How to prove that $\sum\limits_{k=0}^{\infty}\frac{(4k-1)!!}{(2k+1)!\cdot2^{4k+1}}=\frac{\sqrt{3}-1}{\sqrt{2}}$ How to prove that $$\sum\limits_{k=0}^{\infty}\frac{(4k-1)!!}{(2k+1)!\cdot2^{4k+1}}=\frac{\sqrt{3}-1}{\sqrt{2}}$$
I need any hint to start to prove it.
Thanks for any help.
 A: Solution 1. Using the generating function for the central binomial coefficients
$$\frac{1}{\sqrt{1-4x}} = \sum_{n=0}^{\infty}\binom{2n}{n}x^n,$$
we have
$$ \sum_{n=0}^{\infty} \frac{(4n-1)!!}{(2n+1)!2^{4n+1}}
= 4 \int_{0}^{\frac{1}{8}}\sum_{n=0}^{\infty} \binom{4n}{2n} x^{2n} \, \mathrm{d}x
= 2\int_{0}^{\frac{1}{8}} \left( \frac{1}{\sqrt{1-4x}} + \frac{1}{\sqrt{1+4x}} \right) \, \mathrm{d}x, $$
which computes to the desired answer.

Solution 2. Write
\begin{align*}
\sum_{n=0}^{\infty} \frac{(4n-1)!!}{(2n+1)!2^{4n+1}}
&= \sum_{n=0}^{\infty} \frac{\Gamma(2n+\frac{1}{2})}{(2n+1)!2^{2n+1}\sqrt{\pi}} \\
&= \sum_{n=0}^{\infty} \frac{1}{(2n+1)!2^{2n+1}\sqrt{\pi}} \int_{0}^{\infty} x^{2n-\frac{1}{2}} e^{-x} \, \mathrm{d}x \\
&= \frac{1}{\sqrt{\pi}} \int_{0}^{\infty} \frac{\sinh(x/2)}{x^{3/2}} e^{-x} \, \mathrm{d}x \\
&= \frac{1}{\sqrt{2\pi}} \int_{0}^{\infty} \frac{e^{-s^2} - e^{-3s^2}}{s^2} \, \mathrm{d}s \tag{$x=2s^2$}
\end{align*}
Taking integration by parts,
\begin{align*}
\require{cancel}
&\frac{1}{\sqrt{2\pi}} \int_{0}^{\infty} \frac{e^{-s^2} - e^{-3s^2}}{s^2} \, \mathrm{d}s \\
&\qquad = \frac{1}{\sqrt{2\pi}} \cancelto{0}{\left[ - \frac{e^{-s^2} - e^{-3s^2}}{s} \right]_{0}^{\infty}} + \sqrt{\frac{2}{\pi}} \int_{0}^{\infty} \left( 3e^{-3s^2} - e^{-s^2} \right) \, \mathrm{d}s,
\end{align*}
which computes to the desired answer.
A: $$\sum_{k=0}^{\infty}\frac{(4k-1)!!}{(2k+1)!\cdot2^{4k+1}}=\frac 12\sum_{k=0}^{\infty}\frac{(4k-1)!!}{(2k+1)!\cdot16^k}$$
Now, consider
$$\sum_{k=0}^{\infty}\frac{(4k-1)!!}{(2k+1)!}x^k$$ and show that this is the expansion of
$$\frac{\sqrt{2}}{\sqrt{1+\sqrt{1-4 x}}}$$ When done, make $x=\frac 1 {16}$ and divide by $2$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\sum_{k = 0}^{\infty}{\pars{4k - 1}!! \over
\pars{2k + 1}!\ 2^{4k + 1}} = {\root{3}-1 \over \root{2}}:\ {\LARGE ?}}$.

\begin{align}
&\bbox[10px,#ffd]{\sum_{k = 0}^{\infty}{\pars{4k - 1}!! \over
\pars{2k + 1}!\ 2^{4k + 1}}} =
\sum_{k = 0}^{\infty}{\prod_{\ell = 0}^{2k - 1}\pars{4k - 2\ell - 1} \over
\pars{2k + 1}!\ 2^{4k + 1}}
\\[5mm] = &\
\sum_{k = 0}^{\infty}{2^{2k}
\prod_{\ell = 0}^{2k - 1}\pars{\ell + 1/2 - 2k} \over
\pars{2k + 1}!\ 2^{4k + 1}} =
\sum_{k = 0}^{\infty}{\pars{1/2 - 2k}^{\,\overline{2k}} \over
\pars{2k + 1}!\ 2^{2k + 1}}
\\[5mm] = &\
\sum_{k = 0}^{\infty}{\Gamma\pars{1/2 - 2k + 2k}/\Gamma\pars{1/2 - 2k} \over
\pars{2k + 1}!\ 2^{2k + 1}}
\\[5mm] = &\
2\sum_{k = 0}^{\infty}{\pars{1/2}! \over
\pars{-1/2 - 2k}!\pars{2k + 1}!}\,\pars{1 \over 2}^{2k + 1} =
2\sum_{k = 0}^{\infty}{1/2 \choose 2k + 1}\pars{1 \over 2}^{2k + 1}
\\[5mm] = &\
2\sum_{k = 1}^{\infty}{1/2 \choose k}\pars{1 \over 2}^{k}\,
{1 - \pars{-1}^{k} \over 2}
\\[5mm] = &\
\sum_{k = 0}^{\infty}{1/2 \choose k}\pars{1 \over 2}^{k} -
\sum_{k = 0}^{\infty}{1/2 \choose k}\pars{-\,{1 \over 2}}^{k}
\\[5mm] = &\
\pars{1 + {1 \over 2}}^{1/2} - \pars{1 - {1 \over 2}}^{1/2} =
\bbx{\root{3} - 1 \over \root{2}}
\end{align}
