<s>[INST] how to show $a_{n}=[\frac{(2n)!!}{(2n-1)!!}]^2 \frac{1}{2n+1}$ converges? Question: $\displaystyle{a_{n} =
\left[{\left(2n\right)!! \over \left(2n - 1\right)!!}\,\right]^{2}
{1 \over 2n + 1}\,,\quad\mbox{prove}\ a_{n}}$ converges.
My thought: I want to prove {$a_{n}$} is an increasing sequence and it has an upper bound. I've figured out $$a_{n} = (\frac{2\cdot 2 }{1\cdot 3})\cdot(\frac{4\cdot4}{3\cdot5})\dots (\frac{(2n-2)(2n-2)}{(2n-3)(2n-1)})(\frac{(2n)^2}{2n-1})(\frac{1}{2n+1})$$ And, $$\frac{a_{n+1}}{a_{n}}=\frac{(2n)^2}{(2n-1)(2n+1)}\frac{(2n+2)^2}{(2n+1)}\frac{(2n-1)}{(2n)^2}(\frac{2n+1}{2n+3}) = \frac{(2n+2)^2}{(2n+1)(2n+3)} \gt 1$$ So it is increasing.
My problem: I'm stuck at proving it has a upper bound. Maybe there are some inequalities that can be used here?
In addition, I've searched this question and find this sequence converges to $\pi/2$ and it is known as Wallis Formula.But here, I only want to show it converges and I'm not going to find its limit.
Thanks for your time!
 [/INST] $$
\begin{align}
& a_m=\left(\frac{(2m)!!}{(2m-1)!!}\right)^2 \frac{1}{2m+1} = \prod_{n=1}^{m}\left(\frac{2n}{2n-1}\cdot\frac{2n}{2n+1}\right) \Rightarrow \\
& \log(a_m)=\sum_{n=1}^{m}\left[2\log(2n)-\log(2n-1)-\log(2n+1)\right] \\
& \text{Let}\space f(x) = -\log(x) \space\colon x\gt1 \space\Rightarrow f'(x) = -\frac{1}{x} \space\Rightarrow f''(x) = \frac{1}{x^2} \space\gt0 \space\Rightarrow f(x) \space\text{convex function} \\[8mm]
& \text{(1).}\space f\left(\frac{x_1+x_2}{2}\right)\le\frac{f(x_1)+f(x_2)}{2}\Rightarrow \\
& \quad -\log\left(\frac{(2n-1)+(2n+1)}{2}\right) \le \frac{-\log(2n-1)-\log(2n-1)}{2} \Rightarrow \\
& \quad -2\log(2n) \le -\log(2n-1)-\log(2n-1) \Rightarrow 0\le 2\log(2n)-\log(2n-1)-\log(2n-1) \\
& \quad \sum_{n=1}^{m}\left[2\log(2n)-\log(2n-1)-\log(2n+1)\right] \ge 0 \Rightarrow \log(a_m) \ge 0 \Rightarrow \quad\color{red}{a_m \ge 1} \\[8mm]
& \text{(2).}\space f(x) \ge f(y)+f'(y)\,(x-y) \Rightarrow \\
& \quad -\log(2n) \ge -\log(2n-1)-\frac{2}{2n-1} \Rightarrow 0 \ge \log(2n)-\log(2n-1)-\frac{2}{2n-1} \quad\space\& \\
& \quad -\log(2n) \ge -\log(2n+1)+\frac{2}{2n+1} \Rightarrow 0 \ge \log(2n)-\log(2n+1)+\frac{2}{2n-1} \quad\Rightarrow \\
& \quad 0 \ge 2\log(2n)-\log(2n-1)-\log(2n+1)-\frac{2}{2n-1}+\frac{2}{2n+1} \Rightarrow \\
& \quad 2\log(2n)-\log(2n-1)-\log(2n+1) \le \frac{2}{2n-1}-\frac{2}{2n+1} = \frac{4}{4n^2-1} \Rightarrow \\
& \quad \sum_{n=1}^{m}\left[2\log(2n)-\log(2n-1)-\log(2n+1)\right] \le \sum_{n=1}^{m}\frac{4}{4n^2-1} \Rightarrow \\
& \quad \lim_{m\rightarrow\infty} \log(a_m) \le \sum_{n=1}^{\infty}\frac{4}{4n^2-1} = 2 \Rightarrow \quad\color{red}{\lim_{m\rightarrow\infty} a_m\le e^2} \\
& \quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\color{white}{\text{.}}
\end{align}
$$
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