<s>[INST] Prove the sequence $a_n =\frac{1\cdot 3\cdot…\cdot(2n-1)}{2\cdot4\cdot…\cdot2n}$ has a limit I have several questions to ask:
1) Show increasing, find the upper bound if you can of
 $\sqrt{(n^2-1)}/n$.
$\sqrt{(n^2-1)}/n= |n|\sqrt{1-1/n^2}/n$ if $n$ is positive than $\sqrt1$ else $-\sqrt1$; 
bound: $\sqrt{(n^2-1)}/n \le \sqrt {n^2}/n = |n|/n=1$
To show if it is increasing should I do $\frac{\sqrt{(n+1)^2-1}}{n+1} \ge \frac{\sqrt{(n)^2-1}}{n}$?
2)Prove the sequence $a_n = \frac{1\cdot 3\cdot…\cdot(2n-1)}{2\cdot4\cdot…\cdot2n}$ has a limit
$a_n$ is a decreasing sequence, so to have a limit it must be bounded from below.
$a_n = (1-1/2)(1-1/4)…(1-1/2n)$
 [/INST] $$\frac{(2n-1)!!}{(2n)!!}=\frac{1}{4^n}\binom{2n}{n}=\frac{2}{\pi}\int_{0}^{\pi/2}\left(\cos\theta\right)^{2n}\,d\theta $$
is quite clearly decreasing and convergent to zero (by the dominated convergence theorem).
Laplace's method gives $\frac{1}{4^n}\binom{2n}{n}\sim\frac{1}{\sqrt{\pi\left(n+\frac{1}{4}\right)}}$.
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