- #1

- 15

- 0

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.

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- Thread starter Set Abominae
- Start date

- #1

- 15

- 0

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

- 3

- 0

- #3

- 391

- 0

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

I haven't thought about a > 1 yet. I'll get back to that when I get back home.

- #5

- 15

- 0

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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