Proving the Relationship between Lim Sup and Lim Inf

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In summary: But I don't know how to use the theorem in this case...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?Well, I think the definition of ##\limsup## is a sort of subsequence, just as ##\lim## is a subsequence. But in the case of ##\limsup## it does not have to be a subsequence of the original sequence
  • #1
fishturtle1
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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##.
 
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  • #2
fishturtle1 said:

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.
 
  • #3
fresh_42 said:
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.
Thanks for the reply,

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##?
 
  • #4
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?
 
  • #5
fresh_42 said:
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,

fresh_42 said:
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?

fresh_42 said:
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.
 
  • #6
fishturtle1 said:
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\}##.
 
  • #7
fishturtle1 said:
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?
 
  • #8
fresh_42 said:
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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1. What is the definition of "lim sup proof"?

"Lim sup proof" is a mathematical concept that deals with the limit superior of a sequence of real numbers. It is defined as the largest limit point of the sequence, or the supremum of the set of all accumulation points of the sequence.

2. How is "lim sup proof" different from "lim inf proof"?

While "lim sup proof" deals with the largest limit point of a sequence, "lim inf proof" deals with the smallest limit point. In other words, "lim sup proof" looks at the upper bound of a sequence, while "lim inf proof" looks at the lower bound.

3. What is the importance of "lim sup proof" in mathematics?

"Lim sup proof" is important in mathematics because it helps to determine the behavior of a sequence as it approaches infinity. It provides a way to analyze the long-term behavior of a sequence and can be used to prove the convergence or divergence of a series.

4. How do you prove a "lim sup proof"?

To prove a "lim sup proof", you need to show that the sequence has an upper bound and that the supremum of the set of accumulation points is equal to the lim sup of the sequence. This can be done using the definition of "lim sup" and properties of limits.

5. Can "lim sup proof" be used in other branches of science?

Yes, "lim sup proof" can be used in other branches of science such as physics, engineering, and economics. It is a fundamental concept in calculus and analysis, and its applications can be found in various fields that deal with continuous functions and sequences.

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