Solving $\lim_{n\to \infty}\frac{(1)(2)(3)\cdots (n)}{(n+1)(n+2)(n+3)\cdots(2n)}$

Question

I am trying to solve this limit : $$\lim_{n\to \infty}\frac{(1)(2)(3)\cdots (n)}{(n+1)(n+2)(n+3)\cdots(2n)}.$$ Any direction ? I think i should i use the Sandwich theorem or something else.

Answer 1

Here is a silly proof: Let $a_n=\dfrac{(n!)^2}{(2n)!}$. Then

$$\lim_{n\to\infty}\frac{a_{n+1}}{a_n} = \lim_{n\to\infty}\dfrac{((n+1)!)^2}{(2n+2)!}\dfrac{(2n)!}{(n!)^2}=\lim_{n\to\infty}\frac{(n+1)^2}{(2n+2)(2n+1)}=\frac{1}{4}$$ which has magnitude less than $1$. Hence the series $\sum_{n=1}^\infty a_n$ converges by the ratio test. But such a series only converges if $\lim_{n\to\infty} a_n=0$.

Answer 2

$$(n+1)(n+2)(n+3)\cdots2n\ge n^n$$ $$\frac{1}{(n+1)(n+2)(n+3)\cdots2n}\le \frac{1}{n^n}$$ $$0\le\lim_{n\to \infty}\frac{1\cdot2\cdot3\cdots n }{(n+1)(n+2)(n+3)\cdots 2n}\le \lim_{n\to \infty}\frac{n!}{n^n}$$

If you are not familiar with $\lim_{n\to \infty}\frac{n!}{n^n}$ this isn't very helpful.