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.
 A: One possible approach is to notice the term inside the root is $\frac{(2n)!}{n!}$ and apply Stirling's approximation.
A: 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 $[0,1]$.
A: 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 $$\lim\limits_{n\rightarrow\infty} \root n\of {a_n} =
\lim\limits_{n\rightarrow\infty}{1\over n}\root n \of {(n+1)(n+2)\cdots 2n }.
$$
A: 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.
A: There's a theorem that is very helpful for these kind of questions.
Let $a_n$ be a sequence of positive real numbers. If $a_{n+1}/a_n$ converges, then $a_n^{1/n}$ converges to the same limit.
Continuing from here is pretty straightforward.
