What value does $\frac{1}{n} \sqrt[n]{n(n+1)(n+2)\cdots(2n)}$ tend to? I need to find where this sequence tends to:
$$\frac{1}{n} \sqrt[n]{n(n+1)(n+2)...(2n)}$$
My answer is $2$, by using a trick for the $n^{th}$ root: I take the inside and find where $\dfrac{A_{n+1}}{A_n}$ goes, and this is where the original sequence goes.

$$ a_n 
= \frac{1}{n} \sqrt[n]{n(n+1)(n+2)\cdots (2n)} 
= \sqrt[n]{1(1+\frac{1}{n})(1+\frac{2}{n})\cdots (1+\frac{n}{n})}.
$$
$$
\log a_n
= \frac{1}{n} \{ \log 1 + \log (1+\frac{1}{n}) + \log (1+\frac{2}{n}) 
+ \cdots + \log (1+\frac{n}{n}) \}.
$$
We know the above equations. Hence, 
$$
\lim\limits_{n \rightarrow \infty} \log a_n = \int_1^2 \log x dx
= [x\log x -x ]_1^2 = 2\log 2 - 1 = \log \frac{4}{e}.
$$
Thus, the limit of $a_n$ is $\frac{4}{e}$.
