# Examine the convergence and find the sum of series $\sum_{n=1}^{\infty}(-1)^n\frac{(2n-1)!!}{(2n)!!}.$

Examine the convergence and find the sum of series $$\sum_{n=1}^{\infty}(-1)^n\frac{(2n-1)!!}{(2n)!!}.$$

How to examine the convergence of this series before finding the sum? Is it the right method to find the radius of convergence of equivalent power series $$\sum_{n=1}^{\infty}(-1)^n\frac{(2n-1)!!}{(2n)!!}x^n$$ and to conclude from the radius that the series is convergent or divergent? What is the easiest method to find the sum? Could the perturbation method be used?

Please don't use "Consider the function..." method for finding the sum, because I find that method very hard.

## 2 Answers

Hint. As regards the convergence show that the sequence $a_n:=\frac{(2n-1)!!}{(2n)!!}$ decreases to zero and apply Leibniz's rule. To evaluate the sum show that $$a_n=\frac{(2n)!}{4^n(n!)^2}=\frac{1}{4^n}\binom{2n}{n}$$ and recall (see How to show that $1 \over \sqrt{1 - 4x}$ generate $\sum_{n=0}^\infty \binom{2n}{n}x^n$) that for $x\in [-1/4,1/4)$, $$\sum_{n=0}^{\infty}\binom{2n}{n}x^n= \frac{1}{\sqrt{1-4x}}.$$

Since $$\frac{(2n-1)!!}{(2n)!!}=\frac{1}{4^n}\binom{2n}{n} = \frac{1}{\pi}\int_{0}^{\pi/2}\left(\sin x\right)^{2n}\,dx$$ by De Moivre's identity $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$ and $\int_{0}^{2\pi}e^{mi\theta}\,d\theta = 2\pi\,\delta(m)$, the given series (which is conditionally convergent by Leibniz' test) equals $$\sum_{n\geq 1}\binom{2n}{n}\frac{(-1)^n}{4^n} = \frac{1}{\pi}\int_{0}^{\pi/2}\sum_{n\geq 1}(-1)^n\left(\sin x\right)^{2n}\,dx=-\frac{1}{\pi}\int_{0}^{\pi/2}\frac{\sin^2(x)}{1+\sin^2(x)}\,dx$$ where the last integral is straightforward to compute through the substitution $x=\arctan t$, leading to:

$$\sum_{n\geq 1}\binom{2n}{n}\frac{(-1)^n}{4^n} = -\frac{1}{\pi}\left(\frac{\pi}{2}-\int_{0}^{+\infty}\frac{dt}{1+2t^2}\right)=\color{red}{-1+\frac{1}{\sqrt{2}}}.$$