A Sum that came up while solving a integral While evaluating $I$, I did the following-
$$\begin{align}I= \int_{0}^{1} \log \left(\dfrac{1+x}{1-x}\right) \dfrac{1}{x\sqrt{1-x^2}} \ \mathrm{d}x &= 2 \int_{0}^{1}\sum_{n=0}^{\infty} \dfrac{x^{2n+1}}{2n+1} \dfrac{1}{x\sqrt{1-x^2}} \ \mathrm{d}x\\ &=2\sum_{n=0}^{\infty} \int_{0}^{1} \dfrac{x^{2n}}{(2n+1)\sqrt{1-x^2}}  \ \mathrm{d}x \end{align}$$ Then I used the substitution $x \mapsto \sin \theta $.
$$\begin{align} \therefore I &=2\sum_{n=0}^{\infty} \int_{0}^{\pi/2} \dfrac{\sin^{2n} {\theta}}{2n+1} \ \mathrm{d}\theta\\ &=\pi \sum_{n=0}^{\infty} \dfrac{(2n)!}{2^{2n}(n!)^2(2n+1)} \end{align}$$ The last step is due to Wallis' formula. 
However, I couldn't solve the last series. My question is that how do we prove that $$\displaystyle\sum_{n=0}^{\infty} \dfrac{(2n)!}{2^{2n}(n!)^2(2n+1)}=\dfrac{\pi}{2} \ ?$$
 A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\sf\mbox{If you evaluates the sum you'll return to the original integral.}$
$\sf\mbox{In any case the evaluation requires to consider the original integral.}$

So,
\begin{align}&\color{#66f}{\large%
{1 \over \pi}\int_{0}^{1}\log\pars{1 + x \over 1 - x}\,{1 \over x\root{1 - x^{2}}}\,\dd x}
={2 \over \pi}\
\overbrace{\int_{0}^{1}\,{\rm arctanh}\pars{x}\,{1 \over x\root{1 - x^{2}}}\,\dd x}
^{\dsc{x}\ \equiv\ \dsc{\tanh\pars{t}}}
\\[5mm]&={2 \over \pi}\int_{0}^{\infty}t\,{1 \over \tanh\pars{t}\root{1 - \tanh^{2}\pars{t}}}\,\sech^{2}\pars{t}\,\dd t
={2 \over \pi}\int_{0}^{\infty}{t \over \sinh\pars{t}}\,\dd t
\\[5mm]&={4 \over \pi}\int_{0}^{\infty}{t\expo{-t} \over 1 - \expo{-2t}}\,\dd t
={4 \over \pi}\sum_{n\ =\ 0}^{\infty}\int_{0}^{\infty}t\expo{-\pars{2n + 1}t}\,\dd t
={4 \over \pi}\sum_{n\ =\ 0}^{\infty}{1 \over \pars{2n + 1}^{2}}\
\overbrace{\int_{0}^{\infty}t\expo{-t}\,\dd t}^{\ds{=}\ \dsc{1}}
\\[5mm]&={4 \over \pi}\bracks{\sum_{n\ =\ 1}^{\infty}{1 \over n^{2}}
-\sum_{n\ =\ 1}^{\infty}{1 \over \pars{2n}^{2}}}
={4 \over \pi}\pars{{3 \over 4}\
\overbrace{\sum_{n\ =\ 1}^{\infty}{1 \over n^{2}}}^{\ds{=}\ \dsc{\pi^{2} \over 6}}}
=\color{#66f}{\Large{\pi \over 2}}
\end{align}
A: Instead of series expansion, you can do this:
$$\int_0^1 \ln\left(\frac{1+x}{1-x}\right)\frac{1}{x\sqrt{1-x^2}}\,dx=\frac{1}{2}\int_{-1}^1 \ln\left(\frac{1+x}{1-x}\right)\frac{1}{x\sqrt{1-x^2}}\,dx$$
$$=\int_{-1}^1 \frac{\ln(1+x)}{x\sqrt{1-x^2}}\,dx=\int_{-\pi/2}^{\pi/2} \frac{\ln(1+\sin\theta)}{\sin\theta}\,d\theta$$
Consider
$$I(a)=\int_{-\pi/2}^{\pi/2} \frac{\ln(1+a\sin\theta)}{\sin\theta}\,d\theta \Rightarrow I'(a)=\int_{-\pi/2}^{\pi/2} \frac{1}{1+a\sin\theta}\,d\theta=\frac{\pi}{\sqrt{1-a^2}}$$
$$\Rightarrow I(a)=\sin^{-1}a+C$$
Since $I(0)=0$, we have $C=0$ and hence,
$$\boxed{I(1)=\dfrac{\pi}{2}}$$
A: Hint
For $|x| \leq 1$,
$$\displaystyle\sum_{n=0}^{\infty} \dfrac{(2n)!}{2^{2n}(n!)^2(2n+1)}x^{2n}=\frac 1x\sum_{n=0}^{\infty} \dfrac{(2n)!}{2^{2n}(n!)^2(2n+1)}x^{2n+1}=\frac{\sin ^{-1}(x)}{x}$$
