Divergent series

  1. Hello, I'm working on this problem:
    Prove that for any real x, the series SUM n=2 to infinity of 1/(log n)^x diverges.

    So far, I have applied the test that says that if SUM 2^n*a_2n converges then the series converges. I got:

    1/log2*SUM 2^n/n^x I know that 1/n^x converges if x>1, but 2^n will explode, so I'd have convergent (sometimes) times divergent = divergent. Can I do that?

    Thanks,
    CC
     
  2. jcsd
  3. quasar987

    quasar987 4,774
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    No, that is not a propery of series. A counter exemple is

    [tex]\sum \frac{1}{n}\frac{1}{n^2}[/tex]

    In this case, we have divergent times convergeant = convergeant.

    What you can do to prove the desired result however, is simply show that for all x in R, the limit as n-->infty of the argument of the serie is not 0. Hence it does not converges, hence it diverges. (You'll have to do the 3 separate cases x<0, x=0 and x>0 [the later being the most difficult]).
     
  4. I tried it using the ratio test. I got
    [tex]\2(\frac{n}{n+1})^x[\tex]
    and the limit of that is 2>1 so it diverges.
    can I do it that way?
    CC
     
  5. let me try again...I'm still learning latex....
    [tex]\mbox{2}(\frac{n}{n+1})^x[/tex]
    there. And the limit of THAT is 2.
    Is this correct?
     
  6. NateTG

    NateTG 2,537
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    Yes.
    [tex]\lim_{n \rightarrow \infty} 2 \left( \frac{n}{n+1}\right)^x=2[/itex]

    But
    [tex] \frac{(\log n)^x}{(\log n+1)^x} \neq 2 \left( \frac{n}{n+1}\right)^x [/tex]
     
    Last edited: Nov 10, 2005
  7. quasar987

    quasar987 4,774
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    You should do it my way. It's much more instructive then mechanically using a test anyway.
     
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