Calculate the limit $\lim_{n\to\infty} \left(\prod_{k=1}^{n}(1+\frac{k}{n})\right)^{\frac{1}{n}}$ I have to calculate this limit $$  \lim_{n\to\infty} \left(\prod_{k=1}^{n}\left(1+\frac{k}{n}\right)\right)^{\frac{1}{n}}$$
I tried it just first to calculate the limit inside the product, but I think I got the answer 1. Any help?

As an alternative, we have that
$$\left(\prod_{k=1}^{n}\left(1+\frac{k}{n}\right)\right)^{\frac{1}{n}}=e^{\frac{\sum_{k=1}^{n} \log\left(1+\frac{k}{n}\right) }{n}}=e^{\sum_{k=1}^{n} \left(\frac{k}{n^2}-\frac12\frac{k^2}{n^3}+\frac13\frac{k^3}{n^4}+\ldots\right) }\to \frac4e$$
indeed by Faulhaber's formula
$$\sum_{k=1}^{n} \left(\frac{k}{n^2}-\frac12\frac{k^2}{n^3}+\frac13\frac{k^3}{n^4}+\ldots\right)=\sum_{k=1}^n \frac{(-1)^{(k+1)}}{k(k+1)}+O\left(\frac1n\right)\to \ln 4-1$$
indeed by alternating harmonic series
$$\sum_{k=1}^n \frac{(-1)^{(k+1)}}{k(k+1)}=\sum_{k=1}^n \frac{(-1)^{(k+1)}}{k}-\sum_{k=1}^n \frac{(-1)^{(k+1)}}{k+1}\to\ln 2-(1-\ln 2)=2\ln 2-1$$
