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.

  • $\begingroup$ 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
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    $\begingroup$ 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

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 $n$.


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