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Another series problem

  1. Jan 24, 2007 #1
  2. jcsd
  3. Jan 24, 2007 #2

    Gib Z

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    Each term gets smaller and smaller, and converges to zero. It is absolutely convergent.

    The ratio test will tell you it converges as well.
     
    Last edited: Jan 24, 2007
  4. Jan 24, 2007 #3

    HallsofIvy

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    Is that what you mean to say? Each term of
    [tex]\Sigma_{n\rightarrow \infty}\frac{1}{n}[/itex]
    "gets smaller and smaller, and converges to zero" but the series doesn't converge at all.
     
  5. Jan 24, 2007 #4

    mjsd

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    when it is an alternating series you can use Leibniz test
    your pic is not very clear... but my guess is that the hint is to help you establish one of the condition in the Leibniz test namely, the terms are getting smaller

    Leibniz test:
    If [tex]\sum_1^{\infty} (-1)^{n+1} b_n[/tex] such that all [tex]b_n>0[/tex] (ie alternating series) and [tex]b_{n+1} < b_n\; \forall\,n[/tex] and [tex]b_n\rightarrow 0[/tex], then series converges to S and [tex]|S-S_k|\leq b_{k+1}[/tex]
     
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