Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Infinite Limit question

  1. Feb 21, 2010 #1
    1. The problem statement, all variables and given/known data
    Ok. For this sequence: [tex]a_{n} = n^2\left|cosn\pi\right|[/tex], Show/Prove that [tex]a_{n} \rightarrow\infty[/tex]

    2. Relevant equations

    3. 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.
    Last edited: Feb 21, 2010
  2. jcsd
  3. Feb 21, 2010 #2


    Staff: Mentor

    I think something is missing in the problem statement. What is the exact wording of this problem?
  4. Feb 21, 2010 #3
    I just added it above.
  5. Feb 21, 2010 #4


    Staff: Mentor

    Prove that it's greater than what? That still doesn't make any sense.
  6. Feb 21, 2010 #5
    That's what I have to figure out. The example with the fraction I showed is what I have to do somehow.
  7. Feb 21, 2010 #6


    Staff: Mentor

    Assuming n is an integer, |cos(n*pi)| = 1, so n2|cos(n*pi)| = n2. What does that do if n gets large?
  8. Feb 21, 2010 #7
    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.
  9. Feb 21, 2010 #8


    Staff: Mentor

    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?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook