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

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.

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There is a limit missing there. –  Pedro Tamaroff Jun 12 at 2:04
Thinking that a limit is missing, you can apply the Cesaro-Stolz theorem. –  sos440 Jun 12 at 2:05
How could I apply the Cesaro-Stolz theorem? –  Simple Jun 12 at 2:08
See my answer. Telescoping. –  Pedro Tamaroff Jun 12 at 2:19

Hints:

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

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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 $\omega_n=\Delta x_{n}-\Delta x_{n-1}$

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

What is the above?

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What is the Cesaro? I cannot find it in the link. –  Simple Jun 12 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 at 2:22
@Sanchez True. It has some "remainders", but they should be killed off by the $n^{-1}$. –  Pedro Tamaroff Jun 12 at 2:30
Using the given limit you can show: $$\lim_{n \to \infty} (x_n - x_{n - 2m}) = \lim_{n \to \infty} ((x_n - x_{n - 2}) + (x_{n - 2} - x_{n - 4}) + \dotsb + (x_{n - 2(m - 1)} - x_{n - 2m})) = \lim_{n \to \infty} (x_n - x_{n - 2}) + \lim_{n \to \infty} (x_{n - 2} - x_{n - 4}) + \dotsb + \lim_{n \to \infty} (x_{n - 2(m - 1)} - x_{n - 2m}) = 0 \\$$ for any $n$ and $m$ such that $1 \leq n - 2m \leq n$. Now consider the two subsequences $\{x_{2n}\}$ and $\{x_{2n + 1}\}$ (subsequences formed by taking the terms with even subscript and the terms with odd subscript). These sequences converge since the limit proven above show that these are Cauchy sequences. Thus, these two sequences are bounded. So the original sequence is also bounded. Hence, $x_n - x_{n - 1}$ is bounded: say $|x_n - x_{n - 1}| < B$ (where B is some positive constant). So clearly $\frac{x_n - x_{n - 1}}{n}$ converges absolutely to $0$.
To show it is Cauchy, you need some uniformity: given $\epsilon > 0$, you need $x_n - x_{n'}$ to be less than $\epsilon$ for all sufficiently large $n,n'$. What you've done is that when $n-n'$ is a fixed number, this is bounded, but there is no uniformity (with respect to $m$). In fact, this is false - $x_n- x_{n-1}$ may not be bounded. Take for example $x_{even} = 0$, and $x_{odd \, n} = 1 + \cdots + 1/n$. –  Sanchez Jun 12 at 19:11