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Here is an exercise:

Suppose that $\{x_n\}$ is a sequence such that $\lim \limits_{n\to\infty}(x_n-x_{n-2})=0$. Show that:

$$\lim \limits_{n\to\infty}\frac{x_n-x_{n-1}}{n}=0 $$


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There is a limit missing there. – Pedro Tamaroff Jun 12 '13 at 2:04
Thinking that a limit is missing, you can apply the Cesaro-Stolz theorem. – Sangchul Lee Jun 12 '13 at 2:05
How could I apply the Cesaro-Stolz theorem? – Simple Jun 12 '13 at 2:08
See my answer. Telescoping. – Pedro Tamaroff Jun 12 '13 at 2:19
up vote 2 down vote accepted


Let $y_n=|x_n-x_{n-1}|$. Note that $|y_n-y_{n-1}|\le |x_n-x_{n-2}|$. Then $$ |\frac {x_n-x_{n-1}}{n}|=\frac{y_n}{n} \le \frac{|y_n-y_{n-1}|+|y_{n-1}-y_{n-2}|+\dots+|y_{N+1}-y_N|}{n}+\frac{y_N}{n} $$

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Given a sequence $\langle x_n\rangle $ denote $\Delta x_n=x_{n+1}-x_n$.

Let $$a_n=x_n-x_{n-2}$$

Then $a_n\to 0$ and $a_{n+1}\to 0$ so $$\omega_n =a_{n+1}-a_n\to 0$$ Note that $\omega_n=\Delta x_{n}-\Delta x_{n-1}$

By Cesàro,

$$\lim\limits_{n\to \infty}\frac 1n\sum_{k=1}^n\omega_k=0$$

What is the above?

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What is the Cesaro? I cannot find it in the link. – Simple Jun 12 '13 at 2:21
@Simple I am linking to a proof of the following: If $a_n\to \ell$ then $$\frac 1 n\sum_{k=1}^n a_k\to \ell$$ too. This is usually called "Cesàros Theorem" in honor to Ernesto Cesàro – Pedro Tamaroff Jun 12 '13 at 2:22
@Sanchez True. It has some "remainders", but they should be killed off by the $n^{-1}$. – Pedro Tamaroff Jun 12 '13 at 2:30

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