Limit of a product $(1 + \frac{k}{n})$ I am trying to find the limit of
$$\displaystyle \lim_{n\to\infty}\left(\prod_{k=1}^n\left(1+\frac{k}{n}\right)\right)^{1/n}$$
I have that it is equivalent to the
$$\lim_{n\to\infty}\frac{1}{n}\left(\sum_{k=1}^n\ln\left(1+\frac{k}{n}\right)\right)$$
I can't get any further than this. Any help would be much appreciated.
 A: Let $$\displaystyle L = \lim_{n\to\infty}\left(\prod_{k=1}^n\left(1+\frac{k}{n}\right)\right)^{1/n} = \lim_{n\rightarrow \infty}\left[\left(\frac{n+1}{n}\right)\cdot \left(\frac{n+2}{n}\right)....\left(\frac{n+n}{n}\right)\right]^{\frac{1}{n}}$$
So we get $$\displaystyle L = \lim_{n\to\infty}\left[\frac{(2n)!}{n^n\cdot n!}\right]^{\frac{1}{n}}$$
Now Using Stirling approximation $$\displaystyle n! = \left(\frac{n}{e}\right)^{n}\cdot \sqrt{2\pi n}$$
So we get $$\displaystyle L = \lim_{n\to\infty}\left[\frac{\left(\frac{2n}{e}\right)^{2n}\cdot \sqrt{4\pi n}}{n^n\cdot \left(\frac{n}{e}\right)^{n}\cdot \sqrt{2\pi n}}\right]^{\frac{1}{n}}=\lim_{n\rightarrow \infty}\left[\frac{\frac{4n^2}{e^2}\times \sqrt{2}}{\frac{n^2}{e}}\right]^{\frac{1}{n}}$$
So we get $$\displaystyle L = \frac{4}{e}\cdot \lim_{n\rightarrow \infty}(2)^{\frac{1}{2n}}= \frac{4}{e}.$$
A: Note that $$ \sum_{k\leq n}\log\left(1+\frac{k}{n}\right)=\sum_{k\leq n}\log\left(k+n\right)-n\log\left(n\right)
 $$ and using Abel's summation formula we have $$\sum_{k\leq n}\log\left(k+n\right)=\sum_{n<k\leq2n}\log\left(k\right)=2n\log\left(2n\right)-n\log\left(n\right)-\int_{n}^{2n}\frac{\left\lfloor t\right\rfloor }{t}dt
 $$ where $\left\lfloor t\right\rfloor 
 $ is the floor function. So, using $\left\lfloor t\right\rfloor =t+O\left(1\right)
 $ we get $$=n\log\left(n\right)+2n\log\left(2\right)-n+O\left(\log\left(n\right)\right)
 $$ hence $$\frac{1}{n}\sum_{k\leq n}\log\left(1+\frac{k}{n}\right)=\frac{1}{n}\left(2n\log\left(2\right)-n+O\left(\log\left(n\right)\right)\right)\rightarrow2\log\left(2\right)-1.
 $$ 
A: You have a Riemann sum:
$$ \lim_{n\to +\infty}\frac{1}{n}\sum_{k=1}^{n}\log\left(1+\frac{k}{n}\right)=\int_{0}^{1}\log(1+x)\,dx = \color{red}{-1+2\log 2},$$
hence the original limit equals $\large\color{red}{\frac{4}{e}}$.
