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?

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    $\begingroup$ To complete the question/answer, can you also include how you get to that inequality? (indeed, the inequality shows that it converges to $f'_d(0)/2$ instead) $\endgroup$ – user99914 Feb 17 '16 at 10:56

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

$$\frac{f'_d(0)}{2} \leq \lim_{n \to \infty} \sum_{k=1}^n f\left( \frac{k}{n^2} \right) \leq \frac{f'_d(0)}{2}$$


$$\lim_{n \to \infty} \sum_{k=1}^n f\left( \frac{k}{n^2} \right) = \frac{f'_d(0)}{2}.$$

So either the original statement is false or you forgot a factor of $2$ somewhere in the derivation of the inequality.

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    $\begingroup$ To be precise, you should first take $\limsup$, $\liminf$ and get $$\frac{f'_d(0)-\varepsilon}{2} \leq \liminf_{n \to \infty} \sum_{k=1}^n f\left( \frac{k}{n^2} \right) \le \limsup_{n \to \infty} \sum_{k=1}^n f\left( \frac{k}{n^2} \right) \leq \frac{f'_d(0)+\varepsilon}{2}$$ and then take $\epsilon \to 0$ to conclude that $\lim$ exists. $\endgroup$ – user99914 Feb 17 '16 at 11:19
  • $\begingroup$ @JohnMa You are correct. I kind of overlooked the fact that we have to take $\varepsilon \to 0$ in order to actually get the equality of the outer limits. $\endgroup$ – GenericNickname Feb 17 '16 at 11:21

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