Cool series

By the binomial series, $\sqrt{1-x}=\sum_{k=0}^{\infty}{1/2\choose{k}}(-x)^k=1-\frac{1}{2}x-\sum_{k=2}^{\infty}\frac{(2k-3)!!}{2^kk!}x^k$ evaluated at $x=1$ , we have $\sum_{k=2}^{\infty}\frac{(2k-3)!!}{2^kk!}= 1-\frac{1}{2}=\boxed{\frac{1}{2}}$

Note that $\frac{1}{2\times4} = \frac{1\times 2 \times 3 \times 4}{2^2\times4^2\times 3} = \frac{4!}{(2^2\times 2!)^2 \times 3} = \frac{4!}{(2!)^2 \times 4^2\times 3} = \frac{4\choose2}{4^2\times3}$ So we have $\frac{4\choose2}{3\times 4^2} + \frac{6\choose3}{5\times 4^3} + \frac{8\choose4}{7\times 4^4} + \cdots$ But in fact $\frac{4\choose2}{3} = 2C_1$ where $C_1$ is the first Catalan Number. The pattern continues so using a generating function for the Catalan numbers, we can enumerate the value of the series. So $S = 2(C_1x^2+C_2x^3+\cdots) = 2x(C_1x+C_2x^2 + \cdots) = 2x\left(\frac{1-\sqrt{1-4x}}{2x}-1\right)$ evaluated at $x=\frac14$ .

This gives $S = 2\left(\frac14\right)(2-1) = \boxed{\frac12}$ as required.