1
$\begingroup$

I'm having a little bit of difficulty proving this question:

Prove that $ \limsup_{n->\infty} |S_n| = 0 ~~$iff$ ~\lim_{n->\infty}S_n = 0. $

What I have so far:

$ (\Leftarrow)$

$ $Suppose $ \limsup_{n->\infty} |S_n| = 0. $

$ $Let$~ \epsilon > 0.~ \exists~N \in \mathbb{N} ~$such that$~ n > N \Rightarrow sup${${|S_n|: n>N}$}$ < \epsilon. $

$ \Rightarrow \forall n>N ~$we have$~ |S_n| < \epsilon \Rightarrow \lim_{n->\infty} S_n = 0. $

$(\Rightarrow)$

$ ~$Suppose$~ \lim_{n->\infty} S_n = 0. ~$Hence$,~ \lim_{n->\infty} |S_n| = 0. $

And this is about as far as I'm getting, I know I could use the fact that if lim|Sn| converges then limsup |Sn| and liminf|Sn| must also converge to the same limit and hence this would imply directly that limsup|Sn| must equal 0 but I would like to avoid using this fact and rather form a proof from the definition of lim sup, ie lim sup Sn = lim(sup{Sn: n>N}).

Any help for this would be much appreciated :) Also, if someone could tell if my <= argument is 100% correct that would be great too!

$\endgroup$
0

2 Answers 2

Reset to default
0
$\begingroup$

If you remove the phrase "such that $n > N$" in the second line of the proof of $(\Leftarrow)$, your proof there will make sense and be correct.

You may also remove the statement "Hence, $\lim_{n\to \infty} |S_n| = 0$". To continue the proof of $(\Rightarrow)$, let $s = \limsup |S_n|$. Since $S_n \to 0$, given $\epsilon > 0$, there exists a positive integer $N$ such that $|S_n| < \epsilon$ for all $n \ge N$. Then, for all $k \ge N$, $\sup_{n\ge k} |S_n| \le \epsilon$. Taking the limit as $k\to \infty$ results in $s \le \epsilon$. Since $s \ge 0$ and $\epsilon$ was arbitrary, $s = 0$.

$\endgroup$
0
$\begingroup$

I like to solve these sorts of question in as many ways as I possibly can, i am now attempting a new solution:

$ (\Rightarrow) $ Suppose $ \limsup_{n->\infty} |S_n| = 0. $ Let $ \epsilon > 0. ~\exists N \in \mathbb{N} ~$such that$ ~\forall k\geq n~ \sup_{k\geq n} |S_k| < \epsilon.~ $ So we have $ ~0 \leq |S_n| \leq \sup_{k\geq n} |S_k| < \epsilon. $ Since the limit of $ \sup_{k\geq n} |S_k|$ goes to $0$ as $ n\rightarrow\infty$ then by the squeeze theorem $ |S_n| $ goes to $0$ which implies $S_n$ goes to $0$.

$ (\Leftarrow) $ Suppose $ \lim_{n->\infty} S_n = 0. ~~$Since $ S_n \rightarrow 0 ~$then $|S_n| \rightarrow 0 $ and then we have $ \limsup_{n->\infty} |S_n| = 0$.

Are there any flaws with this argument?

(Couldn't post this as a comment, too many characters)

$\endgroup$
5
  • $\begingroup$ In your proof of $(\Leftarrow)$, how did you deduce from $|S_n| \to 0$ that $\limsup_{n\to \infty} |S_n| = 0$? $\endgroup$
    – kobe
    Apr 14, 2015 at 14:18
  • $\begingroup$ A sequence a_n converges to a limit L iff lim sup a_n = lim inf a_n = lim a_n = L $\endgroup$
    – sho
    Apr 14, 2015 at 21:44
  • $\begingroup$ I thought you wanted to avoid using that fact. $\endgroup$
    – kobe
    Apr 14, 2015 at 21:47
  • $\begingroup$ Only for the previous solution method. For this method I allowed myself to use it. $\endgroup$
    – sho
    Apr 14, 2015 at 22:18
  • $\begingroup$ I see. Then the only two things I'll criticize are the third and fourth sentences in the proof of $(\Rightarrow)$. In the third sentence, the expression $k \ge n$ should be $n \ge N$. So in the fourth sentence, it should read something like "So $0 \le |S_n| \le \sup_{k\ge n} |S_k|$ for all $n \ge N$". $\endgroup$
    – kobe
    Apr 14, 2015 at 22:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .