# For which positive real numbers a does the series converge

by .d9n.
Tags: converge, numbers, positive, real, series
 P: 61 1. The problem statement, all variables and given/known data For which positive real numbers a does the series Ʃo→∞ a^log(n) converge. Here logarithms are to the base e 2. Relevant equations Im afraid I'm not sure where to start, I'm not sure which topics would be applicable to this question. If someone could point me in the right direction or give me a clue, it would be much appreciated. tthanks 3. The attempt at a solution
 P: 61 Is it possibly between -1 and 1? As log(n) will keep getting larger as n does. So if a<-1 and a>1 then it will keep increasing in size so therefore it must be -1<=a=<1? If this is right how would I go about proving it?
 P: 61 thanks i'll have a look at that
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P: 7,404

## For which positive real numbers a does the series converge

 Quote by .d9n. 1. The problem statement, all variables and given/known data For which positive real numbers a does the series Ʃo→∞ a^log(n) converge. Here logarithms are to the base e 2. Relevant equations Im afraid I'm not sure where to start, I'm not sure which topics would be applicable to this question. If someone could point me in the right direction or give me a clue, it would be much appreciated. tthanks 3. The attempt at a solution
Hello .d9n. . Welcome to PF !

Write a as elog(a). Then using properties of logarithms you can show that this is a geometric series.

BTW: The summation needs to be from 1 to ∞ . log(0) is undefined.
 P: 61 so you mean sum 1 to infinity e^(log(a))^(log(n)) I'm not sure what property i am trying to find. is there any need to change log(n) to the integral from 1 to n of 1/x? Or is there something to simplify log(a)^log(n). Otherwise i don't see the benefit of changing a to e^log(a). thanks
 P: 61 is it sum 1 to infinity of e^log(a^n)= sum 1 to infinity of a^n? or is that just rubbish?
 Emeritus Sci Advisor HW Helper PF Gold P: 7,404 Perhaps I should have said properties of exponents and logarithms. $\displaystyle \left(b^k\right)^t=b^{kt}=\left(b^t\right)^k$
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Thanks
P: 25,171
 Quote by .d9n. is it sum 1 to infinity of e^log(a^n)= sum 1 to infinity of a^n? or is that just rubbish?
 P: 61 Thanks so i am left with sum 1 to infinity n^(log (a)) correct? In regards to it converging im assuming a must be between 1 and 9 for this to converge as log1=0 and log9=0.954..., bigger than 9 then it wont converge but increase exponentially?
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Thanks
P: 25,171
 Quote by .d9n. Thanks so i am left with sum 1 to infinity n^(log (a)) correct? In regards to it converging im assuming a must be between 1 and 9 for this to converge as log1=0 and log9=0.954..., bigger than 9 then it wont converge but increase exponentially?
You aren't thinking very clearly about this. If a=1 then log(1)=0 and the series is n^0=1. I.e. 1+1+1+... That does not converge. Try some more values. Don't forget values of a<1. You should start to realize this is a p-series. And usually when you write log, without indicating a base then you assume the base is 'e'.
 Emeritus Sci Advisor HW Helper PF Gold P: 7,404 $\displaystyle\sum_{n=1}^\infty\ n^{p}$ converges for what values of p ?
 P: 61 so ∑n=1,∞ n^p converges when -1
 P: 61 so therefore a=-9,-8,-7,-6,-5,-4,-3,-2,2,3,4,5,6,7,8,9?
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Thanks
P: 25,171
 Quote by .d9n. so ∑n=1,∞ n^p converges when -1
No, that doesn't sound correct.
 P: 61 is it nearly correct? i dont think it will converge if bigger than 1 or -1 as it will increase exponentially, so it must be between them?
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P: 7,404
 Quote by .d9n. so ∑n=1,∞ n^p converges when -1
No.

Perhaps it's more familiar to you in the following form.

For what value of r, will the following converge?

$\displaystyle\sum_{n=1}^\infty\ \frac{1}{\ n^r}$
 P: 61 does it converge if p<-1
 P: 61 so it converges when r>1?

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