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.

Here are hints to show that the series converges. (Actually computing the value will take more work.)
Hint1: The series is $\frac1{2\cdot4} + \frac{1\cdot3}{2\cdot4\cdot6} + \frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot8} + \frac{1\cdot3\cdot5\cdot7}{2\cdot4\cdot6\cdot8\cdot10} + \cdots$.
Hint2: One attempt is to cancel the $1$ with the $2$, the $3$ with the $4$, etc. to get $\frac1{2\cdot4} + \frac{1\cdot3}{2\cdot4\cdot6} + \frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot8} + \frac{1\cdot3\cdot5\cdot7}{2\cdot4\cdot6\cdot8\cdot10} + \cdots \le \frac14 + \frac16 + \frac18 + \frac1{10} +\cdots$. But the right hand side diverges, so this is no good, so you need to be more clever.
Hint3: $\frac{1\cdot3\cdot5\cdot7}{2\cdot4\cdot6\cdot8\cdot10} = \frac{(\sqrt1)^2\cdot(\sqrt3)^2\cdot(\sqrt5)^2\cdot(\sqrt7)^2}{2\cdot4\cdot6\cdot8\cdot10} = \sqrt1 \cdot \frac{\sqrt1\cdot\sqrt3}2 \cdot \frac{\sqrt3\cdot\sqrt5}4 \cdot \frac{\sqrt5\cdot\sqrt7}6 \cdot \frac{\sqrt7}{8\cdot10} \le \frac{\sqrt7}{8\cdot 10} $. So the entries grow on the order of $\frac{\sqrt n}{n^2}=\frac1{n^{1.5}}$, which converges by the $p$-series test.
