# What is the binomial sum $\sum_{n=1}^\infty \frac{1}{n^5\,\binom {2n}n}$ in terms of zeta functions?

We have the following evaluations:

$$\begin{aligned} &\sum_{n=1}^\infty \frac{1}{n\,\binom {2n}n} = \frac{\pi}{3\sqrt{3}}\\ &\sum_{n=1}^\infty \frac{1}{n^2\,\binom {2n}n} = \frac{1}{3}\,\zeta(2)\\ &\sum_{n=1}^\infty \frac{1}{n^3\,\binom {2n}n} = -\frac{4}{3}\,\zeta(3)+\frac{\pi\sqrt{3}}{2\cdot 3^2}\,\left(\zeta(2, \tfrac{1}{3})-\zeta(2,\tfrac{2}{3}) \right) \\&\sum_{n=1}^\infty \frac{1}{n^4\,\binom {2n}n} = \frac{17}{36}\,\zeta(4)\\ &\sum_{n=1}^\infty \frac{1}{n^5\,\binom {2n}n} = \,?\\ \end{aligned}$$

The paper gives the 3rd and 5th in terms of the Dirichlet L-functions, but does anyone know how to evaluate the 5th one in terms of the *Hurwitz zeta function* $\zeta(s,a)$?

* Postscript*: (A few hours later)

After Anon gave his answer, I did a little more sleuthing and found the case *p* = 7 in the Mathworld article on (which also had p = 5). The paper I cited was a bit old (1999) and the authors weren’t aware it was already found a year earlier by Plouffe. Hence,

$$\begin{aligned} &\sum_{n=1}^\infty \frac{1}{n^5 \, \binom{2n}n} = -\frac{19}{3}\zeta(5)+\frac{2}{3}\zeta(2)\zeta(3)+\frac{\pi\sqrt{3}}{2^3\cdot3^2}\left(\zeta(4,\tfrac{1}{3})-\zeta(4, \tfrac{2}{3}) \right)\\ &\sum_{n=1}^\infty \frac{1}{n^7 \, \binom{2n}n} = -\frac{493}{24}\zeta(7)+2\zeta(2)\zeta(5)+\frac{17}{18}\zeta(3)\zeta(4)+\frac{11\pi\sqrt{3}}{2^5\cdot3^4}\left(\zeta(6,\tfrac{1}{3})-\zeta(6, \tfrac{2}{3}) \right)\\ \end{aligned}$$

With this “pattern”, I used an integer relations algorithm to try to find p = 9, 11, 13. No luck so far.

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