<s>[INST] limit of a sequence and principle of induction I try to evaluate this limit:
$$\lim_{n\to+\infty} \frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}$$
I considered this inequality
$$\frac{1}{4n}\le\left [ \frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)} \right]^2\le \frac{1}{2n+1}$$
and so 
$$\lim_{n\to+\infty} \frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}=0$$
my questions are:
1)-  how do I prove the inequality with the principle of induction?
2)-  there is another way to solve this limit?
 [/INST] I will only show the upper inequality as @JavaMan's comment shows. For $n=1$, it's only $\frac 14\leq\frac 13$, and if it's true for $n$ then 
\begin{align*}\left(\prod_{j=1}^{n+1}\frac{2j-1}{2j}\right)^2&=\left(\frac{2n+1}{2(n+1)}\right)^2\left(\prod_{j=1}^n\frac{2j-1}{2j}\right)^2\\
&\leq \left(\frac{2n+1}{2(n+1)}\right)^2\frac 1{2n+1}\\
&=\frac{2n+1}{4(n+1)^2}, 
\end{align*}
which is $\leq \frac 1{2n+3}$ since $\frac{(2n+1)(2n+3)}{4(n+1)^2}=\frac{4n^2+5n+3}{4n^2+8n+4}\leq 1$.
An other way: put $a_n:=\prod_{j=1}^n\frac{2j-1}{2j}$. Then by $\ln(1+x)\leq x$ for $x\geq -1$ we have 
$$\ln a_n =\sum_{j=1}^n\ln\left(1-\frac 1{2j}\right)\leq -\frac 12\sum_{j=1}^n\frac 1j,$$
so $a_n=\exp\left(-\frac 12\sum_{j=1}^n\frac 1j\right)$ and we can conclude since the harmonic series diverges.
</s>