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 

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

 A: Hint:
Using the Taylor series 
$$(1+x)^\alpha=\sum_{n=0}^\infty\binom{\alpha}{n}x^n$$ for $\alpha=-\frac12$ we get 
$$(1+x)^{-\frac12}=\sum_{n=0}^\infty \frac{(2n-1)!!}{n!}2^{-n}(-1)^nx^n$$
Play with it a little.
A: The sum is equal to
$$\frac1{16} \sum_{n=0}^{\infty} \frac{(2 n+1)!}{2^{3 n} n !(n+1)!}  = \frac1{16} \sum_{n=0}^{\infty} \frac{2 n+1}{n+1} \frac1{2^{3 n}} \binom{2 n}{n} $$
The sum may be written as
$$\frac1{8} \sum_{n=0}^{\infty} \frac1{2^{3 n}} \binom{2 n}{n} - \frac1{16}\sum_{n=0}^{\infty} \frac1{n+1} \frac1{2^{3 n}} \binom{2 n}{n}$$
Note that
$$\sum_{n=0}^{\infty} \frac1{2^{3 n}} \binom{2 n}{n} x^{n} = \left (1-\frac{x}{2}\right )^{-1/2} $$
so that
$$\sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1} \frac1{2^{3 n}} \binom{2 n}{n} = \int dx \left (1-\frac{x}{2}\right )^{-1/2} = -4 \left (1-\frac{x}{2}\right )^{1/2}+C$$
This sum is zero when $x=0$, so $C=4$, and therefore the sum in question is
$$\frac{\sqrt{2}}{8} - \frac{2-\sqrt{2}}{8} = \frac{\sqrt{2}-1}{4}$$
which is not the answer provided, but which checks against Mathematica.
A: $$\sum_{n=0}^{\infty }\frac{(2n+1)!!}{(n+1)!}2^{-(2n+4)}= \sum_{n=0}^{\infty }\frac{(2n+1)!}{(n+1)!n!}2^{-(3n+4)} = \dfrac{1}{16}\sum_{n=0}^{\infty }{2n+ 1\choose n}x^{n}$$
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 $x = \dfrac{1}{8}$
