# Prove that $\liminf x_n \le \liminf a_n \le \limsup a_n \le \limsup x_n$

Assume that $\{x_n\}$ is a sequence of real numbers and $a_n=\frac{x_1+\dots+x_n}{n}$ .

a) Prove that $\displaystyle \liminf_{n \to\infty} x_n \le \liminf_{n \to\infty} a_n \le \limsup_{n \to\infty} a_n \le \limsup_{n \rightarrow \infty} x_n$.

b) Give an example such that all of the limits written above are finite and $\displaystyle \liminf_{n \to\infty} x_n < \liminf_{n \to\infty} a_n < \limsup_{n \to\infty} a_n < \limsup_{n \rightarrow \infty} x_n$.

c) Give an example such that some of the limits written above are finite and some of them are not.

Note 1 : For a sequence like $\{b_n\}$ we have $\displaystyle \liminf_{n \to\infty} b_n = \lim_{n\to\infty}(\inf\{b_k:k \ge n\})$ and $\displaystyle \limsup_{n \rightarrow \infty} b_n=\lim_{n\to\infty}(\sup\{b_k:k \ge n\})$

Note 2 : This question is adopted from the book "Real analysis : A first course" written by "Russel Gordon".

Note 3 : A small part of this question is available on this link but my question has a lot more than that.

• You should use \lim, \sup, \inf, \liminf and \limsup. Otherwise, this is very unpleasant to read. – tomasz Nov 4 '16 at 13:56
• @tomasz i didn't know that those things exist in latex ... pardon me :) anyway, thanks to Masacroso, it's fixed now – Maryam Seraj Nov 4 '16 at 13:58
• When something does not exist in latex, you can make it. For example, if you did not know \sin exists to write $\sin(x)$, you could still use \operatorname{sin}(x) to obtain $\operatorname{sin}(x)$. – tomasz Nov 4 '16 at 14:09
• Anyway, what exactly do you want us to help with? What problem are you having? – tomasz Nov 4 '16 at 14:11
• @tomasz thanks ! that's great : – Maryam Seraj Nov 4 '16 at 14:12

First assume that $(x_n)$ is a bounded squence.

Let $L=\limsup_{n\to\infty}x_n<\infty$. By definition of $\limsup$, there exists $K$ such that $x_n<L+\epsilon$ for all $n>K$. (This is the well-known "eventual upperbound" property of limsup.)

Then

$\Large\frac{x_1+\dots+x_n}{n}<\frac{x_1+\dots+x_k+(L+\epsilon)(n-k)}{n}$

Taking limsup on both sides gives

$\limsup a_n\leq L+\epsilon$

Since $\epsilon>0$ is arbitrary, $\limsup_{n\to\infty}a_n\leq\limsup_{n\to\infty} x_n$.

The case of $\liminf$ should be similar.

$\liminf a_n\leq\limsup a_n$ is automatic (always holds) so you get it for free.

The $(x_n)$ unbounded case, both $\limsup a_n$ and $\limsup x_n$ will be infinite.