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.

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,
$$ \sum_{n=0}^{\infty} \frac{\left(\frac{1}{2}\right)_{n}}{ n! \, (2n + 1)^2} = \frac{\pi}{2} \, \ln 2. $$
