Homework: limit of $\frac{1}{n}\sqrt[n]{\prod_{k=1}^n(n+k)}$ as $n\to\infty$ Determine the following limit
$$\lim_{n\to\infty}\frac{1}{n}\sqrt[n]{(n+1)(n+2)\ldots (n+n)} $$
Idea #1
the nth root could be rewritten as
$$\lim_{n\to\infty}\frac{1}{n}\sqrt[n]{\frac{(n+n)!}{n!}} $$
Noticing that $n^{-1}$ is an infinitesimal, if the n-th root were bounded, the limit would be $0$.
But hoping that $\sqrt[n]{\frac{(n+n)!}{n!}}$ is bounded is a pipe dream, this clearly tends to $\infty$.
Idea #2
$$\lim_{n\to\infty}\ln\left (\frac{1}{n}[(n+1)\ldots (n+n)]^\frac{1}{n}\right ) = \lim_{n\to\infty}\left [-\ln n+\frac{1}{n}\left (\ln (n+1)+\ln (n+2)+\ldots +\ln (n+n) \right )\right ]$$
but trying to bring it to a common denominator doesn't help either, what I would have is
$$\lim_{n\to\infty}\frac{1}{n} \left [-n\ln n+\ln (n+1)+\ldots +\ln 2+\ln n\right ]$$
Which doesn't bring me any closer to simplification.
The solution of this must something entirely different. I would like a hint/point me in the direction I should go.
 A: From the idea 1, we can derive the result using the Stirling's approximation $$n!\sim\sqrt{2\pi n}\left(\frac{n}{e}\right)^{n}
 $$ hence $$\frac{1}{n}\sqrt[n]{\frac{\left(2n\right)!}{n!}}\sim\frac{1}{n}\sqrt[n]{\frac{\sqrt{2n}}{\sqrt{n}}\left(\frac{2n}{e}\right)^{2n}\left(\frac{e}{n}\right)^{n}}=\frac{2^{2+1/2n}}{e}\rightarrow\frac{4}{e}.
 $$
A: Start with the expression on the right of your Idea #2. Using $\ln (n+k)=\ln n + \ln (1+k/n)$ for each $k$ shows that expression equals
$$-\ln n+\frac{1}{n}\sum_{k=1}^{n}[\ln n +\ln (1+k/n )]
$$ $$= -\ln n+\frac{1}{n}[n\ln n +  \sum_{k=1}^{n}\ln (1+k/n )]= \frac{1}{n}\sum_{k=1}^{n}\ln (1+k/n )$$ $$ \to \int_0^1 \ln (1+x)\, dx = 2\ln2 -1 = \ln (4/e).$$
Exponentiating back gives $4/e$ for the limit.
A: You are looking for 
$$ \lim_{n\to +\infty}\sqrt[n]{\frac{(2n)!}{n!\cdot n^n}}. $$
By setting $a_n=\frac{(2n)!}{n!\cdot n^n}$ we have:
$$ \frac{a_{n+1}}{a_n}=\frac{(2n+2)(2n+1)n^n}{(n+1)(n+1)^{n+1}}=\frac{4n+2}{n+1}\cdot\frac{1}{\left(1+\frac{1}{n}\right)^n}$$
hence:
$$ \lim_{n\to +\infty}\frac{a_{n+1}}{a_n}=\frac{4}{e} $$
and that implies:
$$ \lim_{n\to+\infty}\sqrt[n]{\frac{(2n)!}{n!\cdot n^n}}=\color{red}{\frac{4}{e}}.$$
