(1+frac1kright)=sum_j=0^inftyfrac22j+1left(frac12k+1right)^2j+1tag4
$$
Furthermore,
$$
beginalign
sum_k=1^inftyleft(frac12k+1right)^2j+1
&=sum_k=1^inftyleft(frac12kright)^2j+1
+left(frac12k+1right)^2j+1-frac12^2j+1left(frac1kright)^2j+1\
&=zeta(2j+1)-1-frac12^2j+1zeta(2j+1)\
&=frac2^2j+1-12^2j+1(zeta(2j+1)-1)-frac12^2j+1tag5
endalign
$$
Then, using $(4)$ in $(2)$, and then applying $(5)$, we get
$$
beginalign
gamma
&=sum_k=1^inftyleft(frac1k-logleft(1+frac1kright)right)\
&=sum_k=1^inftyfrac22k-frac22k+1-sum_j=1^inftyfrac22j+1left(frac12k+1right)^2j+1\
&=2(1-log(2))-sum_j=1^inftyfrac22j+1left(frac2^2j+1-12^2j+1(zeta(2j+1)-1)-frac12^2j+1right)\
&=1-log(4)+log(3)-sum_j=1^inftyfrac2^2j+1-12^2j(2j+1)(zeta(2j+1)-1)tag6
endalign
$$
The last equality in $(6)$ follows from plugging $k=frac12$ into $(4)$ to get
$$
log(3)-1=sum_j=1^inftyfrac2(2j+1)2^2j+1tag7
$$
We can accelerate the convergence of $(6)$ once, using $(3)$, we compute
$$
beginalign
sum_j=1^inftyfraczeta(2j+1)-12j+1
&=sum_j=1^inftysum_k=2^inftyfrac1(2j+1)k^2j+1\
&=lim_ntoinftysum_k=2^nfrac12logleft(frack+1k-1right)-frac1k\
&=lim_ntoinftyfrac12log(n(n+1)/2)-(H_n-1)\
&=1-log(2)/2-gammatag8
endalign
$$
If we add twice the left side of $(8)$ to the right side of $(6)$ and vice versa, we get
$$
2-log(2)-gamma
=1-log(4)+log(3)+sum_j=1^inftyfraczeta(2j+1)-12^2j(2j+1)tag9
$$
From which we get the Euler-Stieltjes series:
$$
gamma=1-log(3/2)-sum_j=1^inftyfraczeta(2j+1)-14^j(2j+1)tag10
$$
Using the following application of $(3)$
$$
sum_j=1^inftyfracfrac1n^2j+14^j(2j+1)=logleft(frac2n+12n-1right)-frac1ntag11
$$
we can accelerate the convergence of $(10)$:
$$
gamma=H_n-log(n+1/2)-sum_j=1^inftyfraczeta(2j+1)-sumlimits_k=1^nfrac1k^2j+14^j(2j+1)tag12
$$
$(12)$ converges about $2log_10(2n+2)$ digits per term. Euler-Stieltjes is the case $n=1$ of $(12)$. Note that $zeta(2j+1)-sumlimits_k=1^nfrac1k^2j+1=zeta(2j+1,n+1)$, the Hurwitz Zeta function.
In this answer, I give a another method for computing $gamma$ that uses the an accelerated function for the sum of the tail of the alternating harmonic series.
edited Apr 13 '17 at 12:20
Community♦
1
answered Nov 12 '13 at 12:51
robjohn♦robjohn
269k27311638
$endgroup$
$begingroup$
To robjohn: Thanks for the nice formula which is almost similar to Euler-Stieltjes. And the best part is the very very elementary proof which is accessible to anyone who knows the definition of $gamma$ and the logarithmic series. I also checked your other answer and found it very informative. Thanks a lot (+1)! I have not yet accepted any answer to check for any more interesting answers.
$endgroup$
– Paramanand Singh
Nov 12 '13 at 14:33
$begingroup$
To Robjohn: Before I could connect the dots from your answer to Euler-Stieltjes, you gave the full derivation.. gr8!
$endgroup$
– Paramanand Singh
Nov 12 '13 at 14:43
$begingroup$
@ParamanandSingh: I have added a faster converging extension of Euler-Stieltjes.
$endgroup$
– robjohn♦
Nov 12 '13 at 16:11
$begingroup$
To Robjohn: I am speechless!
$endgroup$
– Paramanand Singh
Nov 12 '13 at 16:30
add a comment |
$begingroup$
Note that for Harmonic numbers, $H_n$,
$$
sum_k=1^nleft(frac1k-logleft(1+frac1kright)right)=H_n-log(n+1)tag1
$$
Taking $(1)$ to the limit gives
$$
gamma=sum_k=1^inftyleft(frac1k-logleft(1+frac1kright)right)tag2
$$
We have the power series
$$
logleft(frac1+x1-xright)=2left(x+fracx^33+fracx^55+fracx^77dotsright)tag3
$$
If we set $x=frac12k+1$, then $frac1+x1-x=1+frac1k$; that is,
$$
logleft(1+frac1kright)=sum_j=0^inftyfrac22j+1left(frac12k+1right)^2j+1tag4
$$
Furthermore,
$$
beginalign
sum_k=1^inftyleft(frac12k+1right)^2j+1
&=sum_k=1^inftyleft(frac12kright)^2j+1
+left(frac12k+1right)^2j+1-frac12^2j+1left(frac1kright)^2j+1\
&=zeta(2j+1)-1-frac12^2j+1zeta(2j+1)\
&=frac2^2j+1-12^2j+1(zeta(2j+1)-1)-frac12^2j+1tag5
endalign
$$
Then, using $(4)$ in $(2)$, and then applying $(5)$, we get
$$
beginalign
gamma
&=sum_k=1^inftyleft(frac1k-logleft(1+frac1kright)right)\
&=sum_k=1^inftyfrac22k-frac22k+1-sum_j=1^inftyfrac22j+1left(frac12k+1right)^2j+1\
&=2(1-log(2))-sum_j=1^inftyfrac22j+1left(frac2^2j+1-12^2j+1(zeta(2j+1)-1)-frac12^2j+1right)\
&=1-log(4)+log(3)-sum_j=1^inftyfrac2^2j+1-12^2j(2j+1)(zeta(2j+1)-1)tag6
endalign
$$
The last equality in $(6)$ follows from plugging $k=frac12$ into $(4)$ to get
$$
log(3)-1=sum_j=1^inftyfrac2(2j+1)2^2j+1tag7
$$
We can accelerate the convergence of $(6)$ once, using $(3)$, we compute
$$
beginalign
sum_j=1^inftyfraczeta(2j+1)-12j+1
&=sum_j=1^inftysum_k=2^inftyfrac1(2j+1)k^2j+1\
&=lim_ntoinftysum_k=2^nfrac12logleft(frack+1k-1right)-frac1k\
&=lim_ntoinftyfrac12log(n(n+1)/2)-(H_n-1)\
&=1-log(2)/2-gammatag8
endalign
$$
If we add twice the left side of $(8)$ to the right side of $(6)$ and vice versa, we get
$$
2-log(2)-gamma
=1-log(4)+log(3)+sum_j=1^inftyfraczeta(2j+1)-12^2j(2j+1)tag9
$$
From which we get the Euler-Stieltjes series:
$$
gamma=1-log(3/2)-sum_j=1^inftyfraczeta(2j+1)-14^j(2j+1)tag10
$$
Using the following application of $(3)$
$$
sum_j=1^inftyfracfrac1n^2j+14^j(2j+1)=logleft(frac2n+12n-1right)-frac1ntag11
$$
we can accelerate the convergence of $(10)$:
$$
gamma=H_n-log(n+1/2)-sum_j=1^inftyfraczeta(2j+1)-sumlimits_k=1^nfrac1k^2j+14^j(2j+1)tag12
$$
$(12)$ converges about $2log_10(2n+2)$ digits per term. Euler-Stieltjes is the case $n=1$ of $(12)$. Note that $zeta(2j+1)-sumlimits_k=1^nfrac1k^2j+1=zeta(2j+1,n+1)$, the Hurwitz Zeta function.
In this answer, I give a another method for computing $gamma$ that uses the an accelerated function for the sum of the tail of the alternating harmonic series.
edited Apr 13 '17 at 12:20
Community♦
1
answered Nov 12 '13 at 12:51
robjohn♦robjohn
269k27311638
$endgroup$
Note that for Harmonic numbers, $H_n$,
$$
sum_k=1^nleft(frac1k-logleft(1+frac1kright)right)=H_n-log(n+1)tag1
$$
Taking $(1)$ to the limit gives
$$
gamma=sum_k=1^inftyleft(frac1k-logleft(1+frac1kright)right)tag2
$$
We have the power series
$$
logleft(frac1+x1-xright)=2left(x+fracx^33+fracx^55+fracx^77dotsright)tag3
$$
If we set $x=frac12k+1$, then $frac1+x1-x=1+frac1k$; that is,
$$
logleft(1+frac1kright)=sum_j=0^inftyfrac22j+1left(frac12k+1right)^2j+1tag4
$$
Furthermore,
$$
beginalign
sum_k=1^inftyleft(frac12k+1right)^2j+1
&=sum_k=1^inftyleft(frac12kright)^2j+1
+left(frac12k+1right)^2j+1-frac12^2j+1left(frac1kright)^2j+1\
&=zeta(2j+1)-1-frac12^2j+1zeta(2j+1)\
&=frac2^2j+1-12^2j+1(zeta(2j+1)-1)-frac12^2j+1tag5
endalign
$$
Then, using $(4)$ in $(2)$, and then applying $(5)$, we get
$$
beginalign
gamma
&=sum_k=1^inftyleft(frac1k-logleft(1+frac1kright)right)\
&=sum_k=1^inftyfrac22k-frac22k+1-sum_j=1^inftyfrac22j+1left(frac12k+1right)^2j+1\
&=2(1-log(2))-sum_j=1^inftyfrac22j+1left(frac2^2j+1-12^2j+1(zeta(2j+1)-1)-frac12^2j+1right)\
&=1-log(4)+log(3)-sum_j=1^inftyfrac2^2j+1-12^2j(2j+1)(zeta(2j+1)-1)tag6
endalign
$$
The last equality in $(6)$ follows from plugging $k=frac12$ into $(4)$ to get
$$
log(3)-1=sum_j=1^inftyfrac2(2j+1)2^2j+1tag7
$$
We can accelerate the convergence of $(6)$ once, using $(3)$, we compute
$$
beginalign
sum_j=1^inftyfraczeta(2j+1)-12j+1
&=sum_j=1^inftysum_k=2^inftyfrac1(2j+1)k^2j+1\
&=lim_ntoinftysum_k=2^nfrac12logleft(frack+1k-1right)-frac1k\
&=lim_ntoinftyfrac12log(n(n+1)/2)-(H_n-1)\
&=1-log(2)/2-gammatag8
endalign
$$
If we add twice the left side of $(8)$ to the right side of $(6)$ and vice versa, we get
$$
2-log(2)-gamma
=1-log(4)+log(3)+sum_j=1^inftyfraczeta(2j+1)-12^2j(2j+1)tag9
$$
From which we get the Euler-Stieltjes series:
$$
gamma=1-log(3/2)-sum_j=1^inftyfraczeta(2j+1)-14^j(2j+1)tag10
$$
Using the following application of $(3)$
$$
sum_j=1^inftyfracfrac1n^2j+14^j(2j+1)=logleft(frac2n+12n-1right)-frac1ntag11
$$
we can accelerate the convergence of $(10)$:
$$
gamma=H_n-log(n+1/2)-sum_j=1^inftyfraczeta(2j+1)-sumlimits_k=1^nfrac1k^2j+14^j(2j+1)tag12
$$
$(12)$ converges about $2log_10(2n+2)$ digits per term. Euler-Stieltjes is the case $n=1$ of $(12)$. Note that $zeta(2j+1)-sumlimits_k=1^nfrac1k^2j+1=zeta(2j+1,n+1)$, the Hurwitz Zeta function.
In this answer, I give a another method for computing $gamma$ that uses the an accelerated function for the sum of the tail of the alternating harmonic series.
edited Apr 13 '17 at 12:20
Community♦
1
answered Nov 12 '13 at 12:51
robjohn♦robjohn
269k27311638
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
1
answered Nov 12 '13 at 12:51
robjohn♦robjohn
269k27311638
answered Nov 12 '13 at 12:51
robjohn♦robjohn
269k27311638
answered Nov 12 '13 at 12:51
robjohn♦robjohn
269k27311638
269k27311638
$begingroup$
To robjohn: Thanks for the nice formula which is almost similar to Euler-Stieltjes. And the best part is the very very elementary proof which is accessible to anyone who knows the definition of $gamma$ and the logarithmic series. I also checked your other answer and found it very informative. Thanks a lot (+1)! I have not yet accepted any answer to check for any more interesting answers.
$endgroup$
– Paramanand Singh
Nov 12 '13 at 14:33
$begingroup$
To Robjohn: Before I could connect the dots from your answer to Euler-Stieltjes, you gave the full derivation.. gr8!
$endgroup$
– Paramanand Singh
Nov 12 '13 at 14:43
$begingroup$
@ParamanandSingh: I have added a faster converging extension of Euler-Stieltjes.
$endgroup$
– robjohn♦
Nov 12 '13 at 16:11
$begingroup$
To Robjohn: I am speechless!
$endgroup$
– Paramanand Singh
Nov 12 '13 at 16:30
add a comment |
$begingroup$
To robjohn: Thanks for the nice formula which is almost similar to Euler-Stieltjes. And the best part is the very very elementary proof which is accessible to anyone who knows the definition of $gamma$ and the logarithmic series. I also checked your other answer and found it very informative. Thanks a lot (+1)! I have not yet accepted any answer to check for any more interesting answers.
$endgroup$
– Paramanand Singh
Nov 12 '13 at 14:33
$begingroup$
To Robjohn: Before I could connect the dots from your answer to Euler-Stieltjes, you gave the full derivation.. gr8!
$endgroup$
– Paramanand Singh
Nov 12 '13 at 14:43
$begingroup$
@ParamanandSingh: I have added a faster converging extension of Euler-Stieltjes.
$endgroup$
– robjohn♦
Nov 12 '13 at 16:11
$begingroup$
To Robjohn: I am speechless!
$endgroup$
– Paramanand Singh
Nov 12 '13 at 16:30
$begingroup$
To robjohn: Thanks for the nice formula which is almost similar to Euler-Stieltjes. And the best part is the very very elementary proof which is accessible to anyone who knows the definition of $gamma$ and the logarithmic series. I also checked your other answer and found it very informative. Thanks a lot (+1)! I have not yet accepted any answer to check for any more interesting answers.
$endgroup$
– Paramanand Singh
Nov 12 '13 at 14:33
$begingroup$
To robjohn: Thanks for the nice formula which is almost similar to Euler-Stieltjes. And the best part is the very very elementary proof which is accessible to anyone who knows the definition of $gamma$ and the logarithmic series. I also checked your other answer and found it very informative. Thanks a lot (+1)! I have not yet accepted any answer to check for any more interesting answers.
$endgroup$
– Paramanand Singh
Nov 12 '13 at 14:33
$begingroup$
To Robjohn: Before I could connect the dots from your answer to Euler-Stieltjes, you gave the full derivation.. gr8!
$endgroup$
– Paramanand Singh
Nov 12 '13 at 14:43
$begingroup$
To Robjohn: Before I could connect the dots from your answer to Euler-Stieltjes, you gave the full derivation.. gr8!
$endgroup$
– Paramanand Singh
Nov 12 '13 at 14:43
$begingroup$
@ParamanandSingh: I have added a faster converging extension of Euler-Stieltjes.
$endgroup$
– robjohn♦
Nov 12 '13 at 16:11
$begingroup$
@ParamanandSingh: I have added a faster converging extension of Euler-Stieltjes.
$endgroup$
– robjohn♦
Nov 12 '13 at 16:11
$begingroup$
To Robjohn: I am speechless!
$endgroup$
– Paramanand Singh
Nov 12 '13 at 16:30
$begingroup$
To Robjohn: I am speechless!
$endgroup$
– Paramanand Singh
Nov 12 '13 at 16:30
add a comment |
$begingroup$
There are a lot of formulas of this type. Some of them are in the Collection of formulae for Euler's constant $gamma;$ by Xavier Gourdon and Pascal Sebah:
$$gamma = frac32 - ln 2 - sum_nge 2frac1nleft(zeta(n)-1- frac12^nright)$$
$$gamma = frac116 - ln 3 - sum_nge 2frac1nleft(zeta(n)-1
-frac12^n -frac13^nright)$$
$$gamma = 1- lnleft(frac32right) -sum_nge1fraczeta(2n+1)-14^n(2n+1) qquadtext(Euler-Stieltjes)
$$
The first two are derived from the Hurwitz zeta function as special cases. The Euler-Stieltjes formula seems near to your remembrance but is listed without proof.
Edit: You can find a proof in the Expansion of Euler's constant in terms of zeta numbers by M. Prévost.
answered Nov 12 '13 at 9:47
gammatestergammatester
16.8k21733
$endgroup$
$begingroup$
Yes I also think that the third one is what I saw long back. I will try to go through the linked PDF references
$endgroup$
– Paramanand Singh
Nov 12 '13 at 10:05
$begingroup$
The first two series are neat in that they are the series given by lhf with the first two or three terms of $zeta(n)$ taken out. They give better convergence at about $0.477$ and $.602$ places per term. The Euler-Stieltjes Series gives approximately $1.2$ places per term. (+1)
$endgroup$
– robjohn♦
Nov 12 '13 at 15:26
$begingroup$
In the spirit of your first two series (which extend the series given by lhf), I have similarly accelerated the Euler-Stieltjes series in my answer.
$endgroup$
– robjohn♦
Nov 12 '13 at 16:39
add a comment |
$begingroup$
There are a lot of formulas of this type. Some of them are in the Collection of formulae for Euler's constant $gamma;$ by Xavier Gourdon and Pascal Sebah:
$$gamma = frac32 - ln 2 - sum_nge 2frac1nleft(zeta(n)-1- frac12^nright)$$
$$gamma = frac116 - ln 3 - sum_nge 2frac1nleft(zeta(n)-1
-frac12^n -frac13^nright)$$
$$gamma = 1- lnleft(frac32right) -sum_nge1fraczeta(2n+1)-14^n(2n+1) qquadtext(Euler-Stieltjes)
$$
The first two are derived from the Hurwitz zeta function as special cases. The Euler-Stieltjes formula seems near to your remembrance but is listed without proof.
Edit: You can find a proof in the Expansion of Euler's constant in terms of zeta numbers by M. Prévost.
answered Nov 12 '13 at 9:47
gammatestergammatester
16.8k21733
$endgroup$
$begingroup$
Yes I also think that the third one is what I saw long back. I will try to go through the linked PDF references
$endgroup$
– Paramanand Singh
Nov 12 '13 at 10:05
$begingroup$
The first two series are neat in that they are the series given by lhf with the first two or three terms of $zeta(n)$ taken out. They give better convergence at about $0.477$ and $.602$ places per term. The Euler-Stieltjes Series gives approximately $1.2$ places per term. (+1)
$endgroup$
– robjohn♦
Nov 12 '13 at 15:26
$begingroup$
In the spirit of your first two series (which extend the series given by lhf), I have similarly accelerated the Euler-Stieltjes series in my answer.
$endgroup$
– robjohn♦
Nov 12 '13 at 16:39
add a comment |
$begingroup$
There are a lot of formulas of this type. Some of them are in the Collection of formulae for Euler's constant $gamma;$ by Xavier Gourdon and Pascal Sebah:
$$gamma = frac32 - ln 2 - sum_nge 2frac1nleft(zeta(n)-1- frac12^nright)$$
$$gamma = frac116 - ln 3 - sum_nge 2frac1nleft(zeta(n)-1
-frac12^n -frac13^nright)$$
$$gamma = 1- lnleft(frac32right) -sum_nge1fraczeta(2n+1)-14^n(2n+1) qquadtext(Euler-Stieltjes)
$$
The first two are derived from the Hurwitz zeta function as special cases. The Euler-Stieltjes formula seems near to your remembrance but is listed without proof.
Edit: You can find a proof in the Expansion of Euler's constant in terms of zeta numbers by M. Prévost.
answered Nov 12 '13 at 9:47
gammatestergammatester
16.8k21733
$endgroup$
There are a lot of formulas of this type. Some of them are in the Collection of formulae for Euler's constant $gamma;$ by Xavier Gourdon and Pascal Sebah:
$$gamma = frac32 - ln 2 - sum_nge 2frac1nleft(zeta(n)-1- frac12^nright)$$
$$gamma = frac116 - ln 3 - sum_nge 2frac1nleft(zeta(n)-1
-frac12^n -frac13^nright)$$
$$gamma = 1- lnleft(frac32right) -sum_nge1fraczeta(2n+1)-14^n(2n+1) qquadtext(Euler-Stieltjes)
$$
The first two are derived from the Hurwitz zeta function as special cases. The Euler-Stieltjes formula seems near to your remembrance but is listed without proof.
Edit: You can find a proof in the Expansion of Euler's constant in terms of zeta numbers by M. Prévost.
answered Nov 12 '13 at 9:47
gammatestergammatester
16.8k21733
edited Nov 12 '13 at 10:01
answered Nov 12 '13 at 9:47
gammatestergammatester
16.8k21733
answered Nov 12 '13 at 9:47
gammatestergammatester
16.8k21733
answered Nov 12 '13 at 9:47
gammatestergammatester
16.8k21733
16.8k21733
$begingroup$
Yes I also think that the third one is what I saw long back. I will try to go through the linked PDF references
$endgroup$
– Paramanand Singh
Nov 12 '13 at 10:05
$begingroup$
The first two series are neat in that they are the series given by lhf with the first two or three terms of $zeta(n)$ taken out. They give better convergence at about $0.477$ and $.602$ places per term. The Euler-Stieltjes Series gives approximately $1.2$ places per term. (+1)
$endgroup$
– robjohn♦
Nov 12 '13 at 15:26
$begingroup$
In the spirit of your first two series (which extend the series given by lhf), I have similarly accelerated the Euler-Stieltjes series in my answer.
$endgroup$
– robjohn♦
Nov 12 '13 at 16:39
add a comment |
$begingroup$
Yes I also think that the third one is what I saw long back. I will try to go through the linked PDF references
$endgroup$
– Paramanand Singh
Nov 12 '13 at 10:05
$begingroup$
The first two series are neat in that they are the series given by lhf with the first two or three terms of $zeta(n)$ taken out. They give better convergence at about $0.477$ and $.602$ places per term. The Euler-Stieltjes Series gives approximately $1.2$ places per term. (+1)
$endgroup$
– robjohn♦
Nov 12 '13 at 15:26
$begingroup$
In the spirit of your first two series (which extend the series given by lhf), I have similarly accelerated the Euler-Stieltjes series in my answer.
$endgroup$
– robjohn♦
Nov 12 '13 at 16:39
$begingroup$
Yes I also think that the third one is what I saw long back. I will try to go through the linked PDF references
$endgroup$
– Paramanand Singh
Nov 12 '13 at 10:05
$begingroup$
Yes I also think that the third one is what I saw long back. I will try to go through the linked PDF references
$endgroup$
– Paramanand Singh
Nov 12 '13 at 10:05
$begingroup$
The first two series are neat in that they are the series given by lhf with the first two or three terms of $zeta(n)$ taken out. They give better convergence at about $0.477$ and $.602$ places per term. The Euler-Stieltjes Series gives approximately $1.2$ places per term. (+1)
$endgroup$
– robjohn♦
Nov 12 '13 at 15:26
$begingroup$
The first two series are neat in that they are the series given by lhf with the first two or three terms of $zeta(n)$ taken out. They give better convergence at about $0.477$ and $.602$ places per term. The Euler-Stieltjes Series gives approximately $1.2$ places per term. (+1)
$endgroup$
– robjohn♦
Nov 12 '13 at 15:26
$begingroup$
In the spirit of your first two series (which extend the series given by lhf), I have similarly accelerated the Euler-Stieltjes series in my answer.
$endgroup$
– robjohn♦
Nov 12 '13 at 16:39
$begingroup$
In the spirit of your first two series (which extend the series given by lhf), I have similarly accelerated the Euler-Stieltjes series in my answer.
$endgroup$
– robjohn♦
Nov 12 '13 at 16:39
add a comment |
$begingroup$
Do you mean this?
$$sum_k=2^infty zeta(k)-1over k= 1-gamma $$
This formula can be found in MathWorld (eq 123).
(Quoted in What is the fastest/most efficient algorithm for estimating Euler's Constant γ?.)
edited Apr 13 '17 at 12:20
Community♦
1
answered Nov 12 '13 at 10:22
lhflhf
166k10171400
$endgroup$
$begingroup$
No! the sum I had seen is mentioned in answer by gammatester namely $$gamma = 1- lnleft(frac32right) -sum_nge1fraczeta(2n+1)-14^n(2n+1) qquadtext(Euler-Stieltjes)$$ The formula which you mention has an easy proof.
$endgroup$
– Paramanand Singh
Nov 12 '13 at 10:44
$begingroup$
Although it may not be the series that Paramanand was seeking, it is still a valid series, giving about $0.3$ places each term. (+1)
$endgroup$
– robjohn♦
Nov 12 '13 at 15:20
add a comment |
$begingroup$
Do you mean this?
$$sum_k=2^infty zeta(k)-1over k= 1-gamma $$
This formula can be found in MathWorld (eq 123).
(Quoted in What is the fastest/most efficient algorithm for estimating Euler's Constant γ?.)
edited Apr 13 '17 at 12:20
Community♦
1
answered Nov 12 '13 at 10:22
lhflhf
166k10171400
$endgroup$
$begingroup$
No! the sum I had seen is mentioned in answer by gammatester namely $$gamma = 1- lnleft(frac32right) -sum_nge1fraczeta(2n+1)-14^n(2n+1) qquadtext(Euler-Stieltjes)$$ The formula which you mention has an easy proof.
$endgroup$
– Paramanand Singh
Nov 12 '13 at 10:44
$begingroup$
Although it may not be the series that Paramanand was seeking, it is still a valid series, giving about $0.3$ places each term. (+1)
$endgroup$
– robjohn♦
Nov 12 '13 at 15:20
add a comment |
$begingroup$
Do you mean this?
$$sum_k=2^infty zeta(k)-1over k= 1-gamma $$
This formula can be found in MathWorld (eq 123).
(Quoted in What is the fastest/most efficient algorithm for estimating Euler's Constant γ?.)
edited Apr 13 '17 at 12:20
Community♦
1
answered Nov 12 '13 at 10:22
lhflhf
166k10171400
$endgroup$
Do you mean this?
$$sum_k=2^infty zeta(k)-1over k= 1-gamma $$
This formula can be found in MathWorld (eq 123).
(Quoted in What is the fastest/most efficient algorithm for estimating Euler's Constant γ?.)
edited Apr 13 '17 at 12:20
Community♦
1
answered Nov 12 '13 at 10:22
lhflhf
166k10171400
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
1
answered Nov 12 '13 at 10:22
lhflhf
166k10171400
answered Nov 12 '13 at 10:22
lhflhf
166k10171400
answered Nov 12 '13 at 10:22
lhflhf
166k10171400
166k10171400
$begingroup$
No! the sum I had seen is mentioned in answer by gammatester namely $$gamma = 1- lnleft(frac32right) -sum_nge1fraczeta(2n+1)-14^n(2n+1) qquadtext(Euler-Stieltjes)$$ The formula which you mention has an easy proof.
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– Paramanand Singh
Nov 12 '13 at 10:44
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Although it may not be the series that Paramanand was seeking, it is still a valid series, giving about $0.3$ places each term. (+1)
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– robjohn♦
Nov 12 '13 at 15:20
add a comment |
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No! the sum I had seen is mentioned in answer by gammatester namely $$gamma = 1- lnleft(frac32right) -sum_nge1fraczeta(2n+1)-14^n(2n+1) qquadtext(Euler-Stieltjes)$$ The formula which you mention has an easy proof.
$endgroup$
– Paramanand Singh
Nov 12 '13 at 10:44
$begingroup$
Although it may not be the series that Paramanand was seeking, it is still a valid series, giving about $0.3$ places each term. (+1)
$endgroup$
– robjohn♦
Nov 12 '13 at 15:20
$begingroup$
No! the sum I had seen is mentioned in answer by gammatester namely $$gamma = 1- lnleft(frac32right) -sum_nge1fraczeta(2n+1)-14^n(2n+1) qquadtext(Euler-Stieltjes)$$ The formula which you mention has an easy proof.
$endgroup$
– Paramanand Singh
Nov 12 '13 at 10:44
$begingroup$
No! the sum I had seen is mentioned in answer by gammatester namely $$gamma = 1- lnleft(frac32right) -sum_nge1fraczeta(2n+1)-14^n(2n+1) qquadtext(Euler-Stieltjes)$$ The formula which you mention has an easy proof.
$endgroup$
– Paramanand Singh
Nov 12 '13 at 10:44
$begingroup$
Although it may not be the series that Paramanand was seeking, it is still a valid series, giving about $0.3$ places each term. (+1)
$endgroup$
– robjohn♦
Nov 12 '13 at 15:20
$begingroup$
Although it may not be the series that Paramanand was seeking, it is still a valid series, giving about $0.3$ places each term. (+1)
$endgroup$
– robjohn♦
Nov 12 '13 at 15:20
add a comment |
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The following paper has appeared just a few weeks ago: Jeffrey C. Lagarias: Euler's constant – Euler's work and modern developments. Bull. Am. Math. Soc. 50 (2013), 527–628. In this paper Lagarias has done for Euler's constant what Melville did for the whale.
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– Christian Blatter
Nov 12 '13 at 14:27
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Only a note (off-topic). The series with value of $zeta(n)$ are not the best way to evaluate $gamma$. The best two easy ways to calculate Euler's constant numerically are:<br> (1) Euler-Maclaurin sumformula<br> (2) With an asymptotic expansion of li(x) [integrallogarithm]. This Method is going back to Heinrich Wilhelm Brandes (1777–1834) in 1824.
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– skraemer
Mar 11 at 19:17