How to prove that $\lim \frac{1}{n} \sqrt[n]{(n+1)(n+2)... 2n} = \frac{4}{e}$

I'd like a hint to show that:

$$\lim _{n\to\infty}\frac{1}{n} \sqrt[n]{(n+1)(n+2) \cdots 2n} = \frac{4}{e} .$$

Thanks.

Taking $\log$ of the expression you get

$\frac{1}{n}\sum \log (1+\frac{k}{n}) $.

This is a Riemann sum for the function $\log(1+x)$ on the interval $[0,1]$.

This is @Jonas Meyer's idea from the link in his answer:

Let $$a_n={(n+1)(n+2)\cdots 2n\over n^n}.$$

Then $$ \lim_{n\rightarrow\infty} {a_{n+1}\over a_n}= \lim_{n\rightarrow\infty}{(2n+1)(2n+2)\over n+1}\cdot {n^n\over (n+1)^{n+1}}= \lim_{n\rightarrow\infty}{2(2n+1)\over n+1}(1+\textstyle{1\over n})^{-n}={4\over e}. $$

But, for $a_n>0$, if $\lim\limits_{n\rightarrow\infty}{a_{n+1}\over a_n}=L$, then $\lim\limits_{n\rightarrow\infty}\root n\of {a_n}=L$ (see page 3 of Pete Clark's notes here).

In this case $$\lim\limits_{n\rightarrow\infty} \root n\of {a_n} = \lim\limits_{n\rightarrow\infty}{1\over n}\root n \of {(n+1)(n+2)\cdots 2n }. $$

You could rewrite it as

$$4\left(\frac{\sqrt[2n]{(2n)!}}{2n}\right)^2\cdot\frac{n}{\sqrt[n]{n!}}$$ and use the result of this question.