$\lim_{n\to \infty} {1\over n}\sqrt[n]{(n+1)(n+2)\cdots(2n)}$ $$\lim_{n\to \infty} {1\over n}\sqrt[n]{(n+1)(n+2)\cdots(2n)}$$
My attempt:
\begin{align}
\lim_{n\to \infty} {1/ n}\sqrt[n]{(n+1)(n+2)\cdots(2n)} &= \lim_{n\to \infty} \sqrt[n]{(1+{1/ n})(1+{2/ n})\cdots(1+{n/ n})}\\
 &= \lim_{n\to \infty}(1+{1/ n})^{n*{1/ n^2}}(1+{2/ n})^{n*{1/ n^2}}\cdots(1+{n/ n})^{n*{1/ n^2}}\\
 &=\lim_{n\to \infty}e^{{1/ n^2}}e^{{2/ n^2}}\cdots e^{{n/ n^2}}\\
  &= \sqrt{e}
\end{align}
I know that this is wrong answer. Could you please show me where my mistake is?
EDIT  I know the right solution, I just wanted to see where my mistake is. Thanks for your answers anyway! 
EDIT 2 5 beautiful answers received already, but I dont think I can accept any of them, because my question was to show mistake in my own derivations. Anyway I learnt a lot especially that $\lim_{n\to \infty} \frac{b_{n+1}}{b_n}=\lim_{n\to \infty} \sqrt[n]b_n$. Thanks for everyone!
 A: $$\lim_{n\rightarrow+\infty}\ln{a_n}=\lim_{n\rightarrow+\infty}\sum_{k=1}^n\frac{1}{n}\ln\left(1+\frac{k}{n}\right)=\int\limits_{0}^1\ln(1+x)dx=$$
$$=\left((1+x)\ln(1+x)-x\right)_0^1=2\ln2-1,$$
which gives $\frac{4}{e}.$
A: Try using logarithms.
I.e. $$\ln ( \rm A)= \lim_{n \to \infty} \sum_{r=1}^{n} \frac 1n\ln \left( 1+\frac rn \right) =\int_0^1 \ln(1+x) \rm dx=\ln(4)-1$$
Implying $$A=e^{\ln(4)-1}=\frac 4e$$
A: (Other way )
Using $\lim \frac{|x_{n+1}|}{|x_n|}=L$ so $\lim \sqrt[n]{|x_n|}=L .$
take $x_n=\frac{(2n)!}{n!n^n}$, so $\frac{x_{n+1}}{x_n}=\frac{2(2n+1)}{n+1}\frac{1}{(1+\frac{1}{n})^{n}} \rightarrow \frac{4}{e}.$
A: I'm typing this for fun and to practise Stirling's approximation: $n! \sim (n/e)^n \sqrt{2\pi n}$.
\begin{align}
\frac1n \sqrt[n]{\frac{(2n)!}{n!}} &\sim
\frac1n \sqrt[n]{\frac{(2n/e)^{2n} \sqrt{2\pi(2n)}}{(n/e)^{n} \sqrt{2\pi(n)}}} \\
&= \frac1n \sqrt[n]{\frac{2^{2n} n^n \sqrt2}{e^n}} \\
&= \frac1n \frac{2^2n}{e} \sqrt[2n]{2} \\
&= \frac4e \sqrt[2n]{2} \xrightarrow[n\to\infty]{} \frac4e
\end{align}
A: Let indicate
$$a_n={1\over n}\sqrt[n]{ (n+1)(n+2)\cdots(2n) }$$
$$b_n=a_n^n=\frac{{(n+1)(n+2)\cdots(2n)}}{n^n}$$
then
$$\frac{b_{n+1}}{b_n}=\frac{{(n+2)(n+3)\cdots(2n+2)}}{(n+1)^{n+1}}\frac{n^n}{(n+1)(n+2)\cdots(2n)}=\left(\frac{n}{1+n}\right)^n\frac{(2n+2)(2n+1)}{(n+1)^2}=\frac{1}{\left(1+\frac1n\right)^n}\frac{(2n+2)(2n+1)}{(n+1)^2}\to \frac4e$$
thus (see also for reference)
$$\lim_{n\to \infty} \frac{b_{n+1}}{b_n}=\lim_{n\to \infty} \sqrt[n]b_n=\lim_{n\to \infty} a_n=\frac4e$$
