Evaluate $\sum_{n=0}^{\infty} \frac{(2n-1)!!}{2^n n! (2n+1)^2}$. 
Evaluate $\sum_{n=0}^{\infty} \frac{(2n-1)!!}{2^n n! (2n+1)^2}$.

I am attempting to evaluate this sum as an alternate form of an integral I was trying to calculate. Can anyone give any insight? WolframAlpha says it's $\frac{1}{2}\pi\log(2)$ but I'm not sure how to get there.
 A: Possible hint
Start with$$S= \sum_{n=0}^{\infty} \frac{(2n-1)!!}{n! \,(2n+1)^2} x^{2n+1}$$
$$S'=\sum_{n=0}^{\infty} \frac{(2n-1)!!}{n! \,(2n+1)} x^{2n}=\frac 1x\sum_{n=0}^{\infty} \frac{(2n-1)!!}{n! \,(2n+1)} x^{2n+1}=\frac 1x \frac{\sin ^{-1}\left(x\sqrt{2} \right)}{\sqrt{2}}$$
A: Consider using $ (2 n -1)!! = 2^n \, \left(\frac{1}{2}\right)_{n}$, where $(a)_{n}$ is the Pochhammer symbol, to obtain
$$ f(x) = \sum_{n=0}^{\infty} \frac{(2 n -1)!! \, x^{2 n +1}}{2^n \, n! \, (2n + 1)^2} = \sum_{n=0}^{\infty} \frac{\left(\frac{1}{2}\right)_{n} \, x^{2 n +1}}{ n! \, (2n + 1)^2}. $$
With some manipulations the series becomes:
\begin{align}
x \, \frac{d f}{d x} &= \sum_{n=0}^{\infty} \frac{\left(\frac{1}{2}\right)_{n} \, x^{2 n + 1}}{n! \, (2 n + 1)} \\
\frac{d}{dx} \, \left(x \, \frac{d f}{dx} \right) &= \sum_{n=0}^{\infty} \frac{\left(\frac{1}{2}\right)_{n} \, x^{2 n}}{n!} = \frac{1}{\sqrt{1-x^2}} \\
x \, \frac{d f}{dx} &= \int^{x} \frac{d u}{\sqrt{1- u^2}} = \sin^{-1}(x) + c_{0} \\
f(x) &= \int^{x} \frac{\sin^{-1}(u)}{u} \, du + c_{0} \, \ln x + c_{1} \\
&= \sin^{-1}(x) \, \ln\left( 1 - e^{2 \, i \, \sin^{-1}(x)} \right) - \frac{i}{8} \, \left( \left(\sin^{-1}(x)\right)^2 + \text{Li}_{2}\left(e^{2 \, i \, \sin^{-1}(x)} \right) \right) + c_{0} \, \ln(x) + c_{1}.
\end{align}
Since $f(0) = 0$ then $c_{0} = 0$ and $c_{1} = \frac{i \, \zeta(2)}{2}$ which gives
$$ f(x) = \sin^{-1}(x) \, \ln\left( 1 - e^{2 \, i \, \sin^{-1}(x)} \right) - \frac{i}{8} \, \left( \left(\sin^{-1}(x)\right)^2 + \text{Li}_{2}\left(e^{2 \, i \, \sin^{-1}(x)} \right) \right) + \frac{i}{2} \, \zeta(2). $$
When $x = 1$ this becomes
$$ f(1) = \frac{\pi}{2} \, \ln 2. $$
Hence,
$$ \sum_{n=0}^{\infty} \frac{\left(\frac{1}{2}\right)_{n}}{ n! \, (2n + 1)^2} = \frac{\pi}{2} \, \ln 2. $$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
& \color{#44f}{\sum_{n = 0}^{\infty}{\pars{2n - 1}!! \over 2^{n}\, n!\, \pars{2n + 1}^{2}}} =
\sum_{n = 0}^{\infty}{1 \over 2^{n}\, n!\, \pars{2n + 1}^{2}}\
\overbrace{\pars{-1}^{n}\,2^{n}\,n!{-1/2 \choose n}}
^{\ds{\pars{2n - 1}!!}}
\\[5mm] = & \
\sum_{n = 0}^{\infty}{-1/2 \choose n}{\pars{-1}^{n} \over \pars{2n + 1}^{2}} =
\left. -\,{1 \over 4}\partiald{}{a}\sum_{n = 0}^{\infty}{-1/2 \choose n}{\pars{-1}^{n} \over n + a}\right\vert_{a\ =\ 1/2}
\\[5mm] = & \
\left. -\,{1 \over 4}\partiald{}{a}\sum_{n = 0}^{\infty}{-1/2 \choose n}\pars{-1}^{n}
\int_{0}^{1}t^{n\ +\ a\ -\ 1}\,\,\,\,\dd t
\,\right\vert_{\, a\ =\ 1/2}
\\[5mm] = & \
\left. -\,{1 \over 4}\partiald{}{a}\int_{0}^{1}t^{a - 1}
\sum_{n = 0}^{\infty}{-1/2 \choose n}\pars{-t}^{n}\,\dd t
\,\right\vert_{\, a\ =\ 1/2}
\\[5mm] = & \
\left. -\,{1 \over 4}\partiald{}{a}\int_{0}^{1}t^{a - 1}\
\pars{1 - t}^{-1/2}\,\,\dd t
\,\right\vert_{\, a\ =\ 1/2}
\\[5mm] = & \
\left. -\,{1 \over 4}\partiald{}{a}
{\Gamma\pars{a}\Gamma\pars{1/2} \over \Gamma\pars{a + 1/2}}
\,\right\vert_{\, a\ =\ 1/2} =
\bbx{\color{#44f}{{1 \over 2}\,\pi\,\ln\pars{2}}} \approx 1.08879 \\ &
\end{align}
