How do I evaluate this series? How do I evaluate this series: 
\begin{equation}
\sum_{n=2}^\infty \frac{\prod_{k=1}^{n-1} (2k-1) }{2^nn!} = \frac{1}{8} + \frac{1}{16} + \frac{5}{128} + \frac{7}{256} +\ldots
\end{equation}
I wanted to use the Comparison test to show convergence, but I didn't know what to compare it to since my series has a product in the numerator...
I'm lost as to how to evaluate it.

To show convergence, first note that
$$\prod_{k=1}^n(2k-1)=(2n-3)!!=\frac{(2n-2)!}{2^{n-1}(n-1)!}\;,$$
so
$$\frac{(2n-3)!!}{2^nn!}=\frac{(2n-2)!}{2^{2n-1}n!(n-1)!}=\frac1{2^{2n-1}n}\binom{2n-2}{n-1}\;.\tag{1}$$
The ratio test doesn’t help here, but you can use the fact that the central binomial coefficient $\binom{2n}n$ is asymptotically $\frac{4^n}{\sqrt{\pi n}}$, meaning that
$$\lim_{n\to\infty}\frac{\binom{2n}n}{\frac{4^n}{\sqrt{\pi n}}}=1\;.$$
Thus, the righthand side of $(1)$ is asymptotically
$$\frac2{4^nn}\cdot\frac{4^{n-1}}{\sqrt{\pi(n-1)}}=\frac1{2n\sqrt{\pi(n-1)}}\;,$$
which is on the order of $n^{-3/2}$.
