# Limits superior and inferior

• danielakkerma
In summary, Daniel attempted to solve a homework problem, but ran into some difficulty. He asks for help from someone who seems to be more knowledgeable about the topic.
danielakkerma
(Hello everyone!)

## Homework Statement

Given that $$\limsup_{n \rightarrow \infty}(\frac{1}{x_n})\cdot \limsup_{n \rightarrow \infty}(x_n)=1$$
Show that x_n converges.

## Homework Equations

Recalling that:
$$x_n \text{ converges } \iff \liminf(x_n)=\limsup(x_n)$$

## The Attempt at a Solution

Started with this:
Firstly, let's designate the limsup of (x_n) as L;
Therefore, according to the stipulated conditions, limsup(1/x_n) = 1/L;
Now:
$$1. \forall \epsilon >0 \; \exists N \mid \forall n>N \mid x_n \leq L+\epsilon$$
However, for the limit inferior:
$$2. \forall \epsilon >0 \; \exists N^{*} \mid \forall n>N^{*} \mid x_n \geq L-\epsilon$$
Here arises my first question: can I unite both Ns, by finding the *maximum* of both, i.e. N'=Max{N, N*}?(Even though, I am not sure that in fact, higher orders of x_n, would find themselves in the inferior limit region).
And if not, how do I proceed?
(I tried inverting ineq. 2), to arrive at a value for limsup(1/x_n), but, to no avail.
Any help will be much appreciated,
(And I'm sorry if I bungled some of the definitions here, it's been a while since I had last done this ,
Most beholden,
And grateful as always,
Daniel

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What do you think of this(is this valid?); some limited progress:
$$1. \limsup(x_n) = L \Rightarrow \forall \epsilon >0 \mid \exists N_1 \mid \forall n>N_1 \Rightarrow x_n < L+\epsilon \\ 2. \limsup(\frac{1}{x_n}) = \frac{1}{L} \mid \forall \epsilon >0 \Rightarrow \exists N_2 \mid \forall n>N_2 \Rightarrow \frac{1}{x_n} < \frac{1}{L}+\epsilon \\ 3. N^{\prime} = \max(N_1,N_2) \\ 4. \text{Rearranging 2. : if } \frac{1}{x_n} < \frac{1}{L}+\epsilon \Rightarrow x_n > \frac{L}{1+\epsilon L} \\ 5. \text{via 1 & 4(assuming N' >= N1, N2(3.)): } L+\epsilon > x_n > \frac{L}{1+\epsilon L} \\ 6. \text{Setting epsilon: } \epsilon = \frac{a}{n} \mid a \in \mathbf{R} \\ 7. \text{Using the "Squeeze Theorem" :}h(n) = L+\epsilon(n) \rightarrow L \\ g(n) = \frac{L}{1+\epsilon(n) L} \rightarrow L \\ g(n)<x_n<h(n)$$
And so you obtain that x_n converges, and that in fact, as we would expect, to L.
The problem is, can I use the Squeeze theorem, even though I am not certain that x_n, in fact, converges? Or is it simplify sufficient to point out, that since I have already found "N" -- that very N could be translated to a direct proof of the convergence of x_n, via the epsilon/N notation?(i.e. by definition)?
Any thoughts?
Again, very thankful for your attention,
Daniel
(All limits tend to infinity, as stated in the OP)

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danielakkerma said:
What do you think of this(is this valid?); some limited progress:
$$1. \limsup(x_n) = L \Rightarrow \forall \epsilon >0 \mid \exists N_1 \mid \forall n>N_1 \Rightarrow x_n < L+\epsilon \\ 2. \limsup(\frac{1}{x_n}) = \frac{1}{L} \mid \forall \epsilon >0 \Rightarrow \exists N_2 \mid \forall n>N_2 \Rightarrow \frac{1}{x_n} < \frac{1}{L}+\epsilon \\ 3. N^{\prime} = \max(N_1,N_2) \\ 4. \text{Rearranging 2. : if } \frac{1}{x_n} < \frac{1}{L}+\epsilon \Rightarrow x_n > \frac{L}{1+\epsilon L} \\ 5. \text{via 1 & 4(assuming N' >= N1, N2(3.)): } L+\epsilon > x_n > \frac{L}{1+\epsilon L} \\ 6. \text{Setting epsilon: } \epsilon = \frac{a}{n} \mid a \in \mathbf{R} \\ 7. \text{Using the "Squeeze Theorem" :}h(n) = L+\epsilon(n) \rightarrow L \\ g(n) = \frac{L}{1+\epsilon(n) L} \rightarrow L \\ g(n)<x_n<h(n)$$
And so you obtain that x_n converges, and that in fact, as we would expect, to L.
The problem is, can I use the Squeeze theorem, even though I am not certain that x_n, in fact, converges? Or is it simplify sufficient to point out, that since I have already found "N" -- that very N could be translated to a direct proof of the convergence of x_n, via the epsilon/N notation?(i.e. by definition)?
Any thoughts?
Again, very thankful for your attention,
Daniel
(All limits tend to infinity, as stated in the OP)

Haven't figured the problem out yet, but here are some thoughts:
- Where in your proof do you use the given statement?
- In step 6, can you just set epsilon, isn't epsilon supposed to be random >0?

1. I used the initial conditions, when setting the Limsups; That is, I had first declared limsup(x_n) to be some arbitrary(but real) L; then, using the provided relation, I educed limsup(1/x_n) to be 1/L.
2. Setting epsilon thus still provides me with randomness(since "a" could be any real number). However, that was done merely to bound x_n. Since the derived inequality for x_n(5.) applies to all epsilons(greater than 0), then surely it would still hold for diminishing values of epsilon as well.
This gave me the necessary circumscription of x_n, to use the "squeeze theorem"; of the verity of whose usage, however, I am still uncertain.
What do you think, though?
Once again,
Very grateful,
Daniel

danielakkerma said:
1. I used the initial conditions, when setting the Limsups; That is, I had first declared limsup(x_n) to be some arbitrary(but real) L; then, using the provided relation, I educed limsup(1/x_n) to be 1/L.

In the proof, points 1 and 2, I think the N1=N2.
And lim sup(xn)=L => lim sup(1/xn)=1/L even without the given condition.

In the proof, points 1 and 2, I think the N1=N2.
And lim sup(xn)=L => lim sup(1/xn)=1/L even without the given condition.
I am not quite so sure about that;
You see, as far as I understand it, Limsup, and Liminf don't follow the arithmetical rules of (finite) Limits.
In other words, I am not at all aware that Limsup(1/x_n) = 1/Limsup(x_n) = ...
I believe that's part of the reason why that specific condition was given.
However, if you believe this to be the case, I would very much welcome a link to a proof of that particular rule. It could very useful in future.
Also, why should N1 equal N2? could you elaborate on that? in any case, I thought taking the maximum eliminated any need to specify them, or find a particular congruence thereof.
Thanks,
Daniel

danielakkerma said:
I am not quite so sure about that;
You see, as far as I understand it, Limsup, and Liminf don't follow the arithmetical rules of (finite) Limits.
In other words, I am not at all aware that Limsup(1/x_n) = 1/Limsup(x_n) = ...
I believe that's part of the reason why that specific condition was given.
However, if you believe this to be the case, I would very much welcome a link to a proof of that particular rule. It could very useful in future.
Also, why should N1 equal N2? could you elaborate on that? in any case, I thought taking the maximum eliminated any need to specify them, or find a particular congruence thereof.
Thanks,
Daniel

Never mind, what I said is not true.

I think your proof is correct. Now I see how you used the condition.

danielakkerma said:
And so you obtain that x_n converges, and that in fact, as we would expect, to L.
The problem is, can I use the Squeeze theorem, even though I am not certain that x_n, in fact, converges?
Yes, if x_n can be squeezed between two sequences that both converge to the same limit, then that forces x_n to converge, to the same limit.

I think your proof is mostly fine, but you are implicitly assuming (e.g. in step 4) that x_n > 0 for all n. I didn't see that stated as a hypothesis. Isn't the theorem still true if some or all of the x_n's are negative? (I think we can presume that they all have to be nonzero, so that 1/x_n is defined.)

M. Planck, thanks again for all your help; I hope my approach really does check out :)...
***
Jbunn,
Great comment, thanks. As I said, I haven't tinkered with these types of problems for quite a while, and, I completely forgot to generalise the formulae for x_n < 0 (obviously, x_n must be non-zero).
The point is, I'm not really sure how this is supposed to work.
I know from the definition of Limsup that x_n(with n>N) lies to the left of L+eps. But, it is not at all certain, that ALL elements of x_n (n>N) are in fact larger than L-eps.
Therefore, taking the absolute value of x_n is impermissible here.(i.e. |x_n-L|<eps --> which is incidentally the definition for a finite, complete limit).
I also tried to keep L(when manipulating it) as arbitrary and nondescript as possible(save for the fact of course, that it's non-zero). I thought that sufficiently covered it.
Wouldn't negative x_n-s simply invert the direction of the squeezing inequality? meaning:
h(n)>x_n>g(n) in that case?
I am also not so sure what to do, in that case, about "ill-behaved", pathological sets(even though I think the fact that limsup(x_n)*limsup(1/x_n) = 1 > 0 pretty much excludes those sets).
How do you propose to generalise this?
Again,
I remain extremely grateful for all your help!
Daniel

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## What is the definition of limits superior and inferior?

Limits superior and inferior are mathematical concepts used to describe the behavior of a sequence of numbers as it approaches infinity. The limit superior is the largest number that the sequence gets arbitrarily close to, while the limit inferior is the smallest number the sequence gets arbitrarily close to.

## How are the limits superior and inferior calculated?

The limit superior is calculated by finding the greatest accumulation point of the sequence, while the limit inferior is found by finding the smallest accumulation point. An accumulation point is a number that the sequence gets arbitrarily close to an infinite number of times.

## What is the relationship between limits superior and inferior?

The limit superior and inferior are always related in the sense that the limit inferior will always be less than or equal to the limit superior. This means that the limit superior will always be an upper bound for the sequence, while the limit inferior will always be a lower bound.

## How are limits superior and inferior used in real-life applications?

Limits superior and inferior are used in various fields of science and engineering, such as physics, economics, and computer science. They are used to analyze the behavior of systems that involve infinite processes, such as the behavior of a physical system over time or the performance of a computer algorithm as the input size grows.

## What is the difference between limits superior and inferior and regular limits?

The main difference between limits superior and inferior and regular limits is that regular limits describe the behavior of a sequence at a specific point, while limits superior and inferior describe the overall behavior of a sequence as it approaches infinity. Additionally, regular limits can exist for sequences that do not have a limit superior or inferior, while every sequence has a limit superior and inferior.

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