$sum_ n = 1^infty frac1(2n-1)(3n-1)$ The Next CEO of Stack OverflowSum of infinite series with arctan: $sum_n=1^inftyleft(arctanleft(frac14-nright)-arctanleft(-frac14-nright)right)$Prove the Series Converges or Diverges: $sum_n=1^infty frac5+2n(1+2n^2)^2$Why $sum_n=0^infty(-n)^3e^-n^-4$ absolutely converges?Convergence of the series $sum_n=1^inftyfrac12^n n (3n-1)$Hint on computing the series $sum_n=2^infty frac1n^2-1$.What does the series $sum_n=2^infty frac2n^3-n$ converge to?Compute the values of the Double SumHelp summing the telescoping series $sum_n=2^inftyfrac1n^3-n$.Evaluate $int_0^infty fract(t^2+1)(t^2+x^2) mathrmdt$if $f$ is such that : $int_-infty^infty x^2f(x) mathrmdx leq M$ then $f = O(x^3)$
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$sum_ n = 1^infty frac1(2n-1)(3n-1)$
The Next CEO of Stack OverflowSum of infinite series with arctan: $sum_n=1^inftyleft(arctanleft(frac14-nright)-arctanleft(-frac14-nright)right)$Prove the Series Converges or Diverges: $sum_n=1^infty frac5+2n(1+2n^2)^2$Why $sum_n=0^infty(-n)^3e^-n^-4$ absolutely converges?Convergence of the series $sum_n=1^inftyfrac12^n n (3n-1)$Hint on computing the series $sum_n=2^infty frac1n^2-1$.What does the series $sum_n=2^infty frac2n^3-n$ converge to?Compute the values of the Double SumHelp summing the telescoping series $sum_n=2^inftyfrac1n^3-n$.Evaluate $int_0^infty fract(t^2+1)(t^2+x^2) mathrmdt$if $f$ is such that : $int_-infty^infty x^2f(x) mathrmdx leq M$ then $f = O(x^3)$
$begingroup$
I would like to compute the value of the following sum
$$sum_ n = 1^infty frac1(2n-1)(3n-1)$$
Clearly, it converges since $ frac1(2n-1)(3n-1) = O(n^-2)$. I tried to use the partial fraction decomposition to get :
$$frac1(2n-1)(3n-1) = frac22n-1- frac33n-1$$
yet, it doesn't seem to lead anywhere since it's hard to see where the terms cancel out. So I don't really know what I could do in order to to attack this sum.
Thank you for your help!
real-analysis sequences-and-series limits
edited Mar 19 at 11:25
Chipa-Chipa
34
asked Mar 19 at 11:16
mouargmouargmouargmouarg
1145
$endgroup$
add a comment |
$begingroup$
I would like to compute the value of the following sum
$$sum_ n = 1^infty frac1(2n-1)(3n-1)$$
Clearly, it converges since $ frac1(2n-1)(3n-1) = O(n^-2)$. I tried to use the partial fraction decomposition to get :
$$frac1(2n-1)(3n-1) = frac22n-1- frac33n-1$$
yet, it doesn't seem to lead anywhere since it's hard to see where the terms cancel out. So I don't really know what I could do in order to to attack this sum.
Thank you for your help!
real-analysis sequences-and-series limits
edited Mar 19 at 11:25
Chipa-Chipa
34
asked Mar 19 at 11:16
mouargmouargmouargmouarg
1145
$endgroup$
1
$begingroup$
Take a look at the properties of digamma function.
$endgroup$
– Math-fun
Mar 19 at 11:36
$begingroup$
If you just want the answer, Wolfram Alpha gives $$fracpi2sqrt 3+2log 2-frac32log 3$$ But I don't know how that was calculated.
$endgroup$
– TonyK
Mar 19 at 11:37
add a comment |
$begingroup$
I would like to compute the value of the following sum
$$sum_ n = 1^infty frac1(2n-1)(3n-1)$$
Clearly, it converges since $ frac1(2n-1)(3n-1) = O(n^-2)$. I tried to use the partial fraction decomposition to get :
$$frac1(2n-1)(3n-1) = frac22n-1- frac33n-1$$
yet, it doesn't seem to lead anywhere since it's hard to see where the terms cancel out. So I don't really know what I could do in order to to attack this sum.
Thank you for your help!
real-analysis sequences-and-series limits
edited Mar 19 at 11:25
Chipa-Chipa
34
asked Mar 19 at 11:16
mouargmouargmouargmouarg
1145
$endgroup$
I would like to compute the value of the following sum
$$sum_ n = 1^infty frac1(2n-1)(3n-1)$$
Clearly, it converges since $ frac1(2n-1)(3n-1) = O(n^-2)$. I tried to use the partial fraction decomposition to get :
$$frac1(2n-1)(3n-1) = frac22n-1- frac33n-1$$
yet, it doesn't seem to lead anywhere since it's hard to see where the terms cancel out. So I don't really know what I could do in order to to attack this sum.
Thank you for your help!
real-analysis sequences-and-series limits
real-analysis sequences-and-series limits
edited Mar 19 at 11:25
Chipa-Chipa
34
asked Mar 19 at 11:16
mouargmouargmouargmouarg
1145
edited Mar 19 at 11:25
Chipa-Chipa
34
asked Mar 19 at 11:16
mouargmouargmouargmouarg
1145
edited Mar 19 at 11:25
Chipa-Chipa
34
edited Mar 19 at 11:25
Chipa-Chipa
34
edited Mar 19 at 11:25
Chipa-Chipa
34
34
asked Mar 19 at 11:16
mouargmouargmouargmouarg
1145
asked Mar 19 at 11:16
mouargmouargmouargmouarg
1145
asked Mar 19 at 11:16
mouargmouargmouargmouarg
1145
1145
1
$begingroup$
Take a look at the properties of digamma function.
$endgroup$
– Math-fun
Mar 19 at 11:36
$begingroup$
If you just want the answer, Wolfram Alpha gives $$fracpi2sqrt 3+2log 2-frac32log 3$$ But I don't know how that was calculated.
$endgroup$
– TonyK
Mar 19 at 11:37
add a comment |
1
$begingroup$
Take a look at the properties of digamma function.
$endgroup$
– Math-fun
Mar 19 at 11:36
$begingroup$
If you just want the answer, Wolfram Alpha gives $$fracpi2sqrt 3+2log 2-frac32log 3$$ But I don't know how that was calculated.
$endgroup$
– TonyK
Mar 19 at 11:37
1
1
$begingroup$
Take a look at the properties of digamma function.
$endgroup$
– Math-fun
Mar 19 at 11:36
$begingroup$
Take a look at the properties of digamma function.
$endgroup$
– Math-fun
Mar 19 at 11:36
$begingroup$
If you just want the answer, Wolfram Alpha gives $$fracpi2sqrt 3+2log 2-frac32log 3$$ But I don't know how that was calculated.
$endgroup$
– TonyK
Mar 19 at 11:37
$begingroup$
If you just want the answer, Wolfram Alpha gives $$fracpi2sqrt 3+2log 2-frac32log 3$$ But I don't know how that was calculated.
$endgroup$
– TonyK
Mar 19 at 11:37
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
As Math-fun commented, this is related to the digamma function
$$S_p=sum_ n = 1^p frac1(2n-1)(3n-1)=2sum_ n = 1^pfrac12n-1-3sum_ n = 1^p frac13n-1$$
$$S_p=psi ^(0)left(p+frac12right)-psi ^(0)left(p+frac23right)+psi
^(0)left(frac23right)-psi ^(0)left(frac12right)$$ Now, using the asymptotics
$$S_p=psi ^(0)left(frac23right)-psi
^(0)left(frac12right)-frac16 p+Oleft(frac1p^2right)$$ and
$$psi ^(0)left(frac23right)=-gamma +fracpi 2 sqrt3-frac3 log (3)2qquad textand qquad psi
^(0)left(frac12right)=-gamma -2log (2)$$
answered Mar 19 at 11:57
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
$begingroup$
Using they fact that
$$psi(z+1)-psi(s+1)=sum_n=1^infty frac1n+s-frac1n+z$$
Your sum is
$$psi(2/3)-psi(1/2)$$
This simplifies to
$$fracpi2sqrt3-frac3log 32+2log 2$$
answered Mar 19 at 13:06
ZacharyZachary
2,3551214
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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$begingroup$
As Math-fun commented, this is related to the digamma function
$$S_p=sum_ n = 1^p frac1(2n-1)(3n-1)=2sum_ n = 1^pfrac12n-1-3sum_ n = 1^p frac13n-1$$
$$S_p=psi ^(0)left(p+frac12right)-psi ^(0)left(p+frac23right)+psi
^(0)left(frac23right)-psi ^(0)left(frac12right)$$ Now, using the asymptotics
$$S_p=psi ^(0)left(frac23right)-psi
^(0)left(frac12right)-frac16 p+Oleft(frac1p^2right)$$ and
$$psi ^(0)left(frac23right)=-gamma +fracpi 2 sqrt3-frac3 log (3)2qquad textand qquad psi
^(0)left(frac12right)=-gamma -2log (2)$$
answered Mar 19 at 11:57
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
$begingroup$
As Math-fun commented, this is related to the digamma function
$$S_p=sum_ n = 1^p frac1(2n-1)(3n-1)=2sum_ n = 1^pfrac12n-1-3sum_ n = 1^p frac13n-1$$
$$S_p=psi ^(0)left(p+frac12right)-psi ^(0)left(p+frac23right)+psi
^(0)left(frac23right)-psi ^(0)left(frac12right)$$ Now, using the asymptotics
$$S_p=psi ^(0)left(frac23right)-psi
^(0)left(frac12right)-frac16 p+Oleft(frac1p^2right)$$ and
$$psi ^(0)left(frac23right)=-gamma +fracpi 2 sqrt3-frac3 log (3)2qquad textand qquad psi
^(0)left(frac12right)=-gamma -2log (2)$$
answered Mar 19 at 11:57
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
$begingroup$
As Math-fun commented, this is related to the digamma function
$$S_p=sum_ n = 1^p frac1(2n-1)(3n-1)=2sum_ n = 1^pfrac12n-1-3sum_ n = 1^p frac13n-1$$
$$S_p=psi ^(0)left(p+frac12right)-psi ^(0)left(p+frac23right)+psi
^(0)left(frac23right)-psi ^(0)left(frac12right)$$ Now, using the asymptotics
$$S_p=psi ^(0)left(frac23right)-psi
^(0)left(frac12right)-frac16 p+Oleft(frac1p^2right)$$ and
$$psi ^(0)left(frac23right)=-gamma +fracpi 2 sqrt3-frac3 log (3)2qquad textand qquad psi
^(0)left(frac12right)=-gamma -2log (2)$$
answered Mar 19 at 11:57
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
As Math-fun commented, this is related to the digamma function
$$S_p=sum_ n = 1^p frac1(2n-1)(3n-1)=2sum_ n = 1^pfrac12n-1-3sum_ n = 1^p frac13n-1$$
$$S_p=psi ^(0)left(p+frac12right)-psi ^(0)left(p+frac23right)+psi
^(0)left(frac23right)-psi ^(0)left(frac12right)$$ Now, using the asymptotics
$$S_p=psi ^(0)left(frac23right)-psi
^(0)left(frac12right)-frac16 p+Oleft(frac1p^2right)$$ and
$$psi ^(0)left(frac23right)=-gamma +fracpi 2 sqrt3-frac3 log (3)2qquad textand qquad psi
^(0)left(frac12right)=-gamma -2log (2)$$
answered Mar 19 at 11:57
Claude LeiboviciClaude Leibovici
125k1158135
answered Mar 19 at 11:57
Claude LeiboviciClaude Leibovici
125k1158135
answered Mar 19 at 11:57
Claude LeiboviciClaude Leibovici
125k1158135
answered Mar 19 at 11:57
Claude LeiboviciClaude Leibovici
125k1158135
125k1158135
add a comment |
add a comment |
$begingroup$
Using they fact that
$$psi(z+1)-psi(s+1)=sum_n=1^infty frac1n+s-frac1n+z$$
Your sum is
$$psi(2/3)-psi(1/2)$$
This simplifies to
$$fracpi2sqrt3-frac3log 32+2log 2$$
answered Mar 19 at 13:06
ZacharyZachary
2,3551214
$endgroup$
add a comment |
$begingroup$
Using they fact that
$$psi(z+1)-psi(s+1)=sum_n=1^infty frac1n+s-frac1n+z$$
Your sum is
$$psi(2/3)-psi(1/2)$$
This simplifies to
$$fracpi2sqrt3-frac3log 32+2log 2$$
answered Mar 19 at 13:06
ZacharyZachary
2,3551214
$endgroup$
add a comment |
$begingroup$
Using they fact that
$$psi(z+1)-psi(s+1)=sum_n=1^infty frac1n+s-frac1n+z$$
Your sum is
$$psi(2/3)-psi(1/2)$$
This simplifies to
$$fracpi2sqrt3-frac3log 32+2log 2$$
answered Mar 19 at 13:06
ZacharyZachary
2,3551214
$endgroup$
Using they fact that
$$psi(z+1)-psi(s+1)=sum_n=1^infty frac1n+s-frac1n+z$$
Your sum is
$$psi(2/3)-psi(1/2)$$
This simplifies to
$$fracpi2sqrt3-frac3log 32+2log 2$$
answered Mar 19 at 13:06
ZacharyZachary
2,3551214
answered Mar 19 at 13:06
ZacharyZachary
2,3551214
answered Mar 19 at 13:06
ZacharyZachary
2,3551214
answered Mar 19 at 13:06
ZacharyZachary
2,3551214
2,3551214
add a comment |
add a comment |
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1
$begingroup$
Take a look at the properties of digamma function.
$endgroup$
– Math-fun
Mar 19 at 11:36
$begingroup$
If you just want the answer, Wolfram Alpha gives $$fracpi2sqrt 3+2log 2-frac32log 3$$ But I don't know how that was calculated.
$endgroup$
– TonyK
Mar 19 at 11:37