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This series is related to some extent to the previous question of mine, that is Computing $\sum_{n=1}^{\infty} \frac{\psi\left(\frac{n+1}{2}\right)}{ \binom{2n}{n}}$, where an approach by series only is expected. To answer one of the questions
in the previous post, no CAS is able to compute these series, and I suppose this won't be
possible in the near future either.

Prove that $$\sum_{n=1}^{\infty} \left(\frac{\displaystyle \psi(n)-\psi\left(n-\frac{1}{2}\right)}{\displaystyle \binom{4n-2}{2n-1}}+\frac{\displaystyle \psi\left(n+\frac{1}{2}\right)-\psi(n)}{\displaystyle \binom{4n}{2n}}\right) =$$

$$\frac{\pi}{3\sqrt{3}}+\frac{1}{75}\left(\log(1073741824)-3\sqrt{5} \pi^2+\pi\left(\Re \{{4 i \sqrt{5} \log(5- \sqrt{5} + i (\sqrt{15}- \sqrt{3}) ) - 2 i \sqrt{5} \log(5 +\sqrt{5}+ i (\sqrt{3} + \sqrt{15}))\}}+6\sqrt{5}\arctan\left(\sqrt{\frac{5}{3}}\right)-5\sqrt{3}\right)+\sqrt{5}\left(\log(79228162514264337593543950336) \log\left(\frac{3-\sqrt{5}}{2}\right) + \log(324518553658426726783156020576256)\log\left(\frac{3+\sqrt{5}}{2}\right)+24\Re \left\{\operatorname{Li}_2\left(\frac{\sqrt{3}+i}{\sqrt{3}+i\sqrt{5}}\right)-\operatorname{Li}_2\left(\frac{\sqrt{3}+i}{\sqrt{3}-i\sqrt{5}}\right)\right\}\right)\right)$$

A supplementary question - Find the closed form of

$$\sum_{n=1}^{\infty} (-1)^{n+1}\left(\frac{\displaystyle \psi(n)-\psi\left(n-\frac{1}{2}\right)}{\displaystyle \binom{4n-2}{2n-1}}+\frac{\displaystyle \psi\left(n+\frac{1}{2}\right)-\psi(n)}{\displaystyle \binom{4n}{2n}}\right)$$

Hopefully Cleo might help with the alternating version.

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