How to prove $\sum_{n=0}^{\infty} \frac {(2n+1)!} {2^{3n} \; (n!)^2} = 2\sqrt{2} \;$? I found out that the sum
   $$\sum_{n=0}^{\infty} \frac {(2n+1)!} {2^{3n} \; (n!)^2}$$
converges to $2\sqrt{2}$. 
But right now I don't have enough time to figure out how to solve this.
I would really appreciate any help. Just one tiny hint might help too. 
 A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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With $\ds{\verts{\mu} < {1 \over 4}}$:

\begin{align}
\color{#66f}{\large\sum_{n\ =\ 0}^{\infty}\mu^{n}{\pars{2n + 1}! \over \pars{n!}^{2}}}
&=\sum_{n\ =\ 0}^{\infty}\pars{2n + 1}\mu^{n}{2n \choose n}
=\pars{2\mu\,\totald{}{\mu} + 1}
\color{#c00000}{\sum_{n\ =\ 0}^{\infty}\mu^{n}{2n \choose n}}
\end{align}

\begin{align}
\color{#c00000}{\sum_{n\ =\ 0}^{\infty}\mu^{n}{2n \choose n}}
&=\sum_{n\ =\ 0}^{\infty}\mu^{n}
\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{2n} \over z^{n + 1}}
\,{\dd z \over 2\pi\ic}
=\oint_{\verts{z}\ =\ 1}{1 \over z}
\sum_{n\ =\ 0}^{\infty}\bracks{\pars{1 + z}^{2}\mu \over z}^{n}
\,{\dd z \over 2\pi\ic}
\\[5mm]&=\oint_{\verts{z}\ =\ 1}
{1 \over z}{1 \over 1 - \pars{1 + z}^{2}\mu/z}\,{\dd z \over 2\pi\ic}
=-\oint_{\verts{z}\ =\ 1}{1 \over \mu z^{2} + \pars{2\mu - 1}z + \mu}
\,{\dd z \over 2\pi\ic}
\\[3mm]&=-\oint_{\verts{z}\ =\ 1}{1 \over \mu\pars{z - r_{-}}\pars{z - r_{+}}}
\,{\dd z \over 2\pi\ic}
\\[5mm]&\mbox{where}\quad \boxed{\ds{\quad r_{\pm}
\equiv {1 - 2\mu \pm \root{1 - 4\mu} \over 2\mu}\quad}} 
\end{align}

Note that $\ds{\verts{r_{-}}\ <\ 1}$ and $\ds{\verts{r_{+}}\ >\ 1}$ such that:

\begin{align}
\color{#c00000}{\sum_{n\ =\ 0}^{\infty}\mu^{n}{2n \choose n}}
&=-\,{1 \over \mu}\,{1 \over r_{-} - r_{+}}
=-\,{1 \over \mu}{1 \over -2\root{1 - 4\mu}/\pars{2\mu}}
={1 \over \root{1 - 4\mu}}
\end{align}

Then,
  \begin{align}
\color{#66f}{\large\sum_{n\ =\ 0}^{\infty}\mu^{n}{\pars{2n + 1}! \over \pars{n!}^{2}}}
&=\pars{2\mu\,\totald{}{\mu} + 1}{1 \over \root{1 - 4\mu}}
=\color{#66f}{\large{1 \over \root{1 - 4\mu}} + {4\mu \over \pars{1 - 4\mu}^{3/2}}}
\end{align}

Set $\ds{\mu = {1 \over 8}}$ in both members:
\begin{align}
\color{#66f}{\large%
\sum_{n\ =\ 0}^{\infty}{\pars{2n + 1}! \over 2^{3n}\pars{n!}^{2}}}
&=\color{#66f}{\large2\root{2}} \approx {\tt 2.8284}
\end{align}
A: Hint: Observe that (by the binomial theorem for instance)
$$
\sum_{n=0}^{\infty} \frac {(2n)!} {(n!)^2}x^{2n}=\frac{1}{\sqrt{1-4x^2}}, \quad |x|<\frac{1}{2}.
$$ Then multiply by $x$ and perform a termwise differentiation and you readily obtain the desired result with $x^2=\dfrac{1}{2^3}.$
