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

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SUMMARY

The discussion focuses on identifying the positive values of "b" for which the series summation of b^log(n) converges. Participants agree that applying the ratio test is effective, leading to the conclusion that the valid range for b is 0 < b < 1. The challenge arises when attempting to incorporate the limsup condition, which complicates the analysis and introduces contradictions. The key takeaway is that understanding the behavior of the series through logarithmic properties is essential for determining convergence.

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  • Basic knowledge of limits and limsup concepts in mathematical analysis.
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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".
 
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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.
 


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?
 
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