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.

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

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.)



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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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