How to calculate the value of the series $$\sum\limits_{n = 1}^\infty {\frac{1}{{{n^4}}}\left( {\sum\limits_{k = 1}^n {\frac{{{{\left( { - 1} \right)}^{k - 1}}}}{{{k^2}}}} } \right)} \quad\text{and}\quad \sum\limits_{n = 1}^\infty {\frac{1}{{{n^5}}}\left( {\sum\limits_{k = 1}^n {\frac{{{{\left( { - 1} \right)}^{k - 1}}}}{k}} } \right)} .$$
Asked
Active
Viewed 163 times
3
-
see here for the first sum http://math.stackexchange.com/questions/650966/evaluate-sum-infty-n-1-frac1n4-using-parsevals-theorem-fourier-ser – Dr. Sonnhard Graubner Oct 18 '14 at 06:48
-
for the second sum we obtain $1/12,{\frac {12, \left( -1 \right) ^{n}{\it LerchPhi} \left( -1,2,n \right) {n}^{2}+{\pi }^{2}{n}^{2}-12, \left( -1 \right) ^{n}}{{n}^{2 }}} $ – Dr. Sonnhard Graubner Oct 18 '14 at 06:52
-
@Dr.SonnhardGraubner So you stopped answering? – Oct 18 '14 at 06:58
-
$\sum_{n=1}^{\infty}\frac{1}{n^5}=\zeta(5)$ – Dr. Sonnhard Graubner Oct 18 '14 at 06:59
-
for the last sum we obtain $(-1)^{n+1} \Phi (-1,1,n+1)+\log (2)$ – Dr. Sonnhard Graubner Oct 18 '14 at 07:00
-
1@Dr.SonnhardGraubner, did you realize these are two double sums? – andre Oct 18 '14 at 07:38
-
2Once again, may I suggest that you share your attempts at tackling these two sums? I am sure that the users here would be more willing to help you with this problem if you do so. Thanks. – M.N.C.E. Oct 18 '14 at 08:02
-
1The inner sum is already complex to me (even if a CAS gives its expression in terms of two $\zeta$ functions). – Claude Leibovici Oct 18 '14 at 08:18
-
3I managed to get $$\sum^\infty_{n=1}\frac{1}{n^5}\sum^n_{k=1}\frac{(-1)^{k-1}}{k}=-\frac{7\pi^6}{8640}+\frac{31}{16}\zeta(5)\ln{2}+\frac{9}{32}\zeta^2(3)$$ The first sum might not have a closed form though. – M.N.C.E. Oct 18 '14 at 09:17