Find $\lim_{n \to +\infty} \frac{1}{n}\sqrt[n]{\frac{(2n)!}{n!}} $ using Riemann integral Wonder how to determine this limit by the use of Riemann integral. The limit is as follows:
$$\lim_{n \to +\infty} \frac{1}{n}\sqrt[n]{\frac{(2n)!}{n!}} $$
My instructor told me that the usage of Riemann integral gives spectacular result. Checked Rudin, but did not find any valuable references.
I am very interested in seeing how this "spectacular result" emanates.
Help/advices/solutions very, very appreciated!

I will repeat basically the same approach as in this answer: Find the value of : $\lim\limits_{n\to \infty} \sqrt [n]{\frac{(3n)!}{n!(2n+1)!}} $ It is also the same approach as suggested in Paramanand Singh's comment. (I see that the OP asks specifically about a proof using Riemann's integral, but this seems interesting enough to be mentioned, too.)
We will use this fact (see the linked answer for references):

Let $(a_n)$ be a sequence of positive real numbers. If $\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}=L$, then $\sqrt[n]{a_n}$ converges too and $\lim\limits_{n\to\infty}\sqrt[n]{a_n}=L$. 

We use the above for $a_n = \frac{(2n!)}{n!n^n}$ and we get
$$\lim\limits_{n\to\infty} \frac{a_{n+1}}{a_n} = 
\lim\limits_{n\to\infty} \frac{(2n+1)(2n+2)}{(n+1)^2}\cdot\left(\frac{n}{n+1}\right)^n = \frac4e.$$
This implies that also 
$$\lim\limits_{n\to\infty} \frac1n\sqrt[n]{\frac{(2n)!}{n!}} = \lim\limits_{n\to\infty} \sqrt[n]{a_n} = \frac4e.$$ 
