TITLE: Compute $\lim_{n\to \infty }\frac{1}{n}\sum_{k=1}^n\left(1+\frac{k}{n^2}\right)^n$
QUESTION [5 upvotes]: I want to compute $$\lim_{n\to \infty }\frac{1}{n}\sum_{k=1}^n\left(1+\frac{k}{n^2}\right)^n.$$
I really tried several thing, but this $\frac{1}{n^2}$ annoy me very much. It looks like a Riemann sum, but I can't conclude without more information.

REPLY [3 votes]: Hint
$$\left(1+\frac{k}{n^2}\right)^n=\exp\left\{n\ln\left(1+\frac{k}{n^2}\right)\right\}.$$
One can prove that $$x-\frac{x^2}{2}\leq \ln(1+x)\leq x.$$
Therefore, for all $k\in\{1,...,n\}$,
$$\frac{k}{n}-\frac{1}{2n}\leq n\ln\left(1+\frac{k}{n^2}\right)\leq \frac{k}{n}.$$
The claim follow by composing and summing each side.