Prove the following limit is $\lim_{n\rightarrow\infty}\frac{1}{n}\left(\frac{(2n)!}{n!}\right)^{\frac1n}=\frac{4}{e}$ I want to prove the following limit
$$\lim_{n\rightarrow\infty}\frac{1}{n}\left(\frac{(2n)!}{n!}\right)^{\frac1n}=\frac4e$$
In a previous exercise I found that $\int_1^2\ln x dx=2\ln(2)-1$ which is also equal to $\ln(4/e)$. I'm thinking I could use this result somehow?
 A: Hint. One may consider the logarithm of the given sequence,
$$
\begin{align}
\log\left[\frac{1}{n}\left(\frac{(2n)!}{n!}\right)^{\frac1n}\right]&=\log\left(\frac1n\right)+\frac1n \log\left(\prod_{k=1}^n (n+k) \right)
\\\\&=-\log n+\frac1n\sum_{k=1}^n \log\left(n+k \right)
\\\\&=\frac1n\sum_{k=1}^n \log\left(1+\frac kn \right)
\end{align}
$$ then one may conclude with a Riemann sum.
A: Alternatively, we can also use the fact that if $a_n>0$ for all $n\ge1$ and the sequence $\frac{a_{n+1}}{a_n}$ converges in $[0,\infty]$, then
$$
\lim_{n\to\infty}a_n^{1/n}=\lim_{n\to\infty}\frac{a_{n+1}}{a_n}
$$
(see this answer).
We have that
$$
a_n=\frac{(2n)!}{n^nn!}
$$
and
$$
\frac{a_{n+1}}{a_n}
=\frac{(2(n+1))!}{(n+1)^{n+1}(n+1)!}\cdot\frac{n^nn!}{(2n)!}
=\frac{2(n+1)(2n+1)}{(n+1)^2}\cdot\biggl(1-\frac1{n+1}\biggr)^n.
$$
The first term converges to $4$ and the second term converges to $1/e$. Hence,
$$
\lim_{n\to\infty}a_n^{1/n}=\frac4e.
$$
A: Considering $$A=\left(\frac{(2n)!}{n!}\right)^{\frac1n}\implies \log(A)=\frac1n\,\log\left(\frac{(2n)!}{n!}\right)$$ Now, using Stirling approximation, you should get $$\log(A)=\frac1n\left(n (\log (n)-1+2 \log (2))+\frac{\log (2)}{2}+O\left(\frac{1}{n}\right) \right)$$  that is to say $$\log(A)=(\log (n)-1+2 \log (2))+\frac{\log (2)}{2 n}+O\left(\frac{1}{n^2}\right)$$ Now, back to Taylor $$A=e^{\log(A)}=\frac{4 n}{e}+\frac{2 \log (2)}{e}+O\left(\frac{1}{n^2}\right)$$ So, for your expression $$\frac An=\frac{4 }{e}+\frac{2 \log (2)}{en}+O\left(\frac{1}{n^2}\right)$$ which shows the limit and also how it is approached.
