Can anybody find a closed form for this infinite sum?

$$S = \sum _{k=4}^{\infty }{\frac { \left( -\ln \left( 2 \right) \right) ^ {k}\zeta \left( 4-k \right) }{k!}},$$

where $\zeta$ is the Riemann zeta function.

An approximate value of $S$ is

$$S \approx -0.00469807827332540098459248437391306962194656968313196911104278149327118$$

I found nothing with Maple, Mathematica or ISC.

  • $\begingroup$ Wolfram doesn't know one. $\endgroup$ – amcalde Oct 1 '14 at 15:51
  • $\begingroup$ On a related note, $~\displaystyle\sum_{k=0}^\infty\frac{a^k}{k!}\zeta(-k)=\frac1a+\frac1{1-e^a}-1$, for $|a|\le1$. $\endgroup$ – Lucian Oct 1 '14 at 17:15

Starting with the well-known expansion for the polylogarithm in terms of the zeta function, $$ \mathrm{Li}_n(z) = \sum_{m\geq 0, m\neq n-1}\zeta(n-m)\frac{\log^m z}{m!} + \frac{\log^{n-1}z}{(n-1)!}(H_{n-1}-\log\log\tfrac1z), $$ which is valid for $|\log z|<2\pi$, we can substitute $n=4$, $z=\frac12$, and obtain $$ S = \text{Li}_4(\tfrac{1}{2})+\zeta (3) \log2-\zeta (4)-\tfrac{1}{2} \zeta (2) \log^22-\tfrac{1}{6} \log^32 \log \log2+\tfrac{11}{36} \log^32. $$


For any $n\leq 0$ we have: $$\zeta(-n)=-\frac{B_{n+1}}{n+1}$$ hence: $$S=\sum_{n=0}^{+\infty}\frac{(-1)^n (\log 2)^{n+4}\zeta(-n)}{(n+4)!}=(\log 2)^3\sum_{n=0}^{+\infty}\frac{(-\log 2)^{n+1}B_{n+1}}{(n+1)(n+2)(n+3)(n+4)\cdot(n+1)!}$$ but since: $$\sum_{n=0}^{+\infty}\frac{B_{n+1}}{(n+1)!}z^{n+1} = \frac{z}{e^z-1}-1=\frac{1+z-e^z}{e^z-1}=f(z)$$ we get: $$S=-\frac{1}{\log 2}\int_{0}^{-\log 2}\int_{0}^{w}\int_{0}^{v}\int_{0}^{u}f(z)\,dz\,du\,dv\,dw.$$ An explicit integration leads to a complicated expression in terms of powers of $\log 2$ and values of the polylogarithms, up to $\operatorname{Li}_5$, in $1$ (hence values of the $\zeta$-function in the natural numbers greater than one) and $\frac{1}{2}$.

  • 1
    $\begingroup$ Are you sure about $\mathrm{Li}_5$? I got a different answer. $\endgroup$ – Kirill Oct 4 '14 at 19:11
  • $\begingroup$ Yes, it is interesting. The motivation of the problem was $\operatorname{Li}_4(1/2)$. Could you show us the evaluation of $S$? $\endgroup$ – user153012 Oct 5 '14 at 0:28

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