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

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? ]

• How do you define $\limsup_na_n$? Apr 30, 2021 at 18:52
• The largest partial limit of $(a_n)$, but more formally we use in the course the theorem above when trying to prove many theorems regarding $\lim \sup a_n$. : " Let $(a_n)$ be bounded sequence. Then $L = \lim\limits_{n \rightarrow \infty} \sup a_n$ iff for all $\epsilon>0$. $a_n < L+ \epsilon$ for all sufficiently large $n$, and $L-\epsilon < a_n$ infinitely often. " We discussed less about the set interpretation of $\lim \sup a_n$ ( I think that's what you refer to ) so I didn't go in that direction in my attempt of proof above. Apr 30, 2021 at 18:58
• Your negation of the last part is not correct. I think that it should be that for finitely many terms $L-\epsilon \lt a_n$.
– Koro
Apr 30, 2021 at 19:31
• @Koro 1.$a_n < L + \epsilon$ for sufficiently large n $\iff$ $\exists N \forall n>N.a_n < L + \epsilon$. 2. $L - \epsilon < a_n$ infinitely often $\iff$ $\forall N \exists n>N.L - \epsilon < a_n$. Negations: 1.$a_n \geq L + \epsilon$ infinitely often $\iff$ $\forall N \exists n>N.a_n \geq L + \epsilon$. 2.$L - \epsilon \geq a_n$ for sufficiently large n $\iff$ $\exists N \forall n>N.L - \epsilon \geq a_n$. aren't these correct? Apr 30, 2021 at 19:41
• @hazelnut_116: Also include $\epsilon$ in your negations and see the issue.
– Koro
Apr 30, 2021 at 19:49

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 $$L-\frac{\epsilon}{2}. By the definition of supremum, for each $$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$$.

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 $$S=L$$.