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?
 A: 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.
A: We have $$\prod_{k=1}^{n}\left(1+\frac{k}{n^{2}}\right)=\frac{1}{n^{2n}}\prod_{k=1}^{n}\left(n^{2}+k\right)=\frac{\left(n^{2}+1\right)_{n}}{n^{2n}}$$ where $\left(x\right)_{m}$ is the Pochhammer symbol and since $$\left(x\right)_{m}=\frac{\Gamma\left(x+m\right)}{\Gamma\left(x\right)}$$ we have $$\prod_{k=1}^{n}\left(1+\frac{k}{n^{2}}\right)=\frac{\Gamma\left(n^{2}+n+1\right)}{n^{2n}\Gamma\left(n^{2}+1\right)}\rightarrow\color{red}{\sqrt{e}}$$ as $n\rightarrow\infty$, by Stirling's approximation.
A: Hint. Note that for $x>0$,
$$x-\frac{x^2}{2} <\log(1+x)<x.$$
A: Multiply n square on both sides then put it in the log,u will get a limit for e,after simplifying everything you should be getting half
