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Is this series convergent?

  1. Jun 10, 2009 #1
    1. The problem statement, all variables and given/known data

    For what values of alpha is the following series convergent: [tex] \sum_{v=1}^{\infty} \frac{(-1)^{n-1}}{n^{\alpha}} = 1 - \frac{1}{2^{\alpha}} + \frac{1}{3^{\alpha}} + ... [/tex]

    2. Relevant equations



    3. The attempt at a solution

    For negative alpha and alpha = 0 the series obviously diverges, so we need only look at positive alpha. For alpha = 1, the infinite series for ln(2) arises. For alpha > 1, we see that the series is absolutely convergent and so the original series is also convergent.

    Now to prove that the series is divergent for 0 < alpha < 1, notice that [tex] -1/n \le \frac{(-1)^{n-1}}{n^{\alpha}} [/tex] and so [tex] \sum_{v=0}^{n} -1/n \le \sum_{v=0}^{n} \frac{(-1)^{n-1}}{n^{\alpha}} [/tex].

    [tex] \frac{(-1)^{n-1}}{n^{\alpha}} = -1 - 1/(2^{\alpha}) - 1/(3^{\alpha}) - ... - 1/(n^{\alpha}) = -(1 + 1/2^{\alpha} + ... + 1/(n)^{\alpha} ) [/tex]. The series in the brackets diverges for increasing n since alpha is less than 1, and so the entire series converges.

    Thus the original series converges only for alpha >= 1.

    Is this correct?
     
  2. jcsd
  3. Jun 10, 2009 #2
    Ah, I messed up my latex and I can't edit it. The inequality should actually be the sum from 0 to n of -1/n^alpha <= sum from 0 to n of {(-1)^(n-1)}/n^alpha

    And the series the series on the left hand side of the inequality is -(1 + 1/2^alpha + 1/3^alpha + ...), which diverges since 0 < alpha < 1
     
  4. Jun 10, 2009 #3

    Dick

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    Review the alternating series test.
     
  5. Jun 10, 2009 #4
    The test says that if the absolute value of the terms of an alternating series decrease monotonically to zero then the series converges.

    So let a_n be the nth term of the series. Then |a_n| = 1/n^alpha > |a_{n+1}| = 1/(n+1)^alpha

    The terms do not decrease monotonically to 0, but this doesn't mean that the series doesn't converge...so I'm still stuck...
     
  6. Jun 10, 2009 #5

    Office_Shredder

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    Review the definition of decreasing monotonically to zero
     
  7. Jun 10, 2009 #6
    Ha! I can't believe I made that mistake...the absolute value of the terms indeed do go to 0 and so the series converges
     
  8. Jun 10, 2009 #7

    Dick

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    You say |a_n|>|a_{n+1}|. That certainly looks like you are saying the series decreases. But that's not true for every alpha. Certainly not if, for example, alpha=(-1). For which alpha is that true?
     
  9. Jun 10, 2009 #8
    It is true for positive alpha
     
  10. Jun 10, 2009 #9

    Dick

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    Right.
     
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