Convergence of $\sum\limits_{n=2}^\infty \frac{1\cdot 3\cdot \cdot \cdot (2n-3)}{2^n n!}$ 
Convergence of $$\sum_{n=2}^\infty \frac{1\cdot 3\cdot \cdot \cdot (2n-3)}{2^n n!}$$

Well, I have tried almost everything. D'Alembert's criterion doesn't work because the limit is 1. I have tried to bound $\frac{1\cdot 3\cdot \cdot \cdot (2n-3)}{2^n n!}$ in several ways but every bound diverges. Any hint?
P.S: I know this series converges to $\frac{1}{2}$
 A: You can use Gauss test (or Rabee's test, the latter is weaker but still suffices), I will use the Gauss test:

If $u_n >0$ and satisfies
  $$\frac{u_n}{u_{n+1}} = 1 + \frac{h}{n} + O(\frac{1}{n^r})$$ for some $r>1$, then $\sum u_n$ converges iff $h>1$.

In your case, $$u_n = \frac{1\cdot 3\cdots (2n-3)}{2^n n!} = \frac{(2n-2)!}{2^{2n-1}n!(n-1)!}$$ we easily obtain
$$\frac{u_n}{u_{n+1}} = \frac{2n+2}{2n-1} = 1+\frac{3}{2n-1} = 1+\frac{3}{2n}+O(\frac{1}{n^2})$$
so $\sum u_n$ converges.

Alternatively, you can use the Stirling formula
$$n! \sim \sqrt{2\pi n}(\frac{n}{e})^n$$ but I think this is an overkill.

The value of the sum can be derived from the following series (after integrating and shifting terms):
$$\sum_{n=0}^\infty \binom{2n}{n}x^n = \frac{1}{\sqrt{1-4x}}$$
A: $$\sum_{n\geq 2}\frac{(2n-3)!!}{2^n n!} = \sum_{n\geq 1}\frac{(2n-1)!!}{2^{n+1}(n+1)!}=\sum_{n\geq 1}\frac{(2n)!}{2^{2n+1}n!(n+1)!}=\sum_{n\geq 1}\binom{2n}{n}\frac{1}{4^n(2n+2)} $$
and since $\frac{1}{4^n}\binom{2n}{n}\sim\frac{1}{\sqrt{\pi n}}$, the given series is convergent by asymptotic comparison with $\sum_{n\geq 1}\frac{1}{n\sqrt{n}}$, which is convergent by the $p$-test. The given series equals $\frac{1}{2}$ by creative telescoping or probabilistic arguments: have a look at this similar question.
