How to prove the convergence of the sequence?

Question

I need to prove the convergence on $\mathbb R$: $$x_n = \frac{(2n)!!}{(2n+1)!!}$$ I don`t even know from what to start. Help me please.

First, figure out what is $(2n)!!$ and $(2n+1)!!$. Then check: is $(x_n)$ a monotone sequence? Is it bounded? — , Dec 15 '17 at 17:54
See also: Prove that $x_n = (2n)!!/(2n-1)!!$ sequence is divergent. — Martin Sleziak, Jan 08 '20 at 12:21

score 1 · Answer 1 · answered Dec 15 '17 at 18:19

First notice that if $m, r$ are natural numbers and $m>r$ then $\frac{r!}{m!}\leq \frac{1}{m}$ and notice that $x_n>0$ for every $n$ (because both the numerator and the denominator are positive). Then taking $m=(2n+1)!$ and $r=(2n)!$ then $x_n=\frac{r!}{m!}\leq \frac{1}{m}=\frac{1}{(2n+1)!}$ and this proves $x_n$ converges to 0 since it's between two quantities which both converge to 0.

score 1 · Answer 2 · answered Dec 15 '17 at 18:37

1

$x_{n}=\frac{(2^n{n!})^2}{(2n+1)!}$. Then $\frac{x_{n+1}}{x_{n}}=\frac{2(n+1)}{2n+3}<{1}$. So $\{x_{n}\}$ is decreasing and positive. Hence, it converges.

answered Dec 15 '17 at 18:37

M.R. Yegan

672

score 0 · Accepted Answer · answered Dec 15 '17 at 18:23

$$x_n = \frac{(2n)!!}{(2n+1)!!} = \prod_{k=1}^{n}\frac{2k}{2k+1} = \prod_{k=1}^{n}\left(1-\frac{1}{2k+1}\right) $$ and since $\sum_{k\geq 1}\frac{1}{2k+1}=+\infty$, $\lim_{n\to +\infty} x_n=0$. In explicit terms, $$ 0\leq x_n \leq \exp\left(-\sum_{k=1}^{n}\frac{1}{2k+1}\right). $$

How to prove the convergence of the sequence?

3 Answers3