Prove If lim inf sn = lim sup sn, then lim sn is defined and lim sn = lim inf sn = lim sup sn

Question

I need a proof verification or some pointers.

The definition from the book says:

1. $\limsup s_n = \lim\limits_{N\to\infty} \sup \{s_n : n > N\}$

2. $\liminf s_n = \lim\limits_{N\to\infty} \sup \{s_n : n > N\}$.

Case 1. lim inf {$s_n$} = lim sup {$s_n$} = positive infinity

Proof: Let M > 0. Then there exists some n such that n > N implies $s_n$ > M. The question is I'm only given: lim inf {$s_n$} = lim sup {$s_n$}. How do I construct this $s_n$ so that lim inf {$s_n$}= lim $s_n$ = lim sup {$s_n$}. Can I just apply the squeeze theorem here? Then lim sn is defined and lim sn = lim inf sn = lim sup sn by squeeze theorem?. Help?

Answer 1

What you're trying to prove, if you want to use the definition of a limit directly (at least in the $+\infty$ case) is that

For any $M>0$: there is an $N$ such that for any $n > N$, $s_n > M$

From your knowledge of the liminf, you know that

For any $M >0$: there is an $N$ such that $\inf \{s_n:n>N\} > M$

Why does this second statement imply the first?

---

For $\pm \infty$, you will need only the liminf or the limsup. For finite limits, however, you will need both.

The conclusion we want to reach is

For any $\epsilon > 0$: there is an $N$ such that $n>N$ implies $|s_n - L| < \epsilon$.

Using the liminf and limsup, we have the statements:

For any $\epsilon > 0$: there is an $N$ such that $n > N$ implies $\inf \{s_n : n>N\} > L - \epsilon$

For any $\epsilon > 0$: there is an $N$ such that $n > N$ implies $\sup \{s_n : n > N\} < L + \epsilon$

How do these two statements let you deduce the first?

---

You could also use the squeeze theorem if you note that $$ \inf \{s_n: n \geq N\} \leq s_N \leq \sup\{s_n: n \geq N\} $$ and take the limit as $N \to \infty$.