3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
4
$\begingroup$

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 $$ S = \text{Li}_4(\tfrac{1}{2})+\zeta (3) \log2-\zeta (4)-\tfrac{1}{2} \zeta (2) \log^22-\tfrac{1}{6} \log^32 \log \log2+\tfrac{11}{36} \log^32. $$

$\endgroup$
2
$\begingroup$

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 $\frac{1}{2}$.

$\endgroup$
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.