Evaluate $\sum_{n=1}^{\infty} \frac{(2n)!}{2^{2n}(n!)^2 (2n-1)}$ How to evaluate the series
$$S = \sum_{n=1}^{\infty} \frac{1}{2^{2n}(2n-1)} \binom{2n}{n}$$
The original question was to show that for 
$$ a_n=\left(\frac{ 2n-3 }{ 2n }\right)a_{n-1} , a_1 = \frac 1 2, \text{ Show } \; \sum_{k=1}^\infty a_k < 1$$
After simplifying it, I got the above result. But from W|A the above converges to $1$. I hope I haven't made any mistake. Still the I would like to know how to evaluate it.

First, rewrite the summand as $$\frac{1}{4^n (2n-1)} \binom{2n}{n} = \frac{(2n)!}{4^n (2n-1)n! n!} = \frac{2 (2n-2)!}{4^n n! (n-1)!} = \frac{2}{4^n n} \binom{2(n-1)}{n-1}.$$
Switching indices on the sum, you're trying to find
$$\sum_{n=0}^{\infty} \frac{2}{4^{n+1} (n+1)} \binom{2n}{n} = \frac{1}{2}\sum_{n=0}^{\infty} \frac{1}{4^n (n+1)} \binom{2n}{n}$$
Since the $n$th Catalan number is $\binom{2n}{n} \frac{1}{n+1},$ what you need now is the generating function for the Catalan numbers.  This is 
$$\sum_{n=0}^{\infty} \binom{2n}{n} \frac{z^n}{n+1} = \frac{1 - \sqrt{1-4z}}{2z}.$$
With $z = 1/4$, we have
$$\frac{1}{2}\sum_{n=0}^{\infty} \frac{1}{4^n (n+1)} \binom{2n}{n} = \frac{1}{2}\frac{1 - \sqrt{1-1}}{1/2} = 1.$$

Added: There's a probabilistic interpretation here, too.  Imagine two players playing a game in which each flips a fair coin once per round.  They stop when they have accumulated exactly the same number of heads.  Then the probability that the game ends in round $n$ is precisely the summand $$\frac{2}{4^n n} \binom{2(n-1)}{n-1} = \frac{2}{4^n} C_{n-1},$$
with $C_n$ the $n$th Catalan number.  (See, for example, my answer here.)  The fact that the OP's infinite series sums to $1$ indicates that the game will end with probability $1$.
