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Testing convergence of a series

  1. May 8, 2005 #1
    [tex]
    \sum_{n=1}^\infty \frac{cos n}{n^2}
    [/tex]

    Now, which test should i try? It doesn't appear to be anything straightforward like ratio test, comparison test etc and it's not a straightforward limit.
    cos(n) goes negative too, so that complicates things.

    Can anyone offer me any advice on a problem like this? Thanks!
     
  2. jcsd
  3. May 8, 2005 #2

    Pyrrhus

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    Obviously this series has both negative and positive terms, but it's not an alternate series. The signs change is irregular.

    If i were you i would the uniform convergence test.

    [tex] \sum_{n=1}^{\infty} |\frac{cos n}{n^2}| = \sum_{n=1}^{\infty} \frac{|cos n|}{n^2} [/tex]

    then i would try a comparison with a p-series such as [itex] \sum_{n=1}^{\infty} \frac{1}{n^2} [/itex]
     
  4. May 8, 2005 #3
    The convergence of your series follows from Abel's test.
     
  5. May 8, 2005 #4
    maybe this is correct im not sure

    define a sequence [tex]An=\frac{\cos n}{n}[/tex]

    this sequence converges to 0
    [tex]\lim_{n \to \infty} An = 0[/tex]

    [tex]\frac{A1+A2+...+An}{n}=\sum_{n=1}^n \frac{cos n}{n^2}[/tex]
    [tex]\frac{A1+A2+...+An}{n}[/tex] is the arithmetical average of the sequence [tex]An[/tex]
    and it always converges to the same limit as the sequence [tex]An[/tex]
    so:
    [tex]\lim_{n \to \infty} An = \lim_{n \to \infty} \frac{A1+A2+...+An}{n} = \sum_{n=1}^\infty \frac{cos n}{n^2}=0[/tex]
     
  6. May 8, 2005 #5

    saltydog

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    Pretty conclusive Muzza. Thanks for the link to the Abel's test. :smile:
     
  7. May 8, 2005 #6

    Hurkyl

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    IF the respective limits exist.

    Consider the sequence: 0, 1, 0, 1, 0, 1, ...

    The arithmetic mean converges to 1/2, but the sequence clearly does not.
     
  8. May 8, 2005 #7

    matt grime

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    That is wrong, try n=2:

    (cos(1)+cos(2)/2)/2

    which is the LHS, is not equal to

    cos(1)+cos(2)/4

    You are forgetting n is just the variable of the summation, when you're using it as the upper limit as well.

    We could, if this were true, apply it with 1 in the numerator, instead, and you're claiming that the sum of 1/n^2 is zero, when it is clearly not.
     
  9. May 8, 2005 #8
    your right i didn't notice that.
    o well just ignore my post then :smile:
     
  10. May 8, 2005 #9

    Alkatran

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    Let's try for absolute convergence.
    [tex]
    \sum_{n=1}^\infty abs(\frac{cos n}{n^2})
    [/tex]
    is smaller than (comparison test)
    [tex]
    \sum_{n=1}^\infty abs(\frac{1}{n^2})
    [/tex]
    which equals
    [tex]
    \sum_{n=1}^\infty \frac{1}{n^2}
    [/tex]
    which converges by a p-series test.
     
  11. May 8, 2005 #10
    WHOOPS! alketran beat me to it. never mind... :blushing:
     
  12. May 8, 2005 #11

    mathwonk

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    as courant makes so clear, almost the only way to prove convergence (except for the very special alternating test) is to prove absolute convergence by comparison.

    there are essentially only two things to compare with, either the geometric series, or an integral.

    this one follows immediately from comparison with the integral of x^(-2).

    nuff sedd.
     
  13. May 10, 2005 #12

    shmoe

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    For Abel's test the most obvious breakdown will probably involve showing the sums [itex]\sum_{n=1}^{N}\cos n[/itex] are bounded for all N. This may be difficult so a little hint: Euler's formula will reduce this to bounding a geometric series. (showing absolute convergence with a comparison test is probably more straightforward in this case though, and actually yields a stronger result).
     
  14. May 10, 2005 #13

    mathwonk

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    hellooo.... this series has terms bounded in absolute value by 1/n^2, hence converges absolutely by comparison with the integral of 1/x^2.
     
  15. May 10, 2005 #14

    shmoe

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    Of course, I just wanted to clarify Abel's test a little bit as applying it here might be beyond the scope of your typical first yeat calculus course (not that I know the OP's background).
     
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