# Lim sup proof

fishturtle1

## Homework Statement

Prove the ##\limsup \vert s_n \vert = 0## iff ##\lim s_n = 0##.

## Homework Equations

##\limsup s_n = \lim_{N\rightarrow \infty} \sup \lbrace s_n : n > N \rbrace = \sup \text{S}##

##\liminf s_n = \lim_{N\rightarrow \infty} \inf \lbrace s_n : n > N \rbrace = \inf \text{S}##

Definition of limit: ##\lim s_n = L## iff For all ##\varepsilon > 0## there exists ##N(\varepsilon)## such that ##n > N(\varepsilon)## implies ##\vert s_n - L \vert < \varepsilon##.

## The Attempt at a Solution

##(\leftarrow)## Suppose ##\lim s_n = 0##. Then for all ##\varepsilon > 0## there exists ##N(\varepsilon)## such that ##\vert s_n - 0 \vert = \vert s_n \vert < \varepsilon##. So ##\vert \vert s_n \vert \vert < \varepsilon##. So ##\lim \vert s_n \vert = 0##. Since ##\lim \vert s_n \vert = 0 \epsilon \mathbb{R}##, we can say ##\limsup \vert s_n \vert = 0##.

##(\rightarrow)## Suppose ##\limsup \vert s_n \vert = 0##.

Mentor
2022 Award

## Homework Statement

Prove the ##\limsup \vert s_n \vert = 0## iff ##\lim s_n = 0##.

## Homework Equations

##\limsup s_n = \lim_{N\rightarrow \infty} \sup \lbrace s_n : n > N \rbrace = \sup \text{S}##

##\liminf s_n = \lim_{N\rightarrow \infty} \inf \lbrace s_n : n > N \rbrace = \inf \text{S}##

Definition of limit: ##\lim s_n = L## iff For all ##\varepsilon > 0## there exists ##N(\varepsilon)## such that ##n > N(\varepsilon)## implies ##\vert s_n - L \vert < \varepsilon##.

## The Attempt at a Solution

##(\leftarrow)## Suppose ##\lim s_n = 0##. Then for all ##\varepsilon > 0## there exists ##N(\varepsilon)## such that ##\vert s_n - 0 \vert = \vert s_n \vert < \varepsilon##. So ##\vert \vert s_n \vert \vert < \varepsilon##. So ##\lim \vert s_n \vert = 0##. Since ##\lim \vert s_n \vert = 0 \epsilon \mathbb{R}##, we can say ##\limsup \vert s_n \vert = 0##.
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##(\rightarrow)## Suppose ##\limsup \vert s_n \vert = 0##.
The step from ##\lim |s_n|=0## to ##\sup S = 0## is a bit of a cheat. You have ##|s_n| < \varepsilon## for all ##n > N_\varepsilon##. Now can you choose a series of ##\varepsilon## such that ##\{|s_n|\, : \,n> N_\varepsilon\}## is bounded in a way that shows that these boundaries converge to zero?

This should also be your strategy in the other direction: use the definitions and make yourself clear, what has to be shown. Just claiming things isn't a proof, although the result is expected.

fishturtle1
The step from ##\lim |s_n|=0## to ##\sup S = 0## is a bit of a cheat. You have ##|s_n| < \varepsilon## for all ##n > N_\varepsilon##. Now can you choose a series of ##\varepsilon## such that ##\{|s_n|\, : \,n> N_\varepsilon\}## is bounded in a way that shows that these boundaries converge to zero?

This should also be your strategy in the other direction: use the definitions and make yourself clear, what has to be shown. Just claiming things isn't a proof, although the result is expected.

Since ##\lim s_n = 0##, the supremums of ##(s_k)## where ##k \ge n## must be getting smaller and smaller since ##s_k##'s are getting smaller and smaller...

Thm. 11.4 from the text: Every sequence ##(s_n)## has a monotonic subsequence.

So consider a monotonic subsequence ##s_{n_k}## that converges to 0. Then ##\limsup s_{n_k} = 0##. Is this what you meant? Edit: no. This is just the same argument I used before... So should I look for a subsequence of ##N_\varepsilon## or a subsequence of ##s_n##?

Mentor
2022 Award
I'm not sure, if I make it too complicated or not. Say ##S_N := \{s_n\, : \,n>N\}## and ##\varepsilon >0##. Then there is a ##N## with ##|s_n|<\varepsilon ## for all ##n>N##, if we assume ##\lim_{n \to \infty} s_n=0\,.## This means all elements of ##S_N## are between ##\pm \varepsilon##. So ##\sup S_N < \varepsilon##. Now to guarantee that all ##\sup S_n## for ##n>N## are also bounded by ##\varepsilon##, I think we need the observation that ##\sup S_n < \sup S_N## for ##n>N## because ##S_n\subseteq S_N##. Now the sequence ##(S_n)_{n \in \mathbb{N}}## fulfills the ##\varepsilon -##criterion for convergence to zero.

I apologize, if you meant this. I think I was a bit confused. My first thought was to choose ##\varepsilon = \frac{1}{N}##, but then we only have some ##S_{M(N)} < \frac{1}{N}## and it is not immediately obvious, that the convergence of this subsequence ##(S_{M(N)})_N## implies the convergence of ##(S_N)_N## as a whole. E.g. ##a_n=(-1)^n## has also convergent subsequences, but doesn't converge as a whole.

Have you tried the other direction in the meantime?

fishturtle1
I'm not sure, if I make it too complicated or not. Say ##S_N := \{s_n\, : \,n>N\}## and ##\varepsilon >0##. Then there is a ##N## with ##|s_n|<\varepsilon ## for all ##n>N##, if we assume ##\lim_{n \to \infty} s_n=0\,.## This means all elements of ##S_N## are between ##\pm \varepsilon##.
Ok i see how you got here,

So ##\sup S_N < \varepsilon##. Now to guarantee that all ##\sup S_n## for ##n>N## are also bounded by ##\varepsilon##, I think we need the observation that ##\sup S_n < \sup S_N## for ##n>N## because ##S_n\subseteq S_N##. Now the sequence ##(S_n)_{n \in \mathbb{N}}## fulfills the ##\varepsilon -##criterion for convergence to zero.
Why does ##S_n\subseteq S_N## mean that we need the observation that ##\sup S_n < \sup S_N## for ##n>N##?

What would prove ##\limsup s_n = 0##? I think we need to take a sequence of all ##sup (s_n : n > k)## and then see where their limit is, but how?

Have you tried the other direction in the meantime?

I have a theorem from a previous chapter:
Let ##(s_n)## be a sequence in ##\mathbb{R}##.
i) If ##\lim s_n## is defined[ as a real number, ##+\infty, -\infty##], then ##\liminf s_n = \lim s_n = \limsup s_n##.
ii) If ##\liminf s_n = \limsup s_n,## then ##\lim s_n## is defined and ##\lim s_n = \liminf s_n = \limsup s_n##.

So to show the other direction,
Suppose ##\limsup \vert s_n \vert = 0##. We want to show ##\liminf \vert s_n \vert = 0##. This seems reasonable because ##\vert s_n \vert \ge 0## and since the ##\sup (s_n : n > k)## keeps getting closer and closer to 0, then ##\vert s_n \vert##'s should be getting closer and closer to 0, but can't get smaller than 0.

Mentor
2022 Award
Why does ##S_n\subseteq S_N## mean that we need the observation that ##\sup S_n < \sup S_N## for ##n>N##?
For ##n > N## we have ##S_n \subseteq S_N##, because ##S_N = S_n \cup \{s_{N+1},s_{N+2}, \ldots , s_{n-1},s_{n}\}## so the supremum of ##S_N## is eventually greater than the supremum of ##S_n## in case the set ##\{s_{N+1},s_{N+2}, \ldots , s_{n-1},s_{n}\}## contains a greater element. This is the observation. Needed is it, because we have ##a_N := \sup S_N < \varepsilon ##, but we need ##a_n < \varepsilon ## for all ##n> N##. As ##a_n \leq a_N < \varepsilon## this is given.
What would prove ##\lim \sup s_n = 0##?
What I wrote above: For all ##\varepsilon >0 ## there is an ##N \in \mathbb{N}## such that for all ##n > N## we have ##|a_n| < \varepsilon## where our sequence ##(a_n)_{n \in \mathbb{N}} = (\sup S_n)_{n \in \mathbb{N}}##. You see, that the many equal letters ##S_n## for sets and sequence elements is a bit confusing. At least I read your ##\lim \sup s_n = 0## as a typo for ##\lim \sup S_n = 0##. Otherwise I don't know why you want to consider ##\sup s_n##. All that matters is ##\lim s_n = 0## in this part of the proof.
I think we need to take a sequence of all ##sup (s_n : n > k)## and then see where their limit is, but how?
What do the round parentheses now mean? We need ##a_n = \sup S_n = \sup \,\{s_m\, : \,m>n\}##.

Mentor
2022 Award
I have a theorem from a previous chapter:
Let ##(s_n)## be a sequence in ##\mathbb{R}##.
i) If ##\lim s_n## is defined[ as a real number, ##+\infty, -\infty##], then ##\liminf s_n = \lim s_n = \lim \sup s_n##.
ii) If ##\lim \inf s_n = \lim \sup s_n,## then ##\lim s_n## is defined and ##\lim s_n = \lim \inf s_n = \lim \sup s_n##.

So to show the other direction,
Suppose ##\lim \sup \vert s_n \vert = 0##. We want to show ##\lim \inf \vert s_n \vert = 0##. This seems reasonable because ##\vert s_n \vert \ge 0## and since the ##\sup (s_n : n > k)## keeps getting closer and closer to 0, then ##\vert s_n \vert##'s should be getting closer and closer to 0, but can't get smaller than 0.
Shouldn't this be a little more formal than this? I mean, intuitively it is obvious anyway, so a couple of "closer" would not convince me.
Considering this lemma, I wonder what should be shown at all?

fishturtle1
Shouldn't this be a little more formal than this? I mean, intuitively it is obvious anyway, so a couple of "closer" would not convince me.
Considering this lemma, I wonder what should be shown at all?
Yes I meant to say that was my intuition and I couldn't think how to put this into an actual argument.

##(\rightarrow)## Suppose ##\limsup \vert s_n \vert = 0##. Let ##\varepsilon > 0##. By definition of limit, there exists ##N## such that ##n > N## implies ##\vert \sup \vert s_n \vert \vert < \varepsilon##. By definition of supremum and absolute value, we have ##\vert s_n \vert \le \sup \vert s_n \vert \le \vert \sup \vert s_n \vert \vert < \varepsilon##. So ##\vert s_n \vert < \varepsilon##. So for all ##\varepsilon > 0##, there exists an N such that ##n > N## implies ##\vert s_n \vert < \varepsilon##. By definition of limit, ##\lim s_n = 0##.

##(\leftarrow)## Suppose ##\lim s_n = 0##. Let ##\varepsilon_1 > 0##. By definition of limit, there exists ##N_1## such that ##n > N_1## implies ##\vert s_n \vert < \varepsilon_1##. So ##\vert \vert s_n \vert \vert = \vert s_n \vert < \varepsilon_1##. Therefore ##\vert \vert s_n \vert \vert < \varepsilon_1##. Therefore, for all ##\varepsilon_1 > 0## we have ##n > N_1## implies ##\vert \vert s_n \vert \vert < \varepsilon_1##. By definition of limit ##\lim \vert s_n \vert = 0##. By theorem 10.7, ##\lim \sup \vert s_n \vert = 0##.
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