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Struggling to show that this series converges.

  1. Apr 17, 2009 #1
    Hi there,
    I'm trying to show that [tex]\sum\frac{log(r)}{r^{a}}[/tex] (r going from 1 to infinity) converges using the ratio test, but I can't seem to deal with the log(r+1)/log(r) term.
    Any help would be much appreciated.

    Thanks.
     
  2. jcsd
  3. Apr 17, 2009 #2
    Is it not sufficient to note that r^a/(r+1)^a is a null sequence, so anything it is multiplied by will also be a null sequence? (null x bounded = null...etc).
     
  4. Apr 17, 2009 #3
    i would use this (it is supposed that your series goes from n=1 to infinite

    integral criteria , if the integral [tex] \int _{1}^{\infty} dx log(x)x^{-a} [/tex] converges so does your series

    on the other hand an useful trick is to use the expansion of log (r+1) for big r

    log(r+1)=log(r)+log(1+1/r) and since 1/r is small you can expand it into a/r+b/r*r and so on
     
  5. Apr 17, 2009 #4
    It depends on your value of a. If a is negative, the series diverges. If 0<= a <= 1 then for r >=3 we have [tex] \frac{1}{r^a} < \frac{ln(r)}{r^a} [/tex] and so [tex] \sum\frac{1}{r^{a}} < \sum\frac{ln(r)}{r^{a}} [/tex]. The left side of the inequality diverges (because 0<= a <= 1) and so the series on the right side diverges.

    I haven't thought about a > 1 yet. I'll get back to that when I get back home.
     
  6. Apr 17, 2009 #5
    I should have mentioned, I was considering a>1. Thanks.
     
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