TITLE: Evaluate $\lim_{n\to \infty}{\sqrt[n]\frac{(2n)!}{n^n\times{n!}}}$
QUESTION [5 upvotes]: $$\lim_{n\to \infty}{\sqrt[n]\frac{(2n)!}{n^n\times{n!}}}$$
It is a sequence and n is natural
It looks like I should use $\lim_{n\to \infty}{\sqrt[n]{a_n}}=x$ but I don't know how. Does it mean that $a_n=\frac{(2n)!}{n^n\times{n!}}$ and then do it from there or is $a_n=\sqrt[n]\frac{(2n)!}{n^n\times{n!}}$
I have never used this before and I am not sure what to do

REPLY [4 votes]: Let $a_n = \dfrac{(2n)!}{n^n{n!}}$. Then
$\displaystyle
\lim_{n\to \infty}{\sqrt[n]{a_n}}
=
\lim_{n\to \infty}\dfrac{a_{n+1}}{a_n}
$
if this limit exists.
Now
$$
\dfrac{a_{n+1}}{a_n}
=
\dfrac{(2(n+1))!}{(n+1)^{n+1}{(n+1)!}}
\
\dfrac{n^n{n!}}{(2n)!}
=
\dfrac{(2n+2)(2n+1)}{(n+1)(n+1)}
\
\left(\dfrac{n}{n+1}\right)^n
\to
\dfrac{4}{e}
$$