<s>[INST] Prove that $\sum_{n=0}^{\infty }\frac{(2n+1)!!}{(n+1)!}2^{-(2n+4)}=\frac{3-2\sqrt{2}}{4}$ How can I prove that 

$$\sum_{n=0}^{\infty }\frac{(2n+1)!!}{(n+1)!}2^{-(2n+4)}=\frac{3-2\sqrt{2}}{4}$$

 [/INST] $$\sum_{n=0}^{\infty }\frac{(2n+1)!!}{(n+1)!}2^{-(2n+4)}= \sum_{n=0}^{\infty }\frac{(2n+1)!}{(n+1)!n!}2^{-(3n+4)} = \dfrac{1}{16}\sum_{n=0}^{\infty }{2n+ 1\choose n}x^{n}$$
with $x = \dfrac{1}{8}$.
We are going to compute $f(x) = \sum_{n=0}^{\infty }{2n+ 1\choose n}x^{n}$ using the integral presentation of binomial coefficients:
$${2n+ 1\choose n} = \dfrac{1}{2\pi i}\int_{|z| =1}(z+1)^{2n+1}z^{-n-1}dz$$
so we have
\begin{align}
f(x) = &\dfrac{1}{2\pi i}\int_{|z| =1}\sum_{n=0}^\infty \left(\dfrac{(z+1)^2}{z}x\right)^n \dfrac{1+z}{z}dz \\
= & \dfrac{1}{2\pi i}\int_{|z| =1} \dfrac{1}{1-\dfrac{(z+1)^2}{z}x} \dfrac{1+z}{z}dz \\
= & \dfrac{1}{2\pi i}\int_{|z| =1} \dfrac{1+z}{z-(z+1)^2x} dz \\
\end{align}
The last integral is not too hard to evaluate with $x = \dfrac{1}{8}$
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