# The 6-order Euler sums

How to calculate the value of the series $$\sum\limits_{n = 1}^\infty {\frac{1}{{{n^4}}}\left( {\sum\limits_{k = 1}^n {\frac{{{{\left( { - 1} \right)}^{k - 1}}}}{{{k^2}}}} } \right)} \quad\text{and}\quad \sum\limits_{n = 1}^\infty {\frac{1}{{{n^5}}}\left( {\sum\limits_{k = 1}^n {\frac{{{{\left( { - 1} \right)}^{k - 1}}}}{k}} } \right)} .$$

• see here for the first sum math.stackexchange.com/questions/650966/… – Dr. Sonnhard Graubner Oct 18 '14 at 6:48
• @Dr.SonnhardGraubner, did you realize these are two double sums? – andre Oct 18 '14 at 7:38
• Once 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 8:02
• The 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 8:18
• I 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 9:17