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Nth-term Test of Series

  1. Oct 9, 2012 #1
    1. The problem statement, all variables and given/known data
    I attached the problem


    2. Relevant equations



    3. The attempt at a solution
    When I try to apply the nth-term test on the sequence, (-1)^n+1, I get an indeterminate form; so, I try to raise e to the ln[(-1^n+1)], but I run into the same problem. How do I take the limit, in this instance, in order for the nth-term test to decide anything?
     

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  3. Oct 9, 2012 #2

    HallsofIvy

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    What do you mean by the "nth term test"? One theorem says that if the terms do not go to 0, then the series cannot converge.
     
  4. Oct 9, 2012 #3
    Yes, but isn't (-1)^infinity an indeterminate form?
     
  5. Oct 9, 2012 #4
    You are correct that the alternating series of the form (-1)^n diverges.

    So, this series for sine is just a different form of an alternating series.


    Now, the nth-term test is a waste of time. This would be a last resort test, because there are so many other (more simplicities) ways to identify convergence/divergence.

    The nth-term test is when you right out the first 4 or 5 terms and the last few nth-terms.

    Example:
    sin(pi) + sin(3/2 pi) + sin(2 pi) + sin(5/2 pi) + ... + sin([2n-7]/2 pi) + sin([2n-5]/2 pi) + sin([2n-3]/2 pi) + sin([2n-1]/2 pi)

    Simplifying:
    0 - 1 + 0 + 1 + ... + sin([2n-7]/2 pi) + sin([2n-5]/2 pi) + sin([2n-3]/2 pi) + sin([2n-1]/2 pi)

    Therefore, per the nth-term test, the series oscillates between 1 and -1. Therefore, the given series diverges.
     
  6. Oct 9, 2012 #5
    Sorry. I guess I didn't show my last step.

    the series further simplifies to:
    -1 + 1 + ...
    Which is a basic alternating series of (-1)^(n)
    This series is a known to diverge. No further justifications needed
     
  7. Oct 9, 2012 #6

    jbunniii

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    That's not what I would call the nth term test. The usual meaning is described here:

    http://en.wikipedia.org/wiki/Term_test

    and is in fact the FIRST test you should apply when checking whether a series converges or not. Namely, if [itex]\lim_{n \rightarrow \infty} a_n \neq 0[/itex], then there's no way the series [itex]\sum_{n = 1}^{\infty} a_n[/itex] can converge, so you're wasting your time with other, more complicated tests.
     
  8. Oct 9, 2012 #7

    jbunniii

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    To answer Bashyboy's original question:

    No, an indeterminate form is something like [itex]0 / 0[/itex] or [itex]0 \cdot \infty[/itex].

    For [itex](-1)^{n+1}[/itex], it's simple to see that this sequence diverges: it oscillates between +1 and -1, so there can't be any value that all of the sequence terms will be near for arbitrarily large [itex]n[/itex].

    Another way to see this is that the sequence contains a subsequence (the odd values of [itex]n[/itex]) that converges to 1, and another subsequence (the even values of [itex]n[/itex]) that converges to -1. This cannot happen with a convergent sequence.

    A third way to see this is to look at the absolute value: [itex]|(-1)^{n+1}| = 1[/itex] for all [itex]n[/itex], whereas if the sequence converges to zero, so must its absolute value.
     
  9. Oct 9, 2012 #8

    Mark44

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    I heartily agree. It's very easy to apply this test, so if the test is inconclusive, you haven't wasted much time.
     
  10. Oct 9, 2012 #9
    So, could anyone explicitly show me how to apply the nth term test?
     
  11. Oct 9, 2012 #10

    jbunniii

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    I showed you three different, but equivalent, ways in my previous message.
     
  12. Oct 9, 2012 #11

    SammyS

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    [itex]\displaystyle \lim_{n\,\to\,\infty}\,(-1)^{n}[/itex] Does Not Exist, so the limit is certainly not zero.

    Therefore, [itex]\displaystyle \sum_{n=1}^\infty\,(-1)^{n\ \ }[/itex] diverges.
     
    Last edited: Oct 9, 2012
  13. Oct 9, 2012 #12
    Thank you.
     
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