Closed form of $sum_n=1^infty(-1)^n fracln^2(n)n$Closed form for $sum_n=-infty^infty frac1n^4+a^4$Closed form for $sum_n=1^inftyfracpsi(n+frac54)(1+2n)(1+4n)^2$Closed form of specific seriesclosed form sum: $sum_i=1^n frac2^-ii$A closed form of $sum_k=0^inftyfrac(-1)^k+1k!Gamma^2left(frack2right)$Closed form for $ fracH_kk^2 $.Closed form for $sum_n=1^infty frac11+n+n^2+cdots+n^a$Closed form for this particular Taylor seriesIs :$sum_n=2^inftyfrac(-1)^n1-e^n=0.5sum_n=2^inftyfrac11-e^n$?Closed form of $sum_n=1^infty q^- n^2 z^n$
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Closed form of $sum_n=1^infty(-1)^n fracln^2(n)n$
Closed form for $sum_n=-infty^infty frac1n^4+a^4$Closed form for $sum_n=1^inftyfracpsi(n+frac54)(1+2n)(1+4n)^2$Closed form of specific seriesclosed form sum: $sum_i=1^n frac2^-ii$A closed form of $sum_k=0^inftyfrac(-1)^k+1k!Gamma^2left(frack2right)$Closed form for $ fracH_kk^2 $.Closed form for $sum_n=1^infty frac11+n+n^2+cdots+n^a$Closed form for this particular Taylor seriesIs :$sum_n=2^inftyfrac(-1)^n1-e^n=0.5sum_n=2^inftyfrac11-e^n$?Closed form of $sum_n=1^infty q^- n^2 z^n$
$begingroup$
I am working a maths problem and came across this sum. I am unable to evaluates it.
$$sum_n=1^infty(-1)^n fracln^2(n)n$$
I have checked on wolfram and wikipedia maths pages but can't seem to finds it.
Can anyone please point to its closed form? Thank you.
sequences-and-series summation logarithms
edited Mar 21 at 8:07
Andrews
1,2812422
asked Mar 21 at 7:24
coffeeecoffeee
18919
$endgroup$
add a comment |
$begingroup$
I am working a maths problem and came across this sum. I am unable to evaluates it.
$$sum_n=1^infty(-1)^n fracln^2(n)n$$
I have checked on wolfram and wikipedia maths pages but can't seem to finds it.
Can anyone please point to its closed form? Thank you.
sequences-and-series summation logarithms
edited Mar 21 at 8:07
Andrews
1,2812422
asked Mar 21 at 7:24
coffeeecoffeee
18919
$endgroup$
$begingroup$
The Dirichlet eta function $eta(s):=sum_nge 1(-1)^n-1n^-s$, so your sum is $-eta^primeprime(1)$. I'm not sure we can do better than that.
$endgroup$
– J.G.
Mar 21 at 7:44
$begingroup$
@J.G. There is an explicit expression for your result.
$endgroup$
– Claude Leibovici
Mar 21 at 8:51
add a comment |
$begingroup$
I am working a maths problem and came across this sum. I am unable to evaluates it.
$$sum_n=1^infty(-1)^n fracln^2(n)n$$
I have checked on wolfram and wikipedia maths pages but can't seem to finds it.
Can anyone please point to its closed form? Thank you.
sequences-and-series summation logarithms
edited Mar 21 at 8:07
Andrews
1,2812422
asked Mar 21 at 7:24
coffeeecoffeee
18919
$endgroup$
I am working a maths problem and came across this sum. I am unable to evaluates it.
$$sum_n=1^infty(-1)^n fracln^2(n)n$$
I have checked on wolfram and wikipedia maths pages but can't seem to finds it.
Can anyone please point to its closed form? Thank you.
sequences-and-series summation logarithms
sequences-and-series summation logarithms
edited Mar 21 at 8:07
Andrews
1,2812422
asked Mar 21 at 7:24
coffeeecoffeee
18919
edited Mar 21 at 8:07
Andrews
1,2812422
asked Mar 21 at 7:24
coffeeecoffeee
18919
edited Mar 21 at 8:07
Andrews
1,2812422
edited Mar 21 at 8:07
Andrews
1,2812422
edited Mar 21 at 8:07
Andrews
1,2812422
1,2812422
asked Mar 21 at 7:24
coffeeecoffeee
18919
asked Mar 21 at 7:24
coffeeecoffeee
18919
asked Mar 21 at 7:24
coffeeecoffeee
18919
18919
$begingroup$
The Dirichlet eta function $eta(s):=sum_nge 1(-1)^n-1n^-s$, so your sum is $-eta^primeprime(1)$. I'm not sure we can do better than that.
$endgroup$
– J.G.
Mar 21 at 7:44
$begingroup$
@J.G. There is an explicit expression for your result.
$endgroup$
– Claude Leibovici
Mar 21 at 8:51
add a comment |
$begingroup$
The Dirichlet eta function $eta(s):=sum_nge 1(-1)^n-1n^-s$, so your sum is $-eta^primeprime(1)$. I'm not sure we can do better than that.
$endgroup$
– J.G.
Mar 21 at 7:44
$begingroup$
@J.G. There is an explicit expression for your result.
$endgroup$
– Claude Leibovici
Mar 21 at 8:51
$begingroup$
The Dirichlet eta function $eta(s):=sum_nge 1(-1)^n-1n^-s$, so your sum is $-eta^primeprime(1)$. I'm not sure we can do better than that.
$endgroup$
– J.G.
Mar 21 at 7:44
$begingroup$
The Dirichlet eta function $eta(s):=sum_nge 1(-1)^n-1n^-s$, so your sum is $-eta^primeprime(1)$. I'm not sure we can do better than that.
$endgroup$
– J.G.
Mar 21 at 7:44
$begingroup$
@J.G. There is an explicit expression for your result.
$endgroup$
– Claude Leibovici
Mar 21 at 8:51
$begingroup$
@J.G. There is an explicit expression for your result.
$endgroup$
– Claude Leibovici
Mar 21 at 8:51
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
According to a CAS
$$sum_n=1^infty(-1)^n fraclog^2(n)n=2 gamma _1 log (2)-frac13log ^3(2)+gamma log ^2(2)approx 0.0653726$$ where $gamma _1$ is a Stieltjes constant.
If we use, as J.G. commented,
$$eta(s)=sum_n=1^infty(-1)^n-1n^-s=left(2^1-s-1right) zeta (s)$$
$$fracd^2ds^2eta(s)=2^-s left(2 log ^2(2) zeta (s)-4 log (2) zeta '(s)-left(2^s-2right) zeta ''(s)right)$$ and then the result.
Edit
Let us consider the general case
$$S_p =sum _n=1^infty (-1)^nfrac log ^p(n)n$$ The results are
$$left(
beginarraycc
p & fracS_plog(2) \
1 & gamma -fraclog (2)2 \
2 & gamma log (2)-fraclog ^2(2)3+2 gamma _1 \
3 & gamma log ^2(2)-fraclog ^3(2)4+3log (2) gamma _1+3 gamma _2 \
4 & gamma log ^3(2)-fraclog ^4(2)5+4 log ^2(2) gamma _1+6log (2)
gamma _2+4 gamma _3 \
5 & gamma log ^4(2)-fraclog ^5(2)6+5 log ^3(2) gamma _1+10 log ^2(2)
gamma _2+10 log (2) gamma _3+5 gamma _4
endarray
right)$$ where interesting patterns appear.
answered Mar 21 at 8:43
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
$begingroup$
thank you so much @Claude Leibovice. I was about to lose hope.
$endgroup$
– coffeee
Mar 21 at 8:50
$begingroup$
@coffeee. You are very welcome ! I did not do much (a CAS made it). The good point is in J.G.'s comment.
$endgroup$
– Claude Leibovici
Mar 21 at 8:53
$begingroup$
i know, but you give me the result, that is all i want. now i can continue with my work. thank you again.
$endgroup$
– coffeee
Mar 21 at 8:55
$begingroup$
I am sorry to bother @Claude Leibovici, can you help me on the closed form of this sum: $sum_n=1^infty(-1)^n fracln^3(n)n$. I don't have CAS software. Thank you.
$endgroup$
– coffeee
Mar 22 at 6:30
$begingroup$
@coffeee. No problem ! I just added a few more. Do not hesitate to ask. Glad to help.
$endgroup$
– Claude Leibovici
Mar 22 at 6:45
|
show 1 more comment
Your Answer
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1 Answer
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oldest
votes
1 Answer
1
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oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
According to a CAS
$$sum_n=1^infty(-1)^n fraclog^2(n)n=2 gamma _1 log (2)-frac13log ^3(2)+gamma log ^2(2)approx 0.0653726$$ where $gamma _1$ is a Stieltjes constant.
If we use, as J.G. commented,
$$eta(s)=sum_n=1^infty(-1)^n-1n^-s=left(2^1-s-1right) zeta (s)$$
$$fracd^2ds^2eta(s)=2^-s left(2 log ^2(2) zeta (s)-4 log (2) zeta '(s)-left(2^s-2right) zeta ''(s)right)$$ and then the result.
Edit
Let us consider the general case
$$S_p =sum _n=1^infty (-1)^nfrac log ^p(n)n$$ The results are
$$left(
beginarraycc
p & fracS_plog(2) \
1 & gamma -fraclog (2)2 \
2 & gamma log (2)-fraclog ^2(2)3+2 gamma _1 \
3 & gamma log ^2(2)-fraclog ^3(2)4+3log (2) gamma _1+3 gamma _2 \
4 & gamma log ^3(2)-fraclog ^4(2)5+4 log ^2(2) gamma _1+6log (2)
gamma _2+4 gamma _3 \
5 & gamma log ^4(2)-fraclog ^5(2)6+5 log ^3(2) gamma _1+10 log ^2(2)
gamma _2+10 log (2) gamma _3+5 gamma _4
endarray
right)$$ where interesting patterns appear.
answered Mar 21 at 8:43
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
$begingroup$
thank you so much @Claude Leibovice. I was about to lose hope.
$endgroup$
– coffeee
Mar 21 at 8:50
$begingroup$
@coffeee. You are very welcome ! I did not do much (a CAS made it). The good point is in J.G.'s comment.
$endgroup$
– Claude Leibovici
Mar 21 at 8:53
$begingroup$
i know, but you give me the result, that is all i want. now i can continue with my work. thank you again.
$endgroup$
– coffeee
Mar 21 at 8:55
$begingroup$
I am sorry to bother @Claude Leibovici, can you help me on the closed form of this sum: $sum_n=1^infty(-1)^n fracln^3(n)n$. I don't have CAS software. Thank you.
$endgroup$
– coffeee
Mar 22 at 6:30
$begingroup$
@coffeee. No problem ! I just added a few more. Do not hesitate to ask. Glad to help.
$endgroup$
– Claude Leibovici
Mar 22 at 6:45
|
show 1 more comment
$begingroup$
According to a CAS
$$sum_n=1^infty(-1)^n fraclog^2(n)n=2 gamma _1 log (2)-frac13log ^3(2)+gamma log ^2(2)approx 0.0653726$$ where $gamma _1$ is a Stieltjes constant.
If we use, as J.G. commented,
$$eta(s)=sum_n=1^infty(-1)^n-1n^-s=left(2^1-s-1right) zeta (s)$$
$$fracd^2ds^2eta(s)=2^-s left(2 log ^2(2) zeta (s)-4 log (2) zeta '(s)-left(2^s-2right) zeta ''(s)right)$$ and then the result.
Edit
Let us consider the general case
$$S_p =sum _n=1^infty (-1)^nfrac log ^p(n)n$$ The results are
$$left(
beginarraycc
p & fracS_plog(2) \
1 & gamma -fraclog (2)2 \
2 & gamma log (2)-fraclog ^2(2)3+2 gamma _1 \
3 & gamma log ^2(2)-fraclog ^3(2)4+3log (2) gamma _1+3 gamma _2 \
4 & gamma log ^3(2)-fraclog ^4(2)5+4 log ^2(2) gamma _1+6log (2)
gamma _2+4 gamma _3 \
5 & gamma log ^4(2)-fraclog ^5(2)6+5 log ^3(2) gamma _1+10 log ^2(2)
gamma _2+10 log (2) gamma _3+5 gamma _4
endarray
right)$$ where interesting patterns appear.
answered Mar 21 at 8:43
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
$begingroup$
thank you so much @Claude Leibovice. I was about to lose hope.
$endgroup$
– coffeee
Mar 21 at 8:50
$begingroup$
@coffeee. You are very welcome ! I did not do much (a CAS made it). The good point is in J.G.'s comment.
$endgroup$
– Claude Leibovici
Mar 21 at 8:53
$begingroup$
i know, but you give me the result, that is all i want. now i can continue with my work. thank you again.
$endgroup$
– coffeee
Mar 21 at 8:55
$begingroup$
I am sorry to bother @Claude Leibovici, can you help me on the closed form of this sum: $sum_n=1^infty(-1)^n fracln^3(n)n$. I don't have CAS software. Thank you.
$endgroup$
– coffeee
Mar 22 at 6:30
$begingroup$
@coffeee. No problem ! I just added a few more. Do not hesitate to ask. Glad to help.
$endgroup$
– Claude Leibovici
Mar 22 at 6:45
|
show 1 more comment
$begingroup$
According to a CAS
$$sum_n=1^infty(-1)^n fraclog^2(n)n=2 gamma _1 log (2)-frac13log ^3(2)+gamma log ^2(2)approx 0.0653726$$ where $gamma _1$ is a Stieltjes constant.
If we use, as J.G. commented,
$$eta(s)=sum_n=1^infty(-1)^n-1n^-s=left(2^1-s-1right) zeta (s)$$
$$fracd^2ds^2eta(s)=2^-s left(2 log ^2(2) zeta (s)-4 log (2) zeta '(s)-left(2^s-2right) zeta ''(s)right)$$ and then the result.
Edit
Let us consider the general case
$$S_p =sum _n=1^infty (-1)^nfrac log ^p(n)n$$ The results are
$$left(
beginarraycc
p & fracS_plog(2) \
1 & gamma -fraclog (2)2 \
2 & gamma log (2)-fraclog ^2(2)3+2 gamma _1 \
3 & gamma log ^2(2)-fraclog ^3(2)4+3log (2) gamma _1+3 gamma _2 \
4 & gamma log ^3(2)-fraclog ^4(2)5+4 log ^2(2) gamma _1+6log (2)
gamma _2+4 gamma _3 \
5 & gamma log ^4(2)-fraclog ^5(2)6+5 log ^3(2) gamma _1+10 log ^2(2)
gamma _2+10 log (2) gamma _3+5 gamma _4
endarray
right)$$ where interesting patterns appear.
answered Mar 21 at 8:43
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
According to a CAS
$$sum_n=1^infty(-1)^n fraclog^2(n)n=2 gamma _1 log (2)-frac13log ^3(2)+gamma log ^2(2)approx 0.0653726$$ where $gamma _1$ is a Stieltjes constant.
If we use, as J.G. commented,
$$eta(s)=sum_n=1^infty(-1)^n-1n^-s=left(2^1-s-1right) zeta (s)$$
$$fracd^2ds^2eta(s)=2^-s left(2 log ^2(2) zeta (s)-4 log (2) zeta '(s)-left(2^s-2right) zeta ''(s)right)$$ and then the result.
Edit
Let us consider the general case
$$S_p =sum _n=1^infty (-1)^nfrac log ^p(n)n$$ The results are
$$left(
beginarraycc
p & fracS_plog(2) \
1 & gamma -fraclog (2)2 \
2 & gamma log (2)-fraclog ^2(2)3+2 gamma _1 \
3 & gamma log ^2(2)-fraclog ^3(2)4+3log (2) gamma _1+3 gamma _2 \
4 & gamma log ^3(2)-fraclog ^4(2)5+4 log ^2(2) gamma _1+6log (2)
gamma _2+4 gamma _3 \
5 & gamma log ^4(2)-fraclog ^5(2)6+5 log ^3(2) gamma _1+10 log ^2(2)
gamma _2+10 log (2) gamma _3+5 gamma _4
endarray
right)$$ where interesting patterns appear.
answered Mar 21 at 8:43
Claude LeiboviciClaude Leibovici
125k1158135
edited Mar 22 at 6:44
answered Mar 21 at 8:43
Claude LeiboviciClaude Leibovici
125k1158135
answered Mar 21 at 8:43
Claude LeiboviciClaude Leibovici
125k1158135
answered Mar 21 at 8:43
Claude LeiboviciClaude Leibovici
125k1158135
125k1158135
$begingroup$
thank you so much @Claude Leibovice. I was about to lose hope.
$endgroup$
– coffeee
Mar 21 at 8:50
$begingroup$
@coffeee. You are very welcome ! I did not do much (a CAS made it). The good point is in J.G.'s comment.
$endgroup$
– Claude Leibovici
Mar 21 at 8:53
$begingroup$
i know, but you give me the result, that is all i want. now i can continue with my work. thank you again.
$endgroup$
– coffeee
Mar 21 at 8:55
$begingroup$
I am sorry to bother @Claude Leibovici, can you help me on the closed form of this sum: $sum_n=1^infty(-1)^n fracln^3(n)n$. I don't have CAS software. Thank you.
$endgroup$
– coffeee
Mar 22 at 6:30
$begingroup$
@coffeee. No problem ! I just added a few more. Do not hesitate to ask. Glad to help.
$endgroup$
– Claude Leibovici
Mar 22 at 6:45
|
show 1 more comment
$begingroup$
thank you so much @Claude Leibovice. I was about to lose hope.
$endgroup$
– coffeee
Mar 21 at 8:50
$begingroup$
@coffeee. You are very welcome ! I did not do much (a CAS made it). The good point is in J.G.'s comment.
$endgroup$
– Claude Leibovici
Mar 21 at 8:53
$begingroup$
i know, but you give me the result, that is all i want. now i can continue with my work. thank you again.
$endgroup$
– coffeee
Mar 21 at 8:55
$begingroup$
I am sorry to bother @Claude Leibovici, can you help me on the closed form of this sum: $sum_n=1^infty(-1)^n fracln^3(n)n$. I don't have CAS software. Thank you.
$endgroup$
– coffeee
Mar 22 at 6:30
$begingroup$
@coffeee. No problem ! I just added a few more. Do not hesitate to ask. Glad to help.
$endgroup$
– Claude Leibovici
Mar 22 at 6:45
$begingroup$
thank you so much @Claude Leibovice. I was about to lose hope.
$endgroup$
– coffeee
Mar 21 at 8:50
$begingroup$
thank you so much @Claude Leibovice. I was about to lose hope.
$endgroup$
– coffeee
Mar 21 at 8:50
$begingroup$
@coffeee. You are very welcome ! I did not do much (a CAS made it). The good point is in J.G.'s comment.
$endgroup$
– Claude Leibovici
Mar 21 at 8:53
$begingroup$
@coffeee. You are very welcome ! I did not do much (a CAS made it). The good point is in J.G.'s comment.
$endgroup$
– Claude Leibovici
Mar 21 at 8:53
$begingroup$
i know, but you give me the result, that is all i want. now i can continue with my work. thank you again.
$endgroup$
– coffeee
Mar 21 at 8:55
$begingroup$
i know, but you give me the result, that is all i want. now i can continue with my work. thank you again.
$endgroup$
– coffeee
Mar 21 at 8:55
$begingroup$
I am sorry to bother @Claude Leibovici, can you help me on the closed form of this sum: $sum_n=1^infty(-1)^n fracln^3(n)n$. I don't have CAS software. Thank you.
$endgroup$
– coffeee
Mar 22 at 6:30
$begingroup$
I am sorry to bother @Claude Leibovici, can you help me on the closed form of this sum: $sum_n=1^infty(-1)^n fracln^3(n)n$. I don't have CAS software. Thank you.
$endgroup$
– coffeee
Mar 22 at 6:30
$begingroup$
@coffeee. No problem ! I just added a few more. Do not hesitate to ask. Glad to help.
$endgroup$
– Claude Leibovici
Mar 22 at 6:45
$begingroup$
@coffeee. No problem ! I just added a few more. Do not hesitate to ask. Glad to help.
$endgroup$
– Claude Leibovici
Mar 22 at 6:45
|
show 1 more comment
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The Dirichlet eta function $eta(s):=sum_nge 1(-1)^n-1n^-s$, so your sum is $-eta^primeprime(1)$. I'm not sure we can do better than that.
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– J.G.
Mar 21 at 7:44
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@J.G. There is an explicit expression for your result.
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– Claude Leibovici
Mar 21 at 8:51