<s>[INST] 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?
 [/INST] 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 $\lim_{n\to\infty} a_{n+1}/a_n$)
$$\lim_{n\to \infty} a_n = \lim_{n\to \infty} \sqrt[n] b_n =\lim_{n\to \infty} \frac{b_{n+1}}{b_n}=\frac{27}{e^2}$$
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