Struggling to show that this series converges.

  • #1

Main Question or Discussion Point

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.
 

Answers and Replies

  • #2
3
0
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).
 
  • #3
391
0
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
 
  • #4
726
1
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.
 
  • #5
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.
I should have mentioned, I was considering a>1. Thanks.
 

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