Find the limsup and liminf

  • Thread starter Edellaine
  • Start date
In summary: So the limsup is 1 and the liminf is -1.In summary, the limit superior (limsup) is 1 and the limit inferior (liminf) is -1 for the given sequence a_n=(-1)^n(1+1/n).
  • #1
Edellaine
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Homework Statement


find the limsup and liminf of [tex]a_{n}=(-1)^{n}\left( 1+\frac{1}{n} \right)[/tex]


Homework Equations


Given a sequence of real numbers a_n, the supremum limit (also called the limit superior or upper limit), written lim sup is the limit of
[tex]a_{n}=sup_{(k>=n)}a_{k}[/tex]


The Attempt at a Solution


I started by using the identity sup(f+g) = sup(f) + sup(g) to turn the initial equation into a sum after multiplying it out. Then I separated the problem into two different limits. Now it seems obvious that the limsup of the given equation is 1 but I'm not sure how to prove that result.

limsup[tex](-1)^{n}[/tex] is clearly one. I showed this by setting n=2k, for some natural number k, and then saying for some n > k, [tex](-1)^{n}=1[/tex]. Then the sup is 1, and the limsup is 1 also.

I'm running into problems finding the limsup[tex]\frac{(-1)^{n}}{n}[/tex].
Do I have to use the same n = 2k, because the limit here goes to 0, leaving me with the result I want, but I'm not sure that: a) how to go about writing this out and b) if I'm even proceeding correctly.

I think if I can work through the limsup portion of the problem, it will be easy to apply the reasoning to the liminf portion.

Thanks.
 
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  • #2
I'm not sure what you did with setting n=2k, I can't see what the logical argument is exactly, could you please explain it?

Also, a good method to approach the problem could be to use the Squeeze law.

May I also note that It's great to see someone who has given their problem a substantial amount of help. I'm not saying other people around here don't do that as well, but there are far too many who give no effort at all. To everyone read, a good attempt will make us want to help you. Good work Edellaine.
 
  • #3
For n even, that is (1+ 1/n) which goes to 1.

For n odd, that is -(1+1/n) which goes to -1.

Those are the only two subsequential limits.
 

1. What is the definition of limsup and liminf?

The limsup and liminf are mathematical concepts used to describe the behavior of a sequence or a set of numbers. Liminf, also known as the limit inferior, is the smallest value that a sequence approaches as the index approaches infinity. Liminf is the largest value that a sequence approaches as the index approaches infinity.

2. How do you calculate limsup and liminf?

To find the limsup, you need to take the highest value that the sequence approaches as the index approaches infinity. This can be found by taking the supremum (or the least upper bound) of the set of numbers in the sequence. To find the liminf, you need to take the lowest value that the sequence approaches as the index approaches infinity. This can be found by taking the infimum (or the greatest lower bound) of the set of numbers in the sequence.

3. What is the relationship between limsup and liminf?

The limsup and liminf are two different values that describe the behavior of a sequence, but they are closely related. The limsup is always greater than or equal to the liminf. In other words, the liminf is the lower bound and the limsup is the upper bound of the sequence.

4. How do you interpret limsup and liminf graphically?

Graphically, the limsup and liminf can be interpreted as the highest and lowest horizontal asymptotes of the graph of the sequence. As the index approaches infinity, the values of the sequence will approach these asymptotes.

5. What are the practical applications of limsup and liminf?

The limsup and liminf are used in various fields of mathematics, such as analysis, topology, and number theory. They are also used in real-world applications, such as in signal processing, statistics, and finance. These concepts help to understand the long-term behavior and trends of a sequence or a set of numbers.

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