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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.

Community
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Maryam Seraj
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  • You should use \lim, \sup, \inf, \liminf and \limsup. Otherwise, this is very unpleasant to read. – tomasz Nov 04 '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 04 '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 04 '16 at 14:09
  • Anyway, what exactly do you want us to help with? What problem are you having? – tomasz Nov 04 '16 at 14:11
  • @tomasz thanks ! that's great : – Maryam Seraj Nov 04 '16 at 14:12
  • @tomasz i'm stuck i don't know what to do ... no idea ... – Maryam Seraj Nov 04 '16 at 14:13
  • Well, you should put some work into it before asking. Reading and understanding the question you have linked would be a nice start. – tomasz Nov 04 '16 at 14:14
  • @tomasz i'm thinking myself too ... but excuse me i'm new to this website ... i didn't know that rule ... – Maryam Seraj Nov 04 '16 at 14:15
  • Sure. The etiquette here is that it is OK to ask about homework, but just asking for solutions (or, in fact, simply providing them) is frowned upon. The rationale is that a) we don't want this site to be simply a place for people to copy and paste their homework from and that b) we can help you learn better if we know what level you're at, and what you are struggling with. Otherwise, there is risk of pitching too high or too low, which is a waste of everyone's time. – tomasz Nov 04 '16 at 14:26
  • Moreover, you should not ask questions without trying to solve them first -- that would asking strangers to do the work you can do yourself, without pay. People are more likely to want to help you if they see you trying to help yourself. – tomasz Nov 04 '16 at 14:28
  • @tomasz i feel guilty now ... what should i do ? should i delete my question ? – Maryam Seraj Nov 04 '16 at 14:33
  • I don't think you should delete the question. You could improve it (by adding the missing information I have mentioned), but most importantly, you should make sure that any future questions you post are better. Don't worry, posting one or two not so great questions is not bad, as try to improve. What is really bad for the site are people who keep posting their homework verbatim, without any regard for the etiquette. – tomasz Nov 05 '16 at 19:28

2 Answers2

1

Partial answer for (a):

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.

NerdOnTour
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yoyostein
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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?

NerdOnTour
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