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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.

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    $\begingroup$ (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
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    $\begingroup$ 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
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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.

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

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