0
$\begingroup$

I have previously asked a question to prove the converse of the Cesaro theorem under the assumption that $u_{n+1}-u_n=o(\frac{1}{n})$ . This time I have to do it under the assumption that $u_{n+1}-u_n=O(\frac{1}{n})$. A method is "forced upon" us, and I am stuck at a certain step.


Let $|u_{n+1}-u_n|\leq\frac{M}{n}$ for all $n$ and $v_n=\frac{1}{n}\sum\limits_{k=1}^nu_k$. We know that $v_n$ converges to $a\in\mathbb{R}$.

I have already managed to prove that for all $m<n$ we have the following:

$$|u_n-v_n| \leq \frac{m}{m-n}|v_n-v_m|+\frac{1}{n-m}\sum\limits_{k=m+1}^n(u_n-u_k)$$

The point I am stuck on is the one where we are asked to prove that, for big enough $n$, there exists $m<n$ such that for all $\epsilon>0$:

$$\begin{cases} \frac{n-m}{m} \leq \frac{\epsilon}{2M} \\ \frac{m}{n-m} \leq \frac{2M}{\epsilon}+1 \\ |v_n-v_m| \leq \frac{\epsilon}{2(\frac{2M}{\epsilon}+1)} \end{cases}$$


What I've done so far is set $x=\frac{2M}{\epsilon}$ for convenience and translated the first two conditions into:

$$ \frac{x}{x+1} \leq \frac{m}{n} \leq \frac{1+x}{2+x}$$

What I've then tried doing is setting for all n big enough that such an interval contains at least one integer (which does happen since the difference between both sides is strictly positive), $k_n = \lfloor \frac{n(x+1)}{x+2} \rfloor$. $k_n$ is thus the biggest $m$ you can pick, and therefore the one most likely to satisfy the third inequality ($v_n$ converges therefore $v_n-v_m$ converges to $0$ as $m$ goes to $n$).

I'm trying to show that $|v_n-v_{k_n}|$ gets arbitrarily close to 0 as $n\to\infty$, but to no success. I am pretty positive it is the right strategy here given the method I am made to use.

$\endgroup$
  • $\begingroup$ Note that once that these three inequalities are verified, the inequality I already proved immediately immediately solves the problem (it shows $u_n$ converges to $a$). $\endgroup$ – John Do Sep 14 '17 at 17:10
  • 1
    $\begingroup$ Related: math.stackexchange.com/questions/1106820/… $\endgroup$ – Gabriel Romon Sep 15 '17 at 10:13
  • $\begingroup$ @GabrielRomon that pretty much is the same exact problem and outline. Using my notations, that would mean that the correct $k_n$ would be $\lfloor\frac{n-\epsilon}{1+\epsilon})\pm 1\rfloor$ (with the $\pm 1$ being due to his m being one less than mine and some other blurry distinctions). Moreover, the reason why I get some $2M$ in the inequalities is that my problem is calibrated so you get exactly $\leq\epsilon$ in the end whereas he gets $\epsilon M$. I'll get to work on the answers see if I can find anything useful. If I do, it might be a duplicate question... $\endgroup$ – John Do Sep 15 '17 at 12:06
0
$\begingroup$

The answer, from what I've done so far in the question is rather trivial.

As $n\to\infty$, we have $k_n\to\infty$ (due to $k_n\geq\frac{n(x+1)}{x-1}-1$) and therefore $v_{k_n}\to a$.

You therefore have $|v_n-v_{k_n}|\to0$ and thus can be made, with big enough $n$, smaller than any arbitrary real above $0$: specifically, smaller than $\frac{\epsilon}{2(\frac{2M}{\epsilon}+1)}>0$.

Therefore, there exists n such that such an $m$ can be found, $k_n$ being an example of such a value for big enough $n$.

$\endgroup$

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.