Infinite Limit: How to Prove a Sequence Goes to Infinity

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Homework Help Overview

The discussion revolves around proving that the sequence defined by a_{n} = n^2|cos(n\pi)| approaches infinity as n increases. Participants are exploring the implications of the sequence's behavior and the necessary conditions for establishing this limit.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to manipulate the sequence to establish a lower bound, questioning how to express a_{n} in a way that facilitates proving it goes to infinity. There is discussion about the behavior of |cos(n\pi)| and its impact on the sequence as n becomes large.

Discussion Status

Some participants have provided insights regarding the behavior of the sequence when n is an integer, suggesting that |cos(n\pi)| equals 1, which simplifies the analysis. Others are exploring how to formally express the conditions needed to prove the limit, but there is no explicit consensus on the approach yet.

Contextual Notes

There are indications that the original problem statement may lack clarity, as participants are questioning the exact wording and requirements for the proof. This uncertainty may affect the direction of the discussion.

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Homework Statement


Ok. For this sequence: [tex]a_{n} = n^2\left|cosn\pi\right|[/tex], Show/Prove that [tex]a_{n} \rightarrow\infty[/tex]


Homework Equations


N/A


The Attempt at a Solution



I have to manipulate the statement to show that
[tex]n^2\left|cosn\pi\right| > ?[/tex]

I'm having trouble making a statement that's smaller. If it was a fraction I could do it:

Ex: Manipulating [tex]\frac{n^3}{n^2+2}[/tex] gets

[tex]\frac{n^3}{n^2+2} > \frac{n^3}{n^2+n^2} = \frac{n^3}{2n^2} = \frac{n^3}{2}[/tex]

I replace 2 with n^2 to make the denominator bigger thus making it smaller. I somehow have to do something like that with the sequence given but I'm not sure how.

Any help is appreciated.
 
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I think something is missing in the problem statement. What is the exact wording of this problem?
 
Mark44 said:
I think something is missing in the problem statement. What is the exact wording of this problem?

I just added it above.
 
The_Iceflash said:
[tex]a_{n} \rightarrow\infty[/tex] > ?
Prove that it's greater than what? That still doesn't make any sense.
 
Mark44 said:
Prove that it's greater than what? That still doesn't make any sense.

That's what I have to figure out. The example with the fraction I showed is what I have to do somehow.
 
Assuming n is an integer, |cos(n*pi)| = 1, so n2|cos(n*pi)| = n2. What does that do if n gets large?
 
Mark44 said:
Assuming n is an integer, |cos(n*pi)| = 1, so n2|cos(n*pi)| = n2. What does that do if n gets large?

If n gets large then the sequence gets large.

So, would n2|cos(n*pi)| > n|cos(n*pi)| be sufficient?

After I would do that I need to set it greater than M and solve for it.

For example: the fraction above I set [tex]\frac{n^{3}}{2} > M[/tex] and solved for n.

Then I'd put it all together formally.
 
Clearly, from the earlier work, [tex]\lim_{n \to \infty} a_n = \infty[/tex]

Now, to prove that this is so, you want for find a number M s.t., for all n >= M, an > M.

You have an = n2 (reason given earlier) and you want an > M.

Can you combine these and solve for n?
 

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