<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] 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.
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