Closed form for: $\sum_{n=0}^\infty \frac{1}{{2}^{2n}}\frac{\binom{2n}{n}}{{(2n+1)}^2}$?

Question

Does a closed form of the following sum exist?$$\sum_{n=0}^\infty \frac{1}{{2}^{2n}}\frac{\binom{2n}{n}}{{(2n+1)}^2}$$

Answer 1

Yes. Starting with $$ \frac1{\sqrt{1-4x^2}} = \sum_{n=0}^\infty \binom{2n}n x^{2n}, $$ you can perform the following steps to obtain your sum:

• multiply by $x$
• integrate
• divide by $x$
• integrate again
• divide by $x^2$
• set $x=\frac12$

Mathematica says that the ultimate answer is $\frac{\pi\log4}4$.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Note that $\ds{{2n \choose n} = {-1/2 \choose n}\pars{-4}^{n}}$.

\begin{align} \sum_{n = 0}^{\infty}{1 \over 2^{2n}}{{2n \choose n} \over \pars{2n + 1}^{2}} & = \sum_{n = 0}^{\infty}{-1/2 \choose n}\pars{-1}^{n}\ \overbrace{\bracks{-\int_{0}^{1}\ln\pars{x}x^{2n}\,\dd x}} ^{\ds{1 \over \pars{2n + 1}^{2}}} \\[5mm] & = -\int_{0}^{1}\ln\pars{x}\sum_{n = 0}^{\infty}{-1/2 \choose n}\pars{-x^{2}}^{n} \,\dd x = -\int_{0}^{1}\ln\pars{x}\pars{1 - x^{2}}^{-1/2}\,\dd x \\[5mm] & = -\,{1 \over 4}\int_{0}^{1}x^{-1/2}\ln\pars{x}\pars{1 - x}^{-1/2}\,\dd x = \left. -\,{1 \over 4}\partiald{}{\nu}\int_{0}^{1}x^{\nu - 1/2} \pars{1 - x}^{-1/2}\,\dd x\,\right\vert_{\ \nu\ =\ 0} \\[5mm] & = \left. -\,{1 \over 4}\partiald{}{\nu} {\Gamma\pars{\nu + 1/2}\Gamma\pars{1/2} \over \Gamma\pars{\nu + 1}}\right\vert_{\ \nu\ =\ 0} \\[5mm] & = \left. -\,{\root{\pi} \over 4}\partiald{}{\nu} {\Gamma\pars{1/2} + \Gamma\pars{1/2}\Psi\pars{1/2}\nu \over 1 - \gamma\nu}\,\right\vert_{\ \nu\ =\ 0} \\[5mm] & = \left. -\,{\pi \over 4}\partiald{}{\nu} \bracks{1 + \Psi\pars{1 \over 2}\nu}\pars{1 + \gamma\nu} \right\vert_{\ \nu\ =\ 0} \\[5mm] & = -\,{\pi \over 4}\ \overbrace{\bracks{\Psi\pars{1 \over 2} + \gamma}} ^{\ds{-2\ln\pars{2}}}\ = \bbx{{1 \over 2}\,\ln\pars{2}\pi} \approx 1.0888 \end{align}