# Values for b such that sum over b^log n converges

1. Dec 11, 2008

### math8

What are all the positive values of "b" for which the series summation(b^log n) converges.

I think we must find the values of b for which the sequence of partial sums form a bounded sequence. But it gets complicated when I try to solve the inequality
"summation of k from 1 to n of (b^log k)< or equal to some M".

2. Dec 11, 2008

### HallsofIvy

Staff Emeritus
Re: series

Since those are all positive terms, I would be more inclined to use the ratio test: Look at b^log(n)/b^log(n+1) as n goes to infinity. You can use exponential and logarithm properties to simplify that.

3. Dec 11, 2008

### math8

Re: series

Thanks HallsofIvy, I tried that and I got 0<b<1 by just solving b^log(n)/b^log(n+1) <1. I am not sure how to introduce the limsup part. I mean, when I try to solve limsup [b^log(n)/b^log(n+1)] <1; I get lim as n-->inf of b^log(n+1/n) <1 which gives me b^ln1<1 and that gives me 1<1.
How do I get rid of that contradiction?

Last edited: Dec 11, 2008