How do I evaluate $\lim_{n \rightarrow \infty}\sum_{k=1}^{n}\log\left ( 1+\frac{k}{n^{2}} \right )$? My attempt :
$$\begin{align} S &=  \sum_{k=1}^{n} \log\left(1+\frac{k}{n^2}\right)\\ 
&= \log\left(\prod_{k=1}^{n} 1 + \frac{k}{n^2}\right) \\
&= \log\left(\prod_{k=1}^{n} \frac{n^2 + k}{n^2}\right)\\
&= \log\left(\frac{1}{n^2}\prod_{k=1}^{n} {n^2 + k}\right) \end{align}$$
How can I take it from here? 
Any Suggestions?

Hint. An elementary way is to use the inequality
$$
x-\frac{x^2}2 \le \ln(1+x)\le x,\qquad0\le x \le1,
$$ giving
$$
\frac{k}{n^2}-\frac{k^2}{2n^4} \le \ln \left( 1 +\frac{k}{n^2}\right)\le \frac{k}{n^2},\qquad0\le k \le n,
$$ then one may conclude with
$$
\sum_{k=1}^nk=\frac{n(n+1)}2
$$ and the squeeze theorem.
