Struggling to show that this series converges.

[Set Abominae]

Hi there,
I'm trying to show that \sum\frac{log(r)}{r^{a}} (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.

 

[slippers]

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).

 

[zetafunction]

i would use this (it is supposed that your series goes from n=1 to infinite

integral criteria , if the integral \int _{1}^{\infty} dx log(x)x^{-a} 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

 

[JG89]

It depends on your value of a. If a is negative, the series diverges. If 0<= a <= 1 then for r >=3 we have \frac{1}{r^a} &lt; \frac{ln(r)}{r^a} and so \sum\frac{1}{r^{a}} &lt; \sum\frac{ln(r)}{r^{a}}. 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.

 

[Set Abominae]

It depends on your value of a. If a is negative, the series diverges. If 0<= a <= 1 then for r >=3 we have \frac{1}{r^a} &lt; \frac{ln(r)}{r^a} and so \sum\frac{1}{r^{a}} &lt; \sum\frac{ln(r)}{r^{a}}. 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.