If the sequence satisfies the property 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 the sequence satisfies the property lim$_{n\to \infty}(a_n-a_{n-2})=0$, prove that lim$_{n\to \infty}\frac{a_n-a_{n-1}}{n}=0$.
I am experiencing difficulty showing this rather obvious result.
By definition, we know that for any $\epsilon \gt 0$, there is some $N$ such that if $n \gt N$ then $|a_n-a_{n-2}|\lt \epsilon$ My guess is that I need to write $|a_n-a_{n-1}|$ as sums of $|a_n-a_{n-2}|$ from $n$ to $N$. However, I am stuck with forming an inequality here. I would appreciate any solutions or suggestions.
 A: Let $b_k:=a_{2k}-a_{2k-2}$. Then $\sum_{i=1}^kb_i=a_{2k}-a_0$. Since $b_k\rightarrow 0$, you get $\frac{a_{2k}}{k}\rightarrow 0$ by Cesaro limit. You get the same result for the odd limit as well. The result follows.
A: Note that $a_{n}-a_{n-1}=a_{n}-a_{n-2}-(a_{n-1}-a_{n-2})=(a_{n}-a_{n-2})-(a_{n-1}-a_{n-3})+(a_{n-2}-a_{n-4})-\cdots$ so on until you get some $N$ such that for some $\epsilon>0$, $\forall n\ge N, |a_{n}-a_{n-2}|<\epsilon$. Then, $$|a_{n}-a_{n-1}|<(n-N+1)\epsilon\ \forall n\ge N\implies \left|\frac{a_{n}-a_{n-1}}{n}\right|<\left(1-\frac{N-1}{n}\right)\epsilon<\epsilon$$ since $n\ge N$  hence the result follows.
A: Say $x_n=\dfrac{a_n - a_{n-1}}{n}$, for $n\ge 1$ and $y_n=a_n -a_{n-2}$, for $n\ge 2$. Let $y_1=x_1$. Then $n x_n=a_n-a_{n-1}$ and $(n+1)x_{n+1}=a_{n+1}-a_n$. Adding these two, $nx_n+(n+1)x_{n+1}=a_{n+1}-a_{n-1}=y_{n+1}$ for $n\ge 1$. Inductively we have $x_n=\dfrac{b_n-b_{n-1}+\dotsm +{(-1)}^{n-1} b_1}{n}$, for each $n\ge 1$. Choose $\varepsilon >0$. Since $b_n\to 0$, there is a positive integer $n_0$ such that $|b_n|<\dfrac{\varepsilon}{2}$ for $n> n_0$. Choose $a=|b_1|+\dotsm+|b_{n_0}|$. Then there is a positive integer $m_0$ such that $m_0\dfrac{\varepsilon}{2}> a$ i.e. $\dfrac{a}{m_{0}}<\dfrac{\varepsilon}{2}$. Let $n_1>\max\{n_0,m_0\}$. Then for $n>n_1$, we have 
\begin{eqnarray}
|x_n|& \le & \dfrac{1}{n}\left(|b_1|+\dotsm+|b_{n_0}|+|b_{n_0+1}|+\dotsm +|b_n|\right)\\
&=&\dfrac{a}{n}+\dfrac{1}{n}\left(|b_{n_0+1}|+\dotsm +|b_n|\right)\\
&<&\dfrac{a}{m_0}+\dfrac{(n-n_0+1)}{n}\dfrac{\varepsilon}{2}\\
&<& \dfrac{\varepsilon}{2}+\dfrac{\varepsilon}{2}=\varepsilon,
\end{eqnarray} as $\dfrac{(n-n_0+1)}{n}<\dfrac{1}{n}\le 1$. Hence $x_n\to 0$ i.e. $\lim\limits_{n\to\infty}\dfrac{a_n-a_{n-1}}{n}=0$ as required.
