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A hard series Convergence?

  1. Oct 10, 2009 #1
    Hi all,

    A friend of mine asked me if i had any ideas about the following problem, i tackled it but with no success, so i thought i would post it here.
    It is not a homework problem, or a regular textbook problem.


    If we know that a series with positive terms :


    Then what can we say about the nature of the series:



  2. jcsd
  3. Oct 10, 2009 #2
    Try two examples:
    [tex] a_i = \frac{1}{i^2} [/tex]
    [tex] a_i = \frac{1}{i(\log (i+1))^2} [/tex]
  4. Oct 10, 2009 #3
    I appreciate your input. I should've thought about the second example><
  5. Oct 11, 2009 #4
    I'm not sure of how to show this but my mind tells me the second series converges.

    My reasoning is as follows:

    Since the first series converges you can find a p series that is relatively close to ai ( for example i^(1.06) from the p-series we know that 1/i^(p) converges for p >1).

    Assuming your original series if approximated by a p-series that conveges say 1/ (i^p) we can say that the second series is approximately sigma ( 1/ n^((p+1)/2)) ,which converges since P>1, and p+1 > 2 the second series becomes a p-series with p >1 also.

    Thus convergence can be established.

    Unfortunately I can not think of any convergence test that would establish convergence more clearly and nicely. There are probably some test out there that you can use to establish convegence. You may even be to establish some nice inequality that would make convergence more obvious and mathematically sound. I was thinking of the Cauchy -Schwartz but i don't see any immediate benefit of using it.
  6. Oct 11, 2009 #5
    This is a counter example:

    [tex] a_i = \frac{1}{i(\log (i+1))^2} [/tex]

    Since the first series converges while the second diverges.

    Case closed!
  7. Oct 12, 2009 #6


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    It's a pretty sophisticated counterexample, don't think it should have been obvious.
  8. Oct 12, 2009 #7
    It is a standard example for "the integral test" where ratio and root tests fail. If you don't know it, then probably your textbook will do "the integral test" later.
  9. Oct 14, 2009 #8
    Sorry I didn't look at that post.

    My mistake lol.

    I guess I was wrong about convergence.lol
    Last edited: Oct 14, 2009
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