$X_n - X_{n-2}\rightarrow 0$, prove that$ \frac{X_n}{n} \rightarrow 0$. suppose $X_{n}$ is a sequence of real numbers such that $X_{n}-X_{n-2}\rightarrow 0$.
prove that $\frac{X_{n}}{n}\rightarrow 0$.
 A: Let $\{a_n\}_{n\in \mathbb{N}-\{0\}}$ be the sequence defined as
$a_n=X_{2n}-X_{2n-2}$ for all $n\in \mathbb{N}-\{0\}$.
It results that
$a_1+a_2+a_3+...+a_n=$
$=X_2-X_0+X_4-X_2+X_6-X_4+...+X_{2n}-X_{2n-2}=$
$=X_{2n}-X_0$
for all $n\in\mathbb{N}-\{0\}$.
By applying Cesaro Mean Theorem we get
$$\lim_{n\rightarrow\infty} \frac{a_1+a_2+a_3+...+a_n}{n}=\lim_{n\rightarrow\infty} a_n$$
that is
$$\lim_{n\rightarrow\infty}\frac{X_{2n}-X_0}{n}=\lim_{n\rightarrow\infty}\left(X_{2n}-X_{2n-2}\right)=0$$
hence
$$\lim_{n\rightarrow\infty}\frac{X_{2n}}{2n}=\lim_{n\rightarrow\infty}\frac{1}{2}\left(\frac{X_{2n}-X_0}{n}+\frac{X_0}{n}\right)=0.\;\;\;\;\;(*)$$
Let $\{b_n\}_{n\in \mathbb{N}-\{0\}}$ be the sequence defined as
$b_n=X_{2n+1}-X_{2n-1}$ for all $n\in \mathbb{N}-\{0\}$.
It results that
$b_1+b_2+b_3+...+b_n=$
$=X_3-X_1+X_5-X_3+X_7-X_5+...+X_{2n+1}-X_{2n-1}=$
$=X_{2n+1}-X_1$
for all $n\in\mathbb{N}-\{0\}$.
By applying Cesaro Mean Theorem we get
$$\lim_{n\rightarrow\infty} \frac{b_1+b_2+b_3+...+b_n}{n}=\lim_{n\rightarrow\infty} b_n$$
that is
$$\lim_{n\rightarrow\infty}\frac{X_{2n+1}-X_1}{n}=\lim_{n\rightarrow\infty}\left(X_{2n+1}-X_{2n-1}\right)=0$$
hence
$$\lim_{n\rightarrow\infty}\frac{X_{2n+1}}{2n+1}=\lim_{n\rightarrow\infty}\frac{n}{2n+1}\left(\frac{X_{2n+1}-X_1}{n}+\frac{X_1}{n}\right)=0.\;\;\;(**)$$
From $(*)$ and $(**)$ it follows that
$$\lim_{n\rightarrow\infty}\frac{X_n}{n}=0.$$
