2
$\begingroup$

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? ]

$\endgroup$
7
  • $\begingroup$ How do you define $\limsup_na_n$? $\endgroup$ Apr 30, 2021 at 18:52
  • $\begingroup$ 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. $\endgroup$ Apr 30, 2021 at 18:58
  • $\begingroup$ Your negation of the last part is not correct. I think that it should be that for finitely many terms $L-\epsilon \lt a_n$. $\endgroup$
    – Koro
    Apr 30, 2021 at 19:31
  • $\begingroup$ @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? $\endgroup$ Apr 30, 2021 at 19:41
  • $\begingroup$ @hazelnut_116: Also include $\epsilon$ in your negations and see the issue. $\endgroup$
    – Koro
    Apr 30, 2021 at 19:49

2 Answers 2

1
$\begingroup$

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 $limsup$.

$\endgroup$
1
$\begingroup$

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 $S=L$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.