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
 A: Hint, based on Surb:
$$
\log L = \lim_{n\to\infty} \sum_{k=1}^n\frac{1}{k}\log\left(1+\frac{k}{n}\right)
=\frac{1}{n}\sum_{k=1}^n\frac{n}{k}\log\left(1+\frac{k}{n}\right)
=\int_0^1\frac{\log(1+x)}{x}\;dx
$$
by a Riemann sum argument.
A: 
I thought that it would be instructive to present an approach that does not rely on Riemann Sums, but rather makes use of the Taylor Series of $\log(1+x)$.    To that end, we proceed.


The function $\log(1+x)$ can be represented by its Taylor Series, $\log(1+x)=\sum_{\ell=1}\frac{(-1)^{\ell-1}}{\ell}x^\ell$ for $-1<x\le 1$.  Using this representation, we can write
$$\begin{align}
\sum_{k=1}^n\log\left(1+\frac kn\right)^{1/k}&=\sum_{k=1}^n\frac1k\log\left(1+\frac kn\right)\\\\
&=\sum_{k=1}^n\left(\frac1k \sum_{\ell=1}^\infty \frac{(-1)^{\ell-1}}{\ell}\left(\frac kn\right)^\ell\right)\\\\
&=\sum_{\ell=1}^\infty \frac{(-1)^{\ell-1}}{\ell n^\ell}\sum_{k=1}^nk^{\ell-1}\tag1
\end{align}$$

Next, noting that $\displaystyle \sum_{k=1}^n k^{\ell-1}=\frac{n^\ell}{\ell}+O\left(n^{\ell -1}\right)$, we have from $(1)$ that
$$\begin{align}
\sum_{k=1}^n\log\left(1+\frac kn\right)^{1/k}&=\sum_{\ell=1}^\infty \frac{(-1)^{\ell-1}}{\ell^2 }+O\left(\frac1n\right)\tag2
\end{align}$$

Finally, letting $n\to \infty$ in $(2)$ yields the result
$$\bbox[5px,border:2px solid #C0A000]{\sum_{k=1}^\infty \log\left(1+\frac kn\right)^{1/k}=\sum_{k=1}^\infty \frac{(-1)^{k=1}}{k^2}}\tag3$$


The series on the left-hand side of $(3)$ is equal to $\frac{\pi^2}{12}$ (See This).

A: 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 $\lim_{x\to 0^+}\frac{\ln(1+x)}{x}=1$.
