# 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. Nov 4, 2016 at 13:56
• @tomasz i didn't know that those things exist in latex ... pardon me :) anyway, thanks to Masacroso, it's fixed now Nov 4, 2016 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 $\operatorname{sin}(x)$. Nov 4, 2016 at 14:09
• Anyway, what exactly do you want us to help with? What problem are you having? Nov 4, 2016 at 14:11
• @tomasz thanks ! that's great : Nov 4, 2016 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 $$x_n for all $$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.

After @yoyostein's answer has hopefully helped you in tackling exercise a), I want to add a hint for b) and c):

Hint for b): In my opinion, the difficult part about it is to have the middle inequality $$\liminf a_n < \limsup a_n$$. To achieve it, it may be worth understanding what the $$a_n$$ are: the arithmetic mean of $$x_1, x_2, \dots, x_n$$.
It will also be helpful to know when the $$\liminf$$ and $$\limsup$$ of a sequence are the same and when they are not. You have probably read a neat equivalent formulation before.

After considering these thoughts, maybe you know what to look for: A sequence $$(x_n)$$, such that the arithmetic mean of $$x_1, x_2, \dots, x_n$$ oscillates. Can you come up with such a sequence?

Hint for c): If the $$\liminf x_n$$ and $$\limsup x_n$$ are both finite, then you have no change to get what the exercise demands -- we know this because of a), so there is no need to look in this direction.
And although there is a solution where $$\liminf x_n = -\infty$$, $$\limsup x_n = \infty$$, and only $$\liminf a_n = \limsup a_n = 0$$ is finite (which you can come up with after some thinking), I would suggest a more strategic approach:

Try to make $$\liminf x_n$$ finite, for example $$\liminf x_n =0$$, and $$\limsup x_n = \infty$$. Then the exercise is solved regardless of the values of $$\liminf a_n$$ and $$\limsup a_n$$.

Extra: Surely this last approach does not necessarily give you the most insight, so feel encouraged to come up with more examples in b) and c). Maybe you can find one where the strict inequations and c) hold simultaneously?