Infinite Limit: How to Prove a Sequence Goes to Infinity

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Homework Statement


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


Homework Equations


N/A


The Attempt at a Solution



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

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

Ex: Manipulating \frac{n^3}{n^2+2} gets

\frac{n^3}{n^2+2} > \frac{n^3}{n^2+n^2} = \frac{n^3}{2n^2} = \frac{n^3}{2}

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:
a_{n} \rightarrow\infty > ?
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 \frac{n^{3}}{2} > M and solved for n.

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

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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