TITLE: Evaluating the series $\sum\limits_{n=1}^{\infty} \frac{1}{n^{3} \binom{2n}{n}} $
QUESTION [41 upvotes]: Wolfram MathWorld  states that $$ \sum_{n=1}^{\infty} \frac{1}{n^{3} \binom{2n}{n}} = \frac{ \pi \sqrt{3}}{18} \Bigg[ \psi_{1} \left(\frac{1}{3} \right) - \psi_{1} \left(\frac{2}{3} \right) \Bigg]- \frac{4}{3} \zeta(3) \, , $$
where $\psi_{1}(x)$ is the trigamma function.
I'm having difficulty getting the result in that form.
Using the Taylor expansion $$ \arcsin^{2}(x) = \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n^{2} \binom{2n}{n}} (2x)^{2n},$$ I get
$$ \sum_{n=1}^{\infty} \frac{1}{n^{3} \binom{2n}{n}} = 4 \int_{0}^{\frac{1}{2}} \frac{\arcsin^{2}(x)}{x} \, \mathrm dx. $$
Then I integrating by parts, I get
$$ \begin{align} &4 \int_{0}^{\frac{1}{2}} \frac{\arcsin^{2}(x)}{x} \, \mathrm dx \\ &=  \frac{\pi^{2}}{9} \ln \left(\frac{1}{2} \right) - 8 \int_{0}^{\frac{1}{2}} \frac{\arcsin (x) \ln (x)}{\sqrt{1-x^{2}}} \, \mathrm dx \\ &= - \frac{\pi^{2}}{9} \ln 2 - 8 \int_{0}^{\frac{\pi}{6}} u \ln (\sin u ) \, \mathrm du \\ &= - 8 \ln 2 \int_{0}^{\frac{\pi}{6}} u \, \mathrm  du - 8\int_{0}^{\frac{\pi}{6}} u \ln (\sin u ) \, \mathrm du \\ &= - 8 \int_{0}^{\frac{\pi}{6}} u \ln ( 2 \sin u ) \, \mathrm du  \\ &= -8 \ \text{Re} \int_{0}^{\frac{\pi}{6}} u \ln (1-e^{2iu}) \, \mathrm du \\ &= 8 \ \text{Re} \int_{0}^{\frac{\pi}{6}} u \sum_{n=1}^{\infty} \frac{e^{2in u}}{n} \, \mathrm du  \\ &= 8 \ \sum_{n=1}^{\infty}  \frac{1}{n} \int_{0}^{\frac{\pi}{6}} u \cos (2nu) \, du \\ &= \frac{2 \pi}{3} \sum_{n=1}^{\infty} \frac{\sin (\frac{n \pi}{3})}{n^{2}} + 2 \sum_{n=1}^{\infty} \frac{\cos (\frac{n \pi}{3})}{n^{3}} - 2 \zeta(3) \\ &= \frac{2 \pi}{3} \Bigg( \frac{\sqrt{3}}{2} \sum_{n=0}^{\infty} \frac{1}{(6n+1)^{2}} + \frac{\sqrt{3}}{2} \sum_{n=0}^{\infty} \frac{1}{(6n+2)^{2}} - \frac{\sqrt{3}}{2} \sum_{n=0}^{\infty} \frac{1}{(6n+4)^{2}} - \frac{\sqrt{3}}{2} \sum_{n=0}^{\infty} \frac{1}{(6n+5)^{2}} \Bigg) \\ &+ \  2  \Bigg( \frac{1}{2} \sum_{n=0}^{\infty} \frac{1}{(6n+1)^{3}} - \frac{1}{2} \sum_{n=0}^{\infty} \frac{1}{(6n+2)^{3}} - \sum_{n=0}^{\infty} \frac{1}{(6n+3)^{3}} - \frac{1}{2} \sum_{n=0}^{\infty} \frac{1}{(6n+4)^{3}} \\ &+ \frac{1}{2} \sum_{n=0}^{\infty} \frac{1}{(6n+5)^{3}} + \sum_{n=1}^{\infty} \frac{1}{(6n)^{3}} \Bigg) - 2 \zeta(3) \\ &= \frac{\pi \sqrt{3}}{108} \Bigg( \psi_{1} \left(\frac{1}{6} \right) + \psi_{1} \left(\frac{1}{3} \right) - \psi_{1} \left(\frac{2}{3} \right) - \psi_{1}\left(\frac{5}{6} \right) \Bigg) + \frac{1}{432} \Bigg( - \psi_{2} \left(\frac{1}{6} \right) + \psi_{2} \left(\frac{1}{3} \right) \\ &-28 \zeta(3)  + \psi_{2} \left(\frac{2}{3} \right) - \psi_{2} \left(\frac{5}{6} \right) + 4 \zeta(3)\Bigg) - 2 \zeta (3) .\end{align}$$
But from here I've been going in circles trying to get the result in the form given on the Wolfram MathWorld site.

EDIT:
Using the duplication formula for the trigamma function (i.e.,  $ \displaystyle 4 \psi_{1}(2x) = \psi_{1}(x) + \psi_{1} \left(x + \frac{1}{2} \right) $), I get
$$ \begin{align} &\psi_{1} \left(\frac{1}{6} \right) + \psi_{1} \left(\frac{1}{3}  \right) - \psi_{1} \left(\frac{2}{3} \right) - \psi_{1}\left(\frac{5}{6}  \right) \\ &= 4 \psi_{1} \left(\frac{1}{3} \right) -\psi_{1} \left(\frac{2}{3} \right) +  \psi_{1} \left(\frac{1}{3} \right) -\psi_{1} \left(\frac{2}{3} \right) - 4  \psi_{1} \left(\frac{2}{3} \right) + \psi_{1} \left(\frac{1}{3} \right) \\ &= 6 \psi_{1} \left(\frac{1}{3} \right) - 6 \psi_{1} \left(\frac{2}{3} \right). \end{align}$$
Therefore,
$$ \begin{align} \sum_{n=1}^{\infty} \frac{1}{n^{3} \binom{2n}{n}} &= \frac{\sqrt{3} \pi}{18} \Bigg( \psi_{1} \left(\frac{1}{3} \right) - \psi_{1} \left(\frac{2}{3} \right) \Bigg) + 
\frac{1}{432} \Bigg( - \psi_{2} \left(\frac{1}{6} \right) + \psi_{2} \left(\frac{1}{3} \right) \\ &+ \psi_{2} \left(\frac{2}{3} \right) - \psi_{2} \left(\frac{5}{6} \right) - 2 \zeta(3)\Bigg) - 2 \zeta (3). \end{align}$$
So it comes down to somehow showing that  $$-\psi_{2} \left(\frac{1}{6} \right) + \psi_{2} \left(\frac{1}{3} \right) + \psi_{2} \left(\frac{2}{3} \right) - \psi_{2} \left(\frac{5}{6} \right) = 312 \zeta(3) .$$
SECOND EDIT:
Using the duplication formula for $\psi_{2}(x)$, I get
$$ \begin{align} &-\psi_{2} \left(\frac{1}{6} \right) + \psi_{2} \left(\frac{1}{3} \right) + \psi_{2} \left(\frac{2}{3} \right) - \psi_{2} \left(\frac{5}{6} \right) \\ &= -8 \psi_{2} \left(\frac{1}{3} \right) +\psi_{2} \left(\frac{2}{3} \right) +  \psi_{2} \left(\frac{1}{3} \right) +\psi_{2} \left(\frac{2}{3} \right) - 8 \psi_{2} \left(\frac{2}{3} \right) + \psi_{2} \left(\frac{1}{3} \right) \\ &= -6 \psi_{2} \left(\frac{1}{3} \right) - 6 \psi_{2} \left(\frac{2}{3} \right) . \end{align}$$
And then using the triplication forumula , I get
$$ \psi_{2} \left(\frac{1}{3} \right) + \psi_{2} \left( \frac{2}{3} \right) + \psi_{2}(1) = 27 \psi_{2} (1) .$$
Therefore,
$$  -6 \psi_{2} \left(\frac{1}{3} \right) - 6 \psi_{2} \left(\frac{2}{3} \right) = -156 \psi_{2} (1) = 312 \zeta(3) .$$

REPLY [10 votes]: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{\sum_{n=1}^{\infty} \frac{1}{n^{3} \binom{2n}{n}} = \frac{ \pi \sqrt{3}}{18} \Big[ \psi_{1} \left(\frac{1}{3} \right) - \psi_{1} \left(\frac{2}{3} \right) \Big]- \frac{4}{3} \zeta(3)}$

\begin{align}
&\sum_{n = 1}^{\infty}{x^{n} \over n{2n \choose n}}
=\sum_{n = 1}^{\infty}x^{n}\,
{\Gamma\pars{n}\Gamma\pars{n + 1} \over \Gamma\pars{2n + 1}}
=\sum_{n = 1}^{\infty}x^{n}\int_{0}^{1}t^{n - 1}\pars{1 - t}^{n}\,\dd t
\\[3mm] & =
\int_{0}^{1}\sum_{n = 1}^{\infty}\bracks{tx\pars{1 - t}}^{n}
\,{\dd t \over t}
=\int_{0}^{1}\bracks{{1 \over 1 - tx\pars{1 - t}} - 1}\,{\dd t \over t}
\\[3mm] & =
\int_{0}^{1}{x\pars{1 - t} \over 1 - tx\pars{1 - t}}\,\dd t
\end{align}

\begin{align}
&\sum_{n = 1}^{\infty}{x^{n} \over n^{2}{2n \choose n}}
=\int_{0}^{x}\dd y\int_{0}^{1}{1 - t \over 1 - ty\pars{1 - t}}\,\dd t
=-\int_{0}^{1}{\ln\pars{1 - \bracks{1 - t}tx} \over t}\,\dd t
\end{align}

\begin{align}
&\color{#66f}{\large\sum_{n = 1}^{\infty}{1 \over n^{3}{2n \choose n}}}
=-\int_{0}^{1}{\dd x \over x}
\int_{0}^{1}{\ln\pars{1 - \bracks{1 - t}tx} \over t}\,\dd t
\\[3mm] = &\
\int_{0}^{1}
{{\rm Li}_{2}\pars{t\bracks{1 - t}} \over t}\,\dd t
\\[3mm] = &\
\color{#66f}{{\pi \over 36\root{3}}\,  \left[\psi ^{(1)}\left(\frac{1}{3}\right)-\psi ^{(1)}\left(\frac{2}{3}\right)+\psi ^{(1)}\left(\frac{1}{6}\right)-\psi ^{(1)}\left(\frac{5}{6}\right)\right]-\frac{4 \zeta (3)}{3}}
\\[3mm] \approx &\ {\tt 0.5229}
\end{align}