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Infinite series/sequences

  1. Dec 15, 2007 #1
    Hi I have a few problems with regard to infinite series and sequences:

    1.) cos(n) / (n^2) summation from n = 0 to infinity

    2.) cos(pi*n)/n summation from n = 0 to infinity

    3.) sin(n) summation from 0 to infinity

    I have to tell whether they absolutely converge, conditionally converge, or diverge.

    I tried using the ration/root/limit test for them but nothing seemed to help.

    any feedback will be greatly appreciated.
  2. jcsd
  3. Dec 15, 2007 #2


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    The "comparison test" and the "divergence test" will be useful. For the second one, in particular, knowing what cos(pi*n) looks like is also helpful.

    By the way, for the first two the summation can't possibly start at n=0. :wink:
  4. Dec 15, 2007 #3


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    there are two basic limits we an compute, geometric series, and improper integrals. all other series must be compared usually to one of these. a necessary condition is that the terms go to zero, which is sufficient if their size goes to zero monotonically and they have alternating signs.

    compare 1 above to the series 1/n^2 which is then compoared to the integral of 1/x^2.

    2 is the other test , alternating test.

    3 i am not sure of. but suspect it does not converge by the test of whether the terms approach zero.
  5. Dec 16, 2007 #4

    Gib Z

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    My brain is exploding.

    True, the series diverges by several tests, most obviously by the n-term test mathwonk mentioned. But if we go about evaluating the result;

    [tex]S = \sum_{n=1}^{\infty} \sin (n) = \sum_{n=1}^{\infty} \mathRR{Im} (e^{in}) = \mathRR{Im}\sum_{n=1}^{\infty}(e^{in}) = \mathRR{Im} (\frac{1}{1-e^i}) = \frac{1+ \cos 1}{2 \sin^2 1}[/tex].

    What did i do wrong :(
  6. Dec 16, 2007 #5
    The geometric series doesn't converge when |z|=1:

    [tex]s_n = \sum_{i=0}^n z^n = \frac{1}{1-z} - \frac{z^{n+1}}{1-z}[/tex], z not equal 1.

    It's bounded, but is not a Cauchy sequence when |z| = 1.
    ([tex]s_{n+1}-s_n = e^{i \theta n}[/tex]).

    But as mentioned above, the main point is that the terms z^n don't go to zero as n -> infty when |z| = 1.

    [tex]\sum c_n[/tex] converges [itex]\Rightarrow[/itex] [tex]s_n[/tex] is Cauchy [itex]\Rightarrow[/itex] [tex]|c_n| = |s_n-s_{n-1}| \rightarrow 0[/tex].
    Last edited: Dec 16, 2007
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