Calculate $\sum_{n=1}^{\infty}\frac{(2n-1)!!}{(2n)!!\cdot 2^n}$ 
Calculate the sum $$\displaystyle \sum_{n=1}^{\infty}\frac{(2n-1)!!}{(2n)!!\cdot 2^n}$$
  where $(2n-1)!!=1\cdot 3\cdots (2n-1)$, $(2n)!!=2\cdot 4 \cdots 2n$

Using Wolframalpha, the result is $\sqrt{2}-1$. But I don't have any idea to come up with that result.
Thanks a lot.
 A: First note that
$$(2n-1)!!=\frac{(2n)!}{2^nn!}$$
and $(2n)!!=2^nn!$, so
$$\frac{(2n-1)!!}{(2n)!!\cdot2^n}=\frac{(2n)!}{2^nn!\cdot2^nn!\cdot2^n}=\frac1{2^{3n}}\binom{2n}n\;.$$
Now use the power series
$$\frac1{\sqrt{1-4x}}=\sum_{n\ge 0}\binom{2n}nx^n\;,\tag{1}$$
the generating function for the central binomial coefficients. $\binom{2n}n\le 4^n$ for $n\ge 1$, so $(1)$ certainly converges at $x=\frac18$, and we have
$$\sqrt2=\frac1{\sqrt{1-1/2}}=\sum_{n\ge 0}\binom{2n}n\left(\frac18\right)^n=\sum_{n\ge 0}\frac1{2^{3n}}\binom{2n}n=1+\sum_{n\ge 1}\frac{(2n-1)!!}{(2n)!!\cdot2^n}\;.$$
A: Since $(2n)!! = 2^n n!$ and $(2n - 1)!! = (2n-1)(2n-3)\cdots 5\cdot 3\cdot 1$,
\begin{align} \sum_{n = 1}^\infty \frac{(2n-1)!!}{(2n)!!2^n} &= \sum_{n = 1}^\infty \frac{(2n-1)(2n-3)\cdots 5\cdot 3\cdot 1}{2^n n!}\frac{1}{2^n}\\
& = \sum_{n = 1}^\infty \frac{\left(n - \frac{1}{2}\right)\left(n - \frac{3}{2}\right)\cdots \left(\frac{3}{2}\right)\left(\frac{1}{2}\right)}{n!}\frac{1}{2^n}\\
&= \sum_{n = 1}^\infty \frac{\left(-\frac{1}{2} - n  + 1\right)\left(-\frac{1}{2} - n + 2\right)\cdots \left(-\frac{1}{2}\right)}{n!}\left(-\frac{1}{2}\right)^n\\
&= \sum_{n = 0}^\infty \binom{-1/2}{n}\left(-\frac{1}{2}\right)^n - 1\\
&=(1 - 1/2)^{-1/2} - 1 \\
&= \sqrt{2} - 1.
\end{align}
