Prove that a sequence converges and prove that $\limsup\limits_{n \rightarrow \infty}a_n = \lim\limits_{n \rightarrow \infty} S_n $

Question

Problem: Let $ (a_n) $ be bounded. Denote $ S_n := \sup\{ a_k : k \geq n \} $. Prove sequence $ (S_n) $ converges ( to a finite limit ), and also prove $\limsup\limits_{n \rightarrow \infty}a_n = \lim\limits_{n \rightarrow \infty} S_n $.

I have the following theorem I can use: Let $ (a_n) $ be bounded sequence. Then $ L = \limsup\limits_{n \rightarrow \infty} a_n $ iff for all $\epsilon>0 $. $ a_n < L+ \epsilon $ for all sufficiently large $ n $, and $ L-\epsilon < a_n $ infinitely often.

Attempt outline: I showed $ (S_n) $ converges to a finite limit by proving it's monotonic decreasing and utilizing that $ (a_n) $ is bounded. I denoted $ L = \lim\limits_{n \rightarrow \infty} S_n $ Then I went to prove the other part by showing $ L $ satisfies the theorem I provided. I did this by contradiction, that is I assumed $\exists \epsilon>0 $ s.t. $ a_n \geq L+\epsilon $ infinitely often, and managed to find a contradiction. Then I assumed $\exists \epsilon>0 $ s.t. $ L - \epsilon \geq a_n $ for all sufficiently large $ n $. However I got stuck in the last part.
What I managed to do in last part: since $ a_n \leq L - \epsilon $ for all sufficiently large $ n $ and also since $ (a_n) $ is bounded then by Bolzano-Weierstrass there exists $ a_{n_k} \rightarrow L^{'} \in \mathbb{R} $ s.t. $ L^{'} \leq L - \epsilon < L $ so $ L^{'} < L $. Now, since $ S_n \rightarrow L $ then $ \exists N_1 \forall n>N_1.L - \epsilon < S_n < L + \epsilon $ [ Then I think I have to utilize the fact that $ a_n \leq S_n $ but couldn't reach a contradiction, any ideas? ]

Answer 1

Since $S_n$ is a bounded decreasing sequence, we know that $S_n$ has a finite limit, say $L = \lim_{n \to \infty} S_n$.

Now let $M := \lim\sup_{n \to \infty} a_n $. Then $M$ is the largest partial limit of $(a_n)$. Suppose $\{a_{n_k}\}_{k=1}^\infty$ is a subsequence of $a_n$ converging to $M$. We want to show that $M = L$.

$( L \geq M)$ Note that $S_{n_k} = \sup \{ a_i | i \geq n_k\} \geq a_{n_k}$ for all $k$. Since both sides have the limit, we see that $L = \lim S_{n_k} \geq \lim a_{n_k} = M.$

$(L \leq M)$ Let $\epsilon$ be given. Since $L$ is a limit of $S_n$, there is $N\in \mathbb{N}$ such that $n \geq N$ implies that $L-\frac{\epsilon}{2}<S_n<L+\frac{\epsilon}{2}$. By the definition of supremum, for each $n\geq N$ there is $b_n \in \{a_i | i \geq n\}$ such that $S_n - \frac{\epsilon}{2} < b_n \leq S_n$. In all, $n \geq N$ implies that $$L - \epsilon < b_n < L + \epsilon.$$ This implies that $b_n$ converges to $L$. Since $b_n$ is a subsequence of $a_n$ converging to $L$, $L \leq M$ by the definition of $limsup$.

Answer 2

This is further to the comment I dropped above. You may alternatively proceed like this to prove it.

First let's say $S_n\to S\in \mathbb R$ (you have already proven it) and recall that every subsequence of a convergent sequence converges to the same limit.

We have to prove that $L=S$

Suppose on the contrary that $S\ne L$ so we have two possibilities:

Possibility #$(1)$: $S\gt L$
Using the definition, there exists some $N$ such that for all $n\ge N$, we must have $a_n\lt L+\frac{S-L}2=\frac {L+S}2 $
It follows that $\sup_{n\ge N} a_n\le \frac{L+S}2\implies \lim_{N\to \infty}\sup_{n\ge N} a_n\le \frac {L+S}2\implies S\le \frac{L+S}2\implies S\le L$, which is contradiction.

Possibility #$(2)$: $S\lt L$
I leave this to you to get contradiction. Note what we recalled in second sentence above.

Then by contradiction, we must have $S=L$.