Correctness of the proof that $\lim_{n \to \infty}(x_{n}-x_{n-2})=0$ implies $\lim_{n \to \infty}\frac{x_{n}}{n}=0$. I would like to prove that $\lim_{n \to \infty}(x_{n}-x_{n-2})=0$ implies $\lim_{n \to \infty}\frac{x_{n}}{n}=0$. My idea is to use Stolz–Cesàro theorem:
If $n$ runs over all even numbers,we have:
\begin{align}
    \lim_{k \to \infty} \frac{x_{2k}-x_{2(k-1)}}{2k-2(k-1)}&=0\\
    \Rightarrow \lim_{k \to \infty} \frac{x_{2k}}{2k}&=0\\
    \Rightarrow \lim_{k \to \infty} \frac{x_{2k}}{k}&=0 \ (1)
\end{align}
If $n$ instead runs over all odd numbers,we have:
\begin{align}
    \Rightarrow \lim_{k \to \infty} \frac{x_{2k-1}}{2k-1}&=0\\
    \Rightarrow \lim_{k \to \infty} \frac{x_{2k-1}}{k}&=0 \ (2)
\end{align}
Combine (1) and (2)
\begin{align}
    \lim_{n \to \infty} \frac{x_{n}}{n}&=0
\end{align}
Thus in both cases the mentioned statement seems to hold. Is my proof correct? If I not, how can we prove it?
I truly appreciate your help!
 A: If $n$ is odd numbers,we have
\begin{align}
    \lim_{k \to \infty} \frac{x_{2k+1}-x_{2k-1}}{2k+1-(2k-1)}=\lim_{k \to \infty} \frac{x_{2k+1}-x_{2k-1}}2&=0\\
\\
    \Rightarrow \lim_{k \to \infty} \frac{x_{2k+1}}{2k+1}&=0
\end{align}
Another method:
We use the following property:
$$\text{If} ~\lim_{n \to \infty} b_n=L\Rightarrow \lim_{n \to \infty} \frac{b_1+\cdots+b_n}n=L$$
Define $a_n=x_n-x_{n-2}$,
For $n=2k, \lim_{k\to\infty} a_{2k}=0,$ we have $$x_{2k}-x_2=a_2+a_4+\cdots+a_{2k}$$
$$\lim_{k\to\infty} \frac{x_{2k}-x_2}k=\lim_{k\to\infty} \frac{a_2+a_4+\cdots+a_{2k}}k=0$$
$$\Rightarrow \lim_{k\to\infty} \frac{x_{2k}}{k}=0\Rightarrow \lim_{k\to\infty} \frac{x_{2k}}{2k}=0$$
Similar proof for $n=2k+1$ case.
A: I will prove that $\displaystyle\lim_{n \to \infty}(x_{n}-x_{n-1})=0\ \implies\ \displaystyle\lim_{n \to \infty}\frac{x_{n}}{n}=0,\ $ and then the proof can be adapted for the question in the OP by separately considering the two subsequences $(x_{2n})_n$ and $(x_{2n-1})_n.$
So suppose $\displaystyle\lim_{n \to \infty}(x_{n}-x_{n-1})=0.$ Let $\varepsilon > 0.$ Then $\ \exists\ N\ $ such that $\ \vert x_{n}-x_{n-1}\vert < \varepsilon\ $ for all $\ n \geq N.$
Now for all $\ k\geq 1,\ \vert x_{N+k} \vert = \vert\ ( x_{N+k} - x_{N+k-1}) + ( x_{N+k-1} - x_{N+k-2}) + \ldots + (x_{N+1} - x_N) + x_N\ \vert $
$$ \leq\ \vert x_{N+k} - x_{N+k-1}\vert\ + \vert x_{N+k-1} - x_{N+k-2}\vert\ + \ldots + \vert x_{N+1} - x_N \vert + \vert  x_N \vert < k\varepsilon + \vert x_N \vert.$$
So for all $k\geq 1,\ \frac{\vert x_{N+k}\vert}{k} < \varepsilon + \frac{\vert x_{N}\vert}{k},\ \to \varepsilon\ $ as $k\to\infty.$ Therefore, $\ \displaystyle\limsup_{k\to\infty} \frac{\vert x_{N+k}\vert}{k} \leq \varepsilon.$
Therefore, $\ \displaystyle\limsup_{k\to\infty} \frac{\vert x_{N+k}\vert}{N+k} = \limsup_{k\to\infty} \frac{\vert x_{N+k}\vert}{k} \frac{k}{N+k} \leq \varepsilon\cdot1=\varepsilon.$
Since $\ \varepsilon\ $ was arbitrary, we have, $\ \displaystyle\limsup_{n\to\infty} \frac{\vert x_{n}\vert}{n} \leq \varepsilon\ $ for all $\ \varepsilon > 0,\ $ which implies $\lim_{n \to \infty}\frac{x_{n}}{n}=0$.
