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Convergence/Divergence of an infinite series

  1. Oct 19, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex]\sum^{\infty}_{n=1} \frac{(-1)^{n}}{\sqrt{n}+(-1)^{n}}[/tex]

    Prove whether this series converges or diverges using the following analysis:
    Let [tex]b_{k}[/tex] be the sum of terms numbered n=2k-1 and n=2k from the given series.
    By simplifying [tex]b_{k}[/tex], determine if [tex]\sum^{\infty}_{k=1} b_{k}[/tex] converges or diverges.


    2. Relevant equations



    3. The attempt at a solution

    From what I have above, I think:

    [tex]b_{k} = \frac{(-1)^{2k-1}}{\sqrt{2k-1}+(-1)^{2k-1}} + \frac{(-1)^{2k}}{\sqrt{2k}+(-1)^{2k}}[/tex]

    which means,

    [tex]\sum^{\infty}_{k=1} b_{k}= \sum^{\infty}_{k=1} \frac{(-1)^{2k-1}}{\sqrt{2k-1}+(-1)^{2k-1}} + \sum^{\infty}_{k=1} \frac{(-1)^{2k}}{\sqrt{2k}+(-1)^{2k}}[/tex]

    I'm not sure where to go from here as far as simplifying goes. I tried adding the two fractions but I don't think that helps. Also, I can't use the Alternating series test on the original series because the absolute value of the terms is not monotone and decreasing.
     
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  3. Oct 19, 2009 #2

    Dick

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    If k is an integer, then (-1)^(2k)=1 and (-1)^(2k-1)=(-1). Yes, you can't use the alternating series test. But you can learn something by combining the two terms and looking at the asymptotic form of the limit.
     
  4. Oct 20, 2009 #3
    Would you mind clarifying what you said about the asymptotic form? I already tried adding the two terms but it looks a bit messy, and I'm lost as to what I should be looking for next.

    [tex]\sum b_{k} = \sum \frac{-1}{\sqrt{2k-1}-1} + \sum \frac{1}{\sqrt{2k}+1} = \sum \frac{\sqrt{2k-1}-\sqrt{2k}-2}{\sqrt{4k^{2}-2k}+\sqrt{2k-1}-\sqrt{2k}-1}
    [/tex]

    Thank you.
     
  5. Oct 21, 2009 #4

    Dick

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    Can you show lim k->infinity sqrt(2k-1)-sqrt(2k)=0? If you ignore those terms then I would say that what's left behaves like -1/k for large k. That's what I meant by asymptotic form. That suggests that the way to prove it is to show |b_k|>C/k for some C for large k. Can you do that?
     
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