# Confused by limit superior and limit inferior definition.

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.

• (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.
– zhw.
Jul 18, 2017 at 19:39
• So it is necessary that $only$ $1$ term past $x_n$ satisfies that condition? Jul 18, 2017 at 19:43
• take a look at this answer where I try to show it visually. Jul 18, 2017 at 19:46
• Yes, but of course since it holds for any $n$ there will always be infinitely many such terms.
– zhw.
Jul 18, 2017 at 20:35

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.

• 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. Jul 18, 2017 at 20:01

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 $$\limsup_n x_n=\lim_ny_n=\lim_{n}\biggl(\sup_{k\ge n}\,x_k\biggr).$$