If $\lim_{n \to \infty} (a_n - a_{n-2})=0$, prove that $\lim_{n \to \infty} \frac{(a_n - a_{n-1})}{n} = 0$ 
If $\lim\limits_{n \to \infty} (a_n - a_{n-2})=0$, prove that $\lim\limits_{n \to \infty} \frac{(a_n - a_{n-1})}{n} = 0$.

I have no idea about solving it rigorously. I feel that $\lim\limits_{n \to \infty} (a_n - a_{n-1})=0$ holds true as well if all terms are positive or negative. 
 A: Denote, for $n\geq 0$,
$$
\delta_n \stackrel{\rm def}{=} a_{n+2}-a_n
$$
so that by assumption $\lim_{n\to\infty} \delta_n =0$. We have, for all $n\geq 0$,
$$
a_{2n} = a_0 + \sum_{k=0}^{n-1} (a_{2k+2}-a_{2k})
= a_0 + \sum_{k=0}^{n-1} \delta_{2k}\,.
$$
Similarly,
$$
a_{2n-1} = a_1 + \sum_{k=0}^{n-2} (a_{2k+3}-a_{2k+1})
= a_1 + \sum_{k=0}^{n-2} \delta_{2k+1}\,.
$$
Therefore, for $n\geq1$,
$$
\frac{a_{2n}-a_{2n-1}}{2n}
= \frac{a_0-a_1}{2n} + \frac{\sum_{k=0}^{n-1}\delta_{2k}-\sum_{k=0}^{n-2}\delta_{2k+1}}{2n}
= \frac{a_0-a_1}{2n} + \frac{\delta_{2n-2}}{2n} + \frac{1}{2n}\sum_{k=0}^{n-2}(\delta_{2k}-\delta_{2k+1})
$$
The first two terms obviously converge to $0$ (can you see why?) so it remains to show the third does as well. This in turn follows from Cesaro's lemma, as 
$$
\lim_{n\to\infty}(\delta_{2n}-\delta_{2n+1}) = 0-0 = 0.
$$
To summarize: we just showed
$$\boxed{
\lim_{n\to\infty}\frac{a_{2n}-a_{2n-1}}{2n} = 0
}$$
Now, you can easily conclude the same for
$
\lim_{n\to\infty}\frac{a_{2n+1}-a_{2n}}{2n+1}
$, 
as $a_{2n+1}-a_{2n} = \delta_{2n-1}+a_{2n-1}-a_{2n}$.
A: let's denote $a_{n-1}-a_{n}$ by $x_n$
Now given condition states that for any given $\epsilon>0$
|$a_{n}-a_{n-2}$|$<\epsilon$ for all n>N$\in \mathbb{N}$
|$a_{n}-a_{n-2}$|$\geq$||$a_n-a_{n-1}|-|a_{n-1}-a_{n-2}||$
which means $\lim_{n\rightarrow \infty }(|x_n|-|x_{n-1}|)=0$
by stolz ceasaro theorem lim $\frac{|x_n|}{n}$=0
I hope you will stich together these peices 
A: Note that
$$0\le||a_n-a_{n-1}|-|a_{n-1}-a_{n-2}||\le|a_n-a_{n-2}|$$
and also
$$\lim_{n\to\infty}x_n=0\iff\lim_{n\to\infty}|x_n|=0\tag{1}$$
Hence, by applying the Squeeze theorem and $(1)$, we get that
$$\lim_{n\to\infty}\left(|a_n-a_{n-1}|-|a_{n-1}-a_{n-2}|\right)=\lim_{n\to\infty}\frac{|a_n-a_{n-1}|-|a_{n-1}-a_{n-2}|}{n-(n-1)}=0$$
We can then apply the Stolz–Cesàro theorem and $(1)$ to the above limit which directly gives
$$\lim_{n\to\infty}\frac{a_n-a_{n-1}}{n}=0$$
