Compute $\lim_{n\to \infty }\frac{1}{n}\sum_{k=1}^n\left(1+\frac{k}{n^2}\right)^n$

Question

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.

Answer 1

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.