<s>[INST] 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.
 [/INST] Defining$$a_n=\frac{\prod_{k=1}^{n-1} (2k-1) }{2^nn!}$$ you can get rid of the numerator if you notice that $$\prod_{k=1}^{n-1} (2k-1)= \frac{2^{n-1} \Gamma \left(n-\frac{1}{2}\right)}{\sqrt{\pi }}$$ which makes $$a_n=\frac{\Gamma \left(n-\frac{1}{2}\right)}{2 \sqrt{\pi } n!}$$ which makes $$\frac{a_{n+1}}{a_n}=1-\frac{3}{2 (n+1)}$$ As Brian M. Scott answered, the ratio test doesn’t help much here.
But, for large values of $n$, we can use Stirling approximation and get $$a_n=\frac{\left(\frac{1}{n}\right)^{3/2}}{2 \sqrt{\pi }}+\frac{3
   \left(\frac{1}{n}\right)^{5/2}}{16 \sqrt{\pi }}+\frac{25
   \left(\frac{1}{n}\right)^{7/2}}{256 \sqrt{\pi
   }}+O\left(\frac{1}{n^4}\right)$$ Using the first terms $$\sum_{n=2}^\infty \frac{\left(\frac{1}{n}\right)^{3/2}}{2 \sqrt{\pi }}=\frac{\zeta \left(\frac{3}{2}\right)-1}{2 \sqrt{\pi }}\approx 0.454843$$  $$\sum_{n=2}^\infty\frac{3
   \left(\frac{1}{n}\right)^{5/2}}{16 \sqrt{\pi }}=\frac{3 \left(\zeta \left(\frac{5}{2}\right)-1\right)}{16 \sqrt{\pi }}\approx 0.0361244$$ $$\sum_{n=2}^\infty \frac{25
   \left(\frac{1}{n}\right)^{7/2}}{256 \sqrt{\pi
   }}=\frac{25 \left(\zeta \left(\frac{7}{2}\right)-1\right)}{256 \sqrt{\pi }}\approx 0.00698261$$
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