TITLE: How to evaluate $\lim\limits_{n\to+\infty} \prod\limits_{k=1}^n (1+k/n^2)$?
QUESTION [6 upvotes]: 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

REPLY [3 votes]: 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 $P_n\to\sqrt{\mathrm e}$.