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Sequence Abel-Summation

  1. Nov 10, 2012 #1
    1. The problem statement, all variables and given/known data

    Show that the following sequence [itex]\sum\limits_{n=1}^\infty \frac{1}{n^\alpha} [/itex]
    for all real [itex]\alpha > 1[/itex] converges and for all real [itex]\alpha \leq 1[/itex]
    diverges.

    3. The attempt at a solution
    All I know is that the Abel-Summation is the only useful thing here, but I got no clue how to use it the right way and I heard that the common criteria won´t work.

    I would be thankful for any hints or clues to get this proof running. Thank you in advance.
     
    Last edited by a moderator: Feb 6, 2013
  2. jcsd
  3. Nov 10, 2012 #2

    micromass

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    Re: Sequence

    First of all, this is a series and not a sequence.

    Second. Have you seen things like the integral test?? Can you list all the tests which you've seen and think are useful?
     
  4. Nov 10, 2012 #3

    Zondrina

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    Homework Helper

    Re: Sequence

    Break it up into cases. There are 3 cases for you to consider here.
     
  5. Nov 10, 2012 #4
    Re: Sequence

    Thanks micromass and zondrina for your quick help

    I´m sorry for the misconception. I meant series, sorry about that. Well our teacher told us/me that the quotient and/or root-test aren´t helpful here because the first one shows convergence and the second one divergence. He then talked about the summation by parts (Abel) and proved it with Weierstraß. I´m not sure if I´m allowed to use the integral-test on that because newer proofing techniques are forbidden until we have reached and proved them. I already got comfortable with sequences but the pace is so fast that we haven´t even got taught on series and now we gotta prove something which is so new. I apologize for my bad English but I´m from abroad.
     
  6. Nov 10, 2012 #5
    Re: Sequence

    Is the integral test the same like the cauchy-criterion?
     
  7. Nov 10, 2012 #6

    Zondrina

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    Re: Sequence

    The integral test allows you to test for the convergence of your series as long as f(n) is a monotone decreasing function ( Since you're starting at n=1 you don't have much to worry about ).
     
  8. Nov 11, 2012 #7
    Re: Sequence

    Hi Zondrina and all the other helpers/readers. I used the Cauchy-Condensationtest:

    For convergence:

    [itex]s_n = \sum\limits_{n=1}^\infty a_1+a_2+a_3+...+a_n [/itex]

    with the estimate

    [itex]s_N=a_2+a_2+a_4+a_4+a_4+a_4+a_8+...+N*a_N
    =2a_2+4a_4+...+N*a_N
    =2(a_2+2a_4+...+ (\frac N2)*a_N) [/itex]

    with

    [itex]N= 2^k[/itex]


    [itex]s_{2^k} =2(a_2)+2(a_4)+...+(2^{k-1})*a_k)=\sum\limits_{k=1}^n (a^k)a_{2^k}[/itex]

    For divergence:

    [itex]t_n = \sum\limits_{k=1}^n \frac 1k [/itex]

    With the comparison test
    [itex]s_n \leq t_n [/itex]

    [itex]\sum\limits_{k=1}^n (a^k)a_{2^k} \leq \sum\limits_{k=1}^n \frac 1k [/itex]

    mit [itex]a_{2^k} = \frac 1k[/itex]

    The expression [itex] \lim_{n \to \infty} \sum\limits_{k=1}^n 1 [/itex] goes to infinity, so [itex]\sum\limits_{k=1}^n \frac 1k [/itex] goes as well to infinity.


    Is this correct???
     
  9. Nov 12, 2012 #8
    Re: Sequence

    Would be cool if some1 could check it. If I´m close to the solution. Thank you.
     
  10. Nov 13, 2012 #9
    Re: Sequence

    Okay well thanks for your help
     
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