<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] Consider the sequence $(b_n)$ defined by
$$b_n = \frac{2n+2}{2n+1}\cdot a_n.$$
Then $(b_n)$ is decreasing
\begin{align}
\frac{b_{n+1}}{b_n} &= \frac{\frac{2n+4}{2n+3}}{\frac{2n+2}{2n+1}}\cdot\frac{a_{n+1}}{a_n} \\
&= \frac{(2n+4)(2n+1)}{(2n+2)(2n+3)}\cdot\frac{(2n+2)^2}{(2n+1)(2n+3)} \\
&= \frac{(2n+4)(2n+2)}{(2n+3)^2} \\
&= \frac{(2n+3)^2-1}{(2n+3)^2} \\
&< 1
\end{align}
and since $a_n > 0$ and $\frac{2n+2}{2n+1} > 1$ for all $n\geqslant 1$, we have
$$a_1 \leqslant a_n < b_n \leqslant b_1$$
for all $n \geqslant 1$, showing the boundedness of $(a_n)$.
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