Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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:

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

Thanks.

share|improve this question
    
There is a limit missing there. –  Pedro Tamaroff Jun 12 '13 at 2:04
2  
Thinking that a limit is missing, you can apply the Cesaro-Stolz theorem. –  sos440 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

2 Answers 2

up vote 2 down vote accepted

Hints:

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

share|improve this answer

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?

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.