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!

Hint: take the logarithm, and try to make something like $\frac{1}{n}\sum_{k=1}^n f\left(\frac{k}{n}\right)$ appear for $f\colon x\in[0,1] \mapsto \ln (1+x)$. You will find the limit $\ell$ of the logarithm of your quantity, and then by continuity of $\exp$ your answer will be $e^\ell$.

Details. Taking the logarithm,
$$\begin{align}
\ln \sqrt[n]{\frac{(2n)!}{n!}} &= \frac{1}{n} \ln \frac{\prod_{k=1}^{2n} k}{\prod_{k=1}^n k}
= \frac{1}{n} \ln \prod_{k=n+1}^{2n} k
= \frac{1}{n} \sum_{k=n+1}^{2n} \ln  k\\
&= \frac{1}{n} \sum_{k=1}^{n} \ln  (k+n)
= \frac{1}{n} \sum_{k=1}^{n} \ln  \left(1+\frac{k}{n}\right) + n\frac{\ln n}{n}
\end{align}$$
so that the logarithm of your original quantity is
$$\begin{align}
\ln \frac{1}{n}\sqrt[n]{\frac{(2n)!}{n!}} &= 
- \ln n + \frac{1}{n} \sum_{k=1}^{n} \ln  \left(1+\frac{k}{n}\right) + n\frac{\ln n}{n}
= \frac{1}{n} \sum_{k=1}^{n} \ln  \left(1+\frac{k}{n}\right)
\end{align}$$
Can you see how to use Riemann sums now?
