Convergent or divergent $\sum\limits_{n=0}^{\infty }{\frac{1\cdot 3\cdot 5...(2n-1)}{2\cdot 4\cdot 6...(2n+2)}}$

Question

\begin{align} & \sum\limits_{n=0}^{\infty }{\frac{1\cdot 3\cdot 5...(2n-1)}{2\cdot 4\cdot 6...(2n+2)}} \ & \text{ordering} \ & a_{n}=\frac{1\cdot 3\cdot 5...(2n-1)}{2\cdot 4\cdot 6...(2n+2)}=\frac{1\cdot 3\cdot 5...(2n-3)(2n-1)\cdot 1}{2\cdot 4\cdot 6...(2n-2)(2n)(2n+2)}= \ & \underbrace{\left( 1-\frac{1}{2} \right)\left( 1-\frac{1}{4} \right)...\left( 1-\frac{1}{2n} \right)}{n\text{ times}}\frac{1}{(2n+2)} \ & \text{clearly} \ & \left( 1-\frac{1}{2} \right)^{n}\frac{1}{(2n+2)}\le a{n}\le \left( 1-\frac{1}{2n} \right)^{n}\frac{1}{(2n+2)} \ & \text{Root test} \ & \sqrt[n]{\left( 1-\frac{1}{2} \right)^{n}}\frac{1}{\sqrt[n]{(2n+2)}}\le \sqrt[n]{a_{n}}\le \sqrt[n]{\left( 1-\frac{1}{2n} \right)^{n}}\frac{1}{\sqrt[n]{(2n+2)}} \ & n\to \infty \ & \frac{1}{2}\le \underset{n\to \infty }{\mathop{\lim }},\sqrt[n]{a_{n}}\le 1 \ & \text{nothing :(} \ & \text{Ratio test} \ & \frac{a_{n}}{a_{n-1}}=\frac{1\cdot 3\cdot 5...(2n-3)(2n-1)}{2\cdot 4\cdot 6...(2n)(2n+2)}\centerdot \frac{2\cdot 4\cdot 6...(2(n-1)+2)}{1\cdot 3\cdot 5...(2(n-1)-1)}=\frac{2n-1}{2n+2} \ & \underset{n\to \infty }{\mathop{\lim }},\frac{a_{n}}{a_{n-1}}=1 \ & \text{nothing again} \ \end{align}

Any suggestions?

Are you familiar with the $\Gamma$ function ? Please, answer. — Claude Leibovici, Sep 21 '14 at 11:32
See this answer. Your series is dominated by the linked one or you can replace $(2N+1)^{-1}$ by $(2N+2)^{-1}$ . — WimC, Sep 21 '14 at 11:49
@JackD'Aurizio the sum can start at $n = 0$ provided we interpret the product of an empty list of number is 1. — achille hui, Sep 21 '14 at 13:39

Jack D'Aurizio · Answer 1 · 2014-09-22T11:35:41.330

4

An explicit computation is even better. We have: $$\frac{(2n-1)!!}{(2n+2)!!}=\frac{1}{(2n+2)\,4^n}\binom{2n}{n}=\frac{1}{4^n}\binom{2n}{n}-\frac{1}{4^{n+1}}\binom{2n+2}{n+1}$$ hence your series is telescopic and we have: $$\sum_{n=1}^{N}\frac{(2n-1)!!}{(2n+2)!!}=\frac{1}{2}-\frac{1}{4^{N+1}}\binom{2N+2}{N+1}=\frac{1}{2}+O\left(\frac{1}{\sqrt{N}}\right).$$

In order to prove the last asymptotics, notice that: $$\frac{1}{4^n}\binom{2N}{N}=\frac{2}{\pi}\int_{0}^{\pi/2}\cos(x)^{2N}dx\leq\frac{2}{\pi}\int_{0}^{\pi/2}e^{-Nx^2}\,dx\leq\frac{1}{\sqrt{\pi N}}.$$

edited Sep 22 '14 at 11:35

answered Sep 21 '14 at 13:40

Jack D'Aurizio

353,855

How do you prove the last equality? Thanks. – Olivier Oloa Sep 21 '14 at 19:46
@OlivierOloa: I have just updated my answer to prove that asymptotics. – Jack D'Aurizio Sep 22 '14 at 11:36
1

OK Jack D'Aurizio, instead of Stirling's formula, you have used Wallis integrals $$\int_{0}^{\pi/2}\cos(x)^{2N}dx=\frac{1}{4^n}\binom{2N}{N}$$ and the gaussian integral $$\int_{0}^{\infty}e^{-x^2},dx=\frac{\sqrt{\pi}}{ 2}.$$ Thank you! – Olivier Oloa Sep 22 '14 at 18:32

score 3 · Answer 2 · answered Sep 21 '14 at 12:33

You may write $$ \begin{align} \frac{1\cdot 3\cdot 5 \cdots(2n-1)}{2\cdot 4\cdot 6\cdots(2n+2)} &=\frac{1\cdot 2\cdot 3\cdot 4 \cdot 5\cdot6\cdots(2n-1)\cdot 2n}{(2\cdot 4\cdot 6\cdots (2n))^2(2n+2)}\\ &=\frac{(2n)!}{(2^{n} \cdot 1\cdot 2\cdot 3 \cdot 4 \cdots n)^2 \cdot (2n+2)}\\ & =\frac{(2n)!}{2^{2n} (n!)^2 \cdot (2n+2)}\\ & \sim \frac{1}{\sqrt{\pi n}\cdot (2n+2)}\\ & \sim \frac{1}{2\sqrt{\pi}\cdot n^{\Large\frac 32}}, \quad \text{for} \, n \, \text{great} \end{align} $$ where we have use Stirling's approximation, then you easily conclude to the convergence of the series.

Convergent or divergent $\sum\limits_{n=0}^{\infty }{\frac{1\cdot 3\cdot 5...(2n-1)}{2\cdot 4\cdot 6...(2n+2)}}$

2 Answers2