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.

Solution 1:

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.

Solution 2:

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.