Prove that $ \lim\limits_{n \to \infty } \sum\limits_{k=1}^n f \left( \frac{k}{n^2} \right) = \frac 12 f'_d(0). $ Let $I \subset \mathbb{R}$ be an open interval with $0 \in I$ and $f:I \to \mathbb{R}$ a continuous function, with $f(0)=0$, right-differentiable in $0$. Then:
$$ \lim\limits_{n \to \infty } \sum\limits_{k=1}^n f \left( \frac{k}{n^2} \right) =\frac 12 f'_d(0). $$
Well, I got that, $\forall \varepsilon >0, \exists n_\varepsilon \in \mathbb{N}$ such that:
$$ \frac{f'_d(0)-\varepsilon}{2} \cdot \frac{n(n+1)}{n^2} <\sum\limits_{k=1}^n f \left( \frac{k}{n^2} \right) < \frac{f'_d(0)+\varepsilon}{2} \cdot \frac{n(n+1)}{n^2}, \forall n \ge n_\varepsilon. $$
How can I continue from here?
 A: You can use the following. 
If $a_n < b_n$ for all $n$ and $\lim a_n$, $\lim b_n$ exist, then 
$$\lim a_n \leq \lim b_n$$
This also holds for $\liminf$ and $\limsup$.
In your case, since $\frac{n(n+1)}{n} \to 1$, the inequality you wrote down implies (by applying $\lim_{n \to \infty}$ to all three terms) that
$$\frac{f'_d(0)-\varepsilon}{2} \leq \liminf_{n \to \infty} \sum_{k=1}^n f\left( \frac{k}{n^2} \right) \leq \limsup_{n \to \infty} \sum_{k=1}^n f\left( \frac{k}{n^2} \right)\leq \frac{f'_d(0)+\varepsilon}{2}$$
for all $\varepsilon > 0$ (the second inequality is true because $\liminf a_n \leq \limsup a_n$ for all sequences $(a_n)$). 
Hence, it follows (by taking $\varepsilon \to 0$) that
$$\frac{f'_d(0)}{2} \leq \lim_{n \to \infty} \sum_{k=1}^n f\left( \frac{k}{n^2} \right) \leq \frac{f'_d(0)}{2}$$
i.e.
$$\lim_{n \to \infty} \sum_{k=1}^n f\left( \frac{k}{n^2} \right) = \frac{f'_d(0)}{2}.$$
So either the original statement is false or you forgot a factor of $2$ somewhere in the derivation of the inequality.
