Evaluating the $ \lim_{n \to \infty} \prod_{1\leq k \leq n} (1+\frac{k}{n})^{1/k}$ I am really struggling to work out the limit of the following product:
$$ \lim_{n \to \infty} \prod_{1\leq k \leq n} \left (1+\frac{k}{n} \right)^{1/k}.$$
So far, I have spent most of my time looking at the log of the above expression. If we set the desired limit equal to $L$, I end up with:
$$\log L = \lim_{n\to \infty}\log\left(\frac{n+1}{n} \right)+\frac{1}{2}\log\left(\frac{n+2}{n} \right) +\cdots +\frac{1}{n}\log\left(\frac{n+n}{n} \right),$$
which I can simplify to:
$$ \log L = \lim_{n\to \infty} \log(n+1)+\frac{1}{2}\log(n+2)+\cdots \frac{1}{n}\log(2n)-\log(n)\left(1+\frac{1}{2}+\cdots\frac{1}{n}\right). $$
I tried to consider the above expression in a different form with an integral, but was unable to arrive at anything useful.
I have been stuck on this for quite awhile now, and would appreciate any insight.
Thanks

hint
Taking logarithm, the product becomes
$$\frac 1n\sum_{k=1}^n\frac nk\ln(1+\frac kn)=$$
$$\frac 1n\sum_{k=1}^n\frac{\ln(1+\frac kn)}{\frac kn}$$
As a Riemann sum, the limit will be
$$\int_0^1\frac{\ln(1+x)}{x}dx$$
This improper integral is convergent since $\lim_{x\to 0^+}\frac{\ln(1+x)}{x}=1$.
