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.
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 $n$.