Limit of a sequence

Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be a function such that $f(0)=0$ and $f$ is differentiable at $0$ . Let us set

$a_n =\sum_{k=1}^{n} f \left( \frac{k}{n^2} \right)$

Evaluate the limit $\lim \limits_{n \rightarrow +\infty} a_n$ .

Solution

Since $f$ is differentiable at $0$ , there is some $\varepsilon: x \mapsto \varepsilon(x)$ such that

$f(x) = f(0) + x f'(0) + x \varepsilon(x) \quad , \quad \varepsilon(0) =0$

and $\varepsilon$ is of course continuous.

Thus,

$\sum_{k=1}^{n} f\left ( \frac{k}{n^2} \right ) = \frac{f'(0)}{n} \sum_{k=1}^{n} \frac{k}{n} + \sum_{k=1}^{n} \frac{k}{n^2} \varepsilon \left ( \frac{k}{n^2} \right )$

Let $\epsilon>0$ . There exists $\delta>0$ such that $|x| \leq \delta$ which in return means that $|\varepsilon(x)| \leq \epsilon$ . Hence , for $n$ larger than $\frac{1}{\delta}+1$ it holds that