TITLE: Evaluating $\lim\limits_{n\to \infty}\left\{(1+\frac{1}{n})(1+\frac{2}{n})\dots(1+\frac{n}{n})\right\}^{\frac{1}{n}}$
QUESTION [5 upvotes]: Question is to evaluate :
$$\lim_{n\to \infty}\left\{ \left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)\dots\left(1+\frac{n}{n}\right)\right\}^{\frac{1}{n}}$$
I tried to do something like this but it is not getting better...
$(1+\frac{1}{n})(1+\frac{2}{n})\cdots(1+\frac{n}{n})=(\frac{n+1}{n})(\frac{n+2}{n})\cdots (\frac{n+n}{n})=\frac{1}{n^n}(n+1)(n+2)\dots(n+n)$
please give some hint to proceed further.

REPLY [9 votes]: $$(\frac{n+1}{n})(\frac{n+2}{n})\dots (\frac{n+n}{n})=\frac{1}{n^n}(n+1)(n+2)\dots(n+n)$$
$$=\lim_{n\rightarrow \infty}\left[\frac{1}{n^n}\frac{(2n)!}{n!}\right]^{1/n}=\lim_{n\rightarrow \infty}\left[\frac{\sqrt{4\pi n}\cdot 2^{2n}\frac{n^{2n}}{e^{2n}}}{\sqrt{2\pi n}\cdot \frac{n^{2n}}{e^n}}\right]^{1/n}=\lim_{n\rightarrow \infty }[\sqrt{2}]^{1/n}\cdot \frac{4}{e}=\frac{4}{e}$$
Here I've used the Stirling approximation of $n!$,i.e- $n!=\sqrt{2\pi n}\left(\frac{n}{e}\right)^{n}$