1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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]

    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
    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


    User Avatar
    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


    User Avatar
    Homework Helper

    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

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


    [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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook