1
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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
3
$\begingroup$

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.

$\endgroup$
  • 3
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.