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How to prove the convergence?

  1. Dec 3, 2004 #1
    Hi,

    I can't find a way to prove the convergence of the following sum regarding to parameter a:

    [tex]
    \sum_{n=1}^{ \infty } a_{n}
    [/tex]

    where

    [tex]
    a_{n} = \frac { \sqrt{1+ \frac{1}{n}} - 1}{n^{a}}
    [/tex]

    I already proved the neccessary condition for convergence, ie that

    [tex]
    \lim_{n \rightarrow \infty} a_{n} = 0
    [/tex]

    And it showed that a must be in [itex](0, \infty) [/itex].

    But I can't figure out how to prove the convergence. I tried d'Alembert's criterion, comparing criterion, limite comparing criterion but no gave me some useful result (with d'Alembert I got very complicated expression I wasn't able to simplify).

    And one more question: in school I didn't understand, whether there is a equivalency in Abel-Dirichlet's criterion for convergence. I mean if neither condition of the theorem is passed, could we say that the sum diverges?

    Thank you.
     
  2. jcsd
  3. Dec 3, 2004 #2

    arildno

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    Well, but:
    [tex]a_{n}=\frac{1}{(1+\sqrt{1+\frac{1}{n}})n^{(a+1)}}\leq\frac{1}{n^{(a+1)}}[/tex]

    Hence, each [tex]a_{n}[/tex] is bounded by a term in a convergent series.
     
    Last edited: Dec 3, 2004
  4. Dec 3, 2004 #3

    shmoe

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    To apply Abel-Dirichlet's criterea you have to break the terms of your series [tex]a_n[/tex], into a product [tex]a_n=b_{n}c_{n}[/tex]. Depending on how you choose the b and c sequences will affect whether the conditions are satisfied. If you find one such decomposition that works, you have convergence. If you find one that fails, it doesn't mean that they will all fail.
     
  5. Dec 3, 2004 #4
    Clear and simple.. Thank you very much arildno.

    Btw isn't there any list of most common series types and corresponding criterion, which is most usable in the situation?
     
  6. Dec 3, 2004 #5
    Thank you shmoe, that's exactly what I was asking for.
     
  7. Dec 3, 2004 #6

    arildno

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    If there is, it's not inside my head..:wink:
     
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