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.
$a_n = (1-1/2)(1-1/4)…(1-1/2n)$
 A: Cross-multiplication yields, for $k\ge1$,
$$
\left(\frac{2k-1}{2k}\right)^2\le\frac{2k-1}{2k+1}
$$
Therefore,
$$
\begin{align}
\prod_{k=1}^n\left(\frac{2k-1}{2k}\right)^2
&\le\prod_{k=1}^n\frac{2k-1}{2k+1}\\
&=\frac1{2n+1}
\end{align}
$$
Thus,
$$
\prod_{k=1}^\infty\frac{2k-1}{2k}=0
$$

For $n\ge1$,
$$
\begin{align}
\frac{\frac{\sqrt{n^2-1}}n}{\frac{\sqrt{(n+1)^2-1}}{n+1}}
&=\sqrt{\frac{(n+1)^3(n-1)}{n^3(n+2)}}\\
&=\sqrt{\frac{n^4+2n^3-2n-1}{n^4+2n^3}}\\[18pt]
&\lt1
\end{align}
$$
Thus, $\frac{\sqrt{n^2-1}}n$ is increasing. Furthermore,
$$
\begin{align}
\lim_{n\to\infty}\frac{\sqrt{n^2-1}}n
&=\lim_{n\to\infty}\sqrt{1-\frac1{n^2}}\\[6pt]
&=1
\end{align}
$$
A: HINT
We have that
$$a_n=\frac{1\cdot 3\cdot…\cdot(2n-1)}{2\cdot4\cdot…\cdot2n}=\frac{(2n)!}{(2^nn!)^2}$$
then we can use Stirling's approximation.
Refer to the related


*

*What is the limit of $\frac{\prod Odd}{\prod Even}?!$
A: $$\frac{(2n-1)!!}{(2n)!!}=\frac{1}{4^n}\binom{2n}{n}=\frac{2}{\pi}\int_{0}^{\pi/2}\left(\cos\theta\right)^{2n}\,d\theta $$
is quite clearly decreasing and convergent to zero (by the dominated convergence theorem).
Laplace's method gives $\frac{1}{4^n}\binom{2n}{n}\sim\frac{1}{\sqrt{\pi\left(n+\frac{1}{4}\right)}}$.
A: The sequence is decreasing and bounded below by 0. That's why it has a limit.
