# How do I evaluate this series?

How do I evaluate this series:

$$\sum_{n=2}^\infty \frac{\prod_{k=1}^{n-1} (2k-1) }{2^nn!} = \frac{1}{8} + \frac{1}{16} + \frac{5}{128} + \frac{7}{256} +\ldots$$

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.

• Do you want to prove convergence or evaluate the series? Or both? – ClassicStyle Jan 26 '16 at 3:57
• I would like to do both – xkcdFan1011011101111 Jan 26 '16 at 4:01

## 4 Answers

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.

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 $n^{-3/2}$.

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 $$\sum_{n=2}^\infty \frac{\left(\frac{1}{n}\right)^{3/2}}{2 \sqrt{\pi }}=\frac{\zeta \left(\frac{3}{2}\right)-1}{2 \sqrt{\pi }}\approx 0.454843$$ $$\sum_{n=2}^\infty\frac{3 \left(\frac{1}{n}\right)^{5/2}}{16 \sqrt{\pi }}=\frac{3 \left(\zeta \left(\frac{5}{2}\right)-1\right)}{16 \sqrt{\pi }}\approx 0.0361244$$ $$\sum_{n=2}^\infty \frac{25 \left(\frac{1}{n}\right)^{7/2}}{256 \sqrt{\pi }}=\frac{25 \left(\zeta \left(\frac{7}{2}\right)-1\right)}{256 \sqrt{\pi }}\approx 0.00698261$$

$$\sum_{k=1}^{n-1}(2k-1)=\dfrac{(n-1)}\cdot2(1+2n-3)=(n-1)^2=n(n-1)-n+1$$

Now $$\dfrac{\sum_{k=1}^{n-1}(2k-1)}{2^n n!}=\dfrac{n(n-1)-n+1}{2^n n!}=\dfrac14\cdot\dfrac{(1/2)^{n-2}}{(n-2)!}-\dfrac12\cdot\dfrac{(1/2)^{n-1}}{(n-1)!}+\dfrac{(1/2)^n}{n!}$$

Finally use $e^x=\sum_{r=0}^\infty\dfrac{x^r}{r!}$