How to compute $\lim_{n\to \infty}\prod_{k=1}^{2n} \left(1+ \frac{ k}{n} \right)^{1/n} $ 
Find
  $$ \lim_{n\to \infty}\prod_{k=1}^{2n}     \left(1+ \frac{ k}{n} \right)^{1/n}.$$

I tried with ln composition but ineffectively.  Any idea?
 A: Following the suggestion by Paramanand Singh let indicate
$$a_n=\prod_{k=1}^{2n} \left(1+ \frac{ k}{n} \right)^{1/n}=\sqrt[n] b_n \quad b_n=\prod_{k=1}^{2n}     \left(1+ \frac{ k}{n} \right)$$
then
$$\frac{b_{n+1}}{b_n}=\frac{\prod_{k=1}^{2n+2} \left(1+ \frac{ k}{n+1} \right)}{\prod_{k=1}^{2n} \left(1+ \frac{ k}{n} \right)}=\left(1+\frac{2n+2}{n+1}\right)\left(1+\frac{2n+1}{n+1}\right)\prod_{k=1}^{2n} \frac{n(n+k+1)}{(n+1)(n+k)} $$
$$=\left(1+\frac{2n+2}{n+1}\right)\left(1+\frac{2n+1}{n+1}\right)\frac{n^{2n}(3n+1)}{(n+1)^{2n}(n+1)} \to 3\cdot 3\cdot \frac{3}{e^2}=\frac{27}{e^2}$$
and therefore (see Limit of ${a_n}^{1/n}$ is equal to $\lim_{n\to\infty} a_{n+1}/a_n$)
$$\lim_{n\to \infty} a_n = \lim_{n\to \infty} \sqrt[n] b_n =\lim_{n\to \infty} \frac{b_{n+1}}{b_n}=\frac{27}{e^2}$$
A: Hint:
Compute the limit of the logarithm, noticing that
$$\log\biggl(\prod_{k+1}^{2n}\Bigl(1+\frac kn\Bigr)^{\tfrac1n}\biggr)=\sum_{k+1}^{2n}\frac1n\log\Bigl(1+\frac kn\Bigr).$$
Looks very much like an upper Riemann sum.
A: We have that
$$\begin{align}
\prod_{k=1}^{2n} \left(1+ \frac{ k}{n} \right)^{1/n}&=
\frac{1}{n^2} \left(\prod_{k=1}^{2n}(n+ k)\right)^{1/n}=\frac{1}{n^2}\left(\frac{(3n)!}{n!}\right)^{1/n}\\&\sim \frac{1}{n^2}\left(\frac{(3n)^{3n}\sqrt{2\pi (3n)}}{e^{3n}}\cdot
\frac{e^{n}}{n^{n}\sqrt{2\pi n}}\right)^{1/n}=\frac{3^3}{e^2}\left(\sqrt{3}\right)^{1/n}\to\frac{27}{e^2}\end{align} $$
where we used the Stirling approximation for the factorials.
