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Absolute convergence?

  1. Mar 1, 2008 #1
    1. The problem statement, all variables and given/known data
    Determine whether the series is absolutely convergent, conditionally convergent, or divergent.
    summation [cos((n)pi)/3] / n!
    from 1 to inf

    2. Relevant equations
    convergence tests

    3. The attempt at a solution
    I tried using the ratio test but that was ineffective since I ended up with cos(inf) in the numerator and denominator.

    So then I tried using squeeze theorem.

    I did -1/n! <eq a_n <eq 1/n!

    Seeing as how -1/n! and 1/n eventually reach 0, I concluded that a_n is absolutely convergent.

    Other than this, I have 2 other general questions:
    1. Does cos(inf) exist?
    2. Is it valid to use squeeze theorem for series?? (I've only effectively used them for sequences)
  2. jcsd
  3. Mar 1, 2008 #2
    Yes, you can use the squeeze theorem (it's basically the comparison test). [itex]\cos(\pi/3n)[/itex] can be broken up depending on the value of n and you can take the limit in each case.
  4. Mar 1, 2008 #3
    Do I have to prove that summation 1/n! goes to 0?
  5. Mar 1, 2008 #4


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    To test for absolute convergence you want to test if the sum of the absolute value of the series terms converges. To do this you want to use a comparison test with a series you know converges absolutely. That's not the same thing as the 'squeeze theorem'. So far you've only shown that the limit of the terms of the series is zero. Not that the series converges. And don't say things like '1/n! eventually reaches zero'. It doesn't. It approaches zero.
  6. Mar 1, 2008 #5


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    No. You don't show the summation 1/n! goes to 0. You show that the summation is finite. You might think about doing a ratio test on it.
  7. Mar 1, 2008 #6
    The ratio test did not work for me. Did I do it incorrectly?
  8. Mar 1, 2008 #7
    What if I did [cos((n)pi)/3] / n! <eq (1/n)^n?
    Since (1/n)^n converges by geometric series, then summation b_n (and hence summation a_n) must also converge.
  9. Mar 1, 2008 #8


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    Don't use the ratio test on the original series. Use the ratio test on the series you are comparing it to. 1/n!.
  10. Mar 1, 2008 #9


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    i) that's not a geometric series. ii) n^n>=n!. So 1/(n^n)<=1/n!
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