<s>[INST] 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.
 [/INST] 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.
</s>