3
$\begingroup$

I'm trying to grasp the idea behind limit superior and limit inferior. Using the concept of subsequential limits, I understand that for a sequence $(x_n)$, the $\limsup x_n$ is the largest limit that a subsequence $(x_{n_k})$ will approach. However, looking at this alternative definition/theorem, I get confused:

Definition:

Suppose $\limsup x_n \in \mathbb R$. Then $\beta = \limsup x_n$ if and only if for all $\epsilon > 0$,

(i) there exists $n_0 \in \mathbb N$ such that $x_n < \beta + \epsilon$ for all $n \ge n_0$ and

(ii) given $n \in \mathbb N$, there exists $k \in \mathbb N$ with $k \ge n$ such that $x_k > \beta - \epsilon$.

Specifically, I am confused as to how this definition works with the sequence $x_n = (-1)^n$.

Obviously, the limit superior of $(-1)^n$ is $1$. So $\beta = 1$. Thus, (i) is satisfied, since $ 1 < 1 + \epsilon$ and $-1 < 1+ \epsilon$. But what about (ii)? It seems that no matter what tail of $x_n$ I consider, I will always end up with $-1$ in $x_k$. And it is not true that $-1>1-\epsilon$ for small enough $\epsilon$.

Can someone explain where my logic fails? Please and thank you.

$\endgroup$
  • 1
    $\begingroup$ (ii) just says that for every $n$ there is some term among $x_n, x_{n+1},\dots$ that satisfies the condition. In your example, every other term (!) will do that - an embarrassment of riches. $\endgroup$ – zhw. Jul 18 '17 at 19:39
  • $\begingroup$ So it is necessary that $only$ $1$ term past $x_n$ satisfies that condition? $\endgroup$ – BSplitter Jul 18 '17 at 19:43
  • $\begingroup$ take a look at this answer where I try to show it visually. $\endgroup$ – Masacroso Jul 18 '17 at 19:46
  • 1
    $\begingroup$ Yes, but of course since it holds for any $n$ there will always be infinitely many such terms. $\endgroup$ – zhw. Jul 18 '17 at 20:35
3
$\begingroup$

It seems that no matter what tail of $x_n$ I consider, I will always end up with $-1$ in $x_k$. And it is $not$ true that $-1>1-\epsilon$ for small enough $\epsilon$.

Can someone explain where my logic fails?

You will not always end up with $-1$ in $x_k$.

Condition (ii) says that given $n\in{\mathbb N}$, there exists $k\in{\mathbb N}$ with $k\geq n$ such that $$ x_k>1-\epsilon\tag{*}. $$ But $x_k=(-1)^k$ and ($*$) is always true when $k$ is even.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Okay, I think that's what I was getting hung up on. So $x_k$ is $not$ a sequence, in and of itself. $x_k$ is a member of the sequence. $\endgroup$ – BSplitter Jul 18 '17 at 20:01
2
$\begingroup$

Condition (ii) means there exist $x_k$s with indes $k$ as large as we please such that $x_k>1-\varepsilon$.

This is is indeed true since every other $x_k$ is equal to $1$ (namely those with even $k$.

Maybe you would find it more intuitive keeping in mind that the derived sequence $(y_n)\overset{\text{def}}{=}\sup\limits_{k\ge n}\,x_k$ is non-increasing and that $$\limsup_n x_n=\lim_ny_n=\lim_{n}\biggl(\sup_{k\ge n}\,x_k\biggr).$$

| cite | improve this answer | |
$\endgroup$

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.