How to evaluate $\lim\limits_{n\to+\infty} \prod\limits_{k=1}^n (1+k/n^2)$? I've got a limit which puzzle me several days. The question is
$$ \lim_{n\to+\infty} \prod_{k=1}^n\left(1+\frac{k}{n^2}\right).$$
Can you help me? Thank you in advance
 A: As an alternative to @sos440's nice approach, note that $\mathrm e^{x-x^2}\leqslant1+x\leqslant\mathrm e^{x}$ for every $x$ in $[0,1]$. Hence the $n$th product $P_n$ is such that
$S_n-T_n\leqslant\log(P_n)\leqslant S_n$, with
$$
S_n=\sum_{k=1}^n\frac{k}{n^2}=\frac1n\sum_{k=1}^n\frac{k}{n},\qquad T_n=\sum_{k=1}^n\left(\frac{k}{n^2}\right)^2=\frac1{n^2}\sum_{k=1}^n\left(\frac{k}{n}\right)^2.
$$
At this point, either one knows by heart the sum of the $n$ first integers and the sum of the $n$ first squares of integers, or one recognizes $S_n$ as a Riemann sum of the function $x\mapsto x$ on $[0,1]$, whose integral is $\frac12$, and $nT_n$ as a Riemann sum of the function $x\mapsto x^2$ on $[0,1]$. Either way, $S_n\to\frac12$ and $T_n\to0$, hence $\log P_n\to\frac12$ and $P_n\to\sqrt{\mathrm e}$.
A: Intuitively, we have 
$$\log\left( 1 + \frac{k}{n^2} \right) = \frac{k}{n^2} + O\left(\frac{1}{n^2}\right) \quad \Longrightarrow \quad \log \prod_{k=1}^{n} \left( 1 + \frac{k}{n^2} \right) = \frac{1}{2} + O\left(\frac{1}{n}\right)$$
and therefore the log-limit is $\frac{1}{2}$.
Here is a more elementary approach: Let $P_n$ denote the sequence inside the limit. Then just note that
$$ P_n^2 = \left[ \prod_{k=1}^{n} \left( 1 + \frac{k}{n^2} \right) \right]^2 = \prod_{k=1}^{n} \left( 1 + \frac{k}{n^2} \right)\left( 1 + \frac{n-k}{n^2} \right) = \prod_{k=1}^{n} \left( 1 + \frac{1}{n}+\frac{k(n-k)}{n^4} \right). $$
Now fix $m$ and let $n \geq m$. Since $k (n-k) \leq \frac{1}{4}n^2$, we have
$$ \frac{k(n-k)}{n^4} \leq \frac{1}{4n^2} \leq \frac{1}{4mn}.$$
Thus we have
$$ \left( 1 + \frac{1}{n} \right)^n \leq P_n^2 \leq \left( 1 + \frac{1+(1/4m)}{n} \right)^n. $$
Thus taking $n \to \infty$,
$$e \leq \liminf_{n\to\infty} P_n^2 \leq \limsup_{n\to\infty} P_n^2 \leq e^{1+1/(4m)}.$$
Since $m$ is now arbitrary, we have $P_n^2 \to e$, or equivalently, $P_n \to \sqrt{e}$.
