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Series convergence

  1. Nov 20, 2008 #1
    1. The problem statement, all variables and given/known data
    Show uniform convergence

    [tex]\frac{4b}{\pi} \sum^{\inf}_{n=1} \frac{1-(-1)^{n}}{n^{2}}cos(nt)cos(nx) [/tex]

    for fixed t


    2. Relevant equations



    3. The attempt at a solution

    [tex] \left| cos(nt) \right| \leq 1 [/tex]

    [tex] \left| cos(nx) \right| \leq 1 [/tex]

    [tex]lim \left|\frac{1 - (-1)^{n}}{n^{2}}\right| \ = \ 0 [/tex]

    What's next?
     
  2. jcsd
  3. Nov 20, 2008 #2
    Well it's an alternating series isn't it? So... see if it's conditionally convergent and absolutely convergent
     
  4. Nov 20, 2008 #3
    I'd like to use the Weierstrauss M test. Can someone help me in that direction?
     
  5. Nov 20, 2008 #4
    What's b? It seems like there is a lot of stuff they added here to try to confuse you. Let's ignore the constant 4b/pi for now. Then for any n, for all x the absolute value of the nth term is less than 2/n^2. We know that the series 2/n^2 converges, so this series also converges.
     
  6. Nov 20, 2008 #5

    Dick

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    Homework Helper

    Then do so. For each n you want a M_n such the |f_n|<=M_n (where f_n is the nth function in your series) and the series of the M_n converges. It's really pretty straightforward. You've already handled the cos parts. Any suggestions?
     
  7. Nov 20, 2008 #6
    This is not an alternative series! The terms with n even are all zeros and terms with n odd are 2/n^2 * cos ( ) * cos (). So take its absolute value and compare it with the series sum(2/n^2, n=1..infty) (I'm ignoring the constant outside the summation. Hence the series is absolutely convergent!
     
  8. Nov 21, 2008 #7
    Ok thanks everybody! I think I have gleaned enough to get a good answer.
     
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