For which positive real numbers a does the series converge


by .d9n.
Tags: converge, numbers, positive, real, series
.d9n.
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#1
Jan26-12, 07:55 AM
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
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.d9n.
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#2
Jan26-12, 08:07 AM
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?
.d9n.
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#3
Jan26-12, 08:08 AM
P: 61
thanks i'll have a look at that

SammyS
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#4
Jan26-12, 08:09 AM
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For which positive real numbers a does the series converge


Quote Quote by .d9n. View Post
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.
.d9n.
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#5
Jan26-12, 08:29 AM
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
.d9n.
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#6
Jan26-12, 08:51 AM
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?
SammyS
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#7
Jan26-12, 09:15 AM
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Perhaps I should have said properties of exponents and logarithms.

[itex]\displaystyle \left(b^k\right)^t=b^{kt}=\left(b^t\right)^k[/itex]
Dick
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#8
Jan26-12, 09:15 AM
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Quote Quote by .d9n. View Post
is it
sum 1 to infinity of e^log(a^n)= sum 1 to infinity of a^n?
or is that just rubbish?
Rubbish. (e^log(a))^log(n)=e^(log(a)*log(n))=(e^log(n))^log(a). Think about that.
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#9
Jan31-12, 06:27 AM
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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#10
Jan31-12, 09:16 AM
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Quote Quote by .d9n. View Post
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'.
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#11
Jan31-12, 10:37 AM
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[itex]\displaystyle\sum_{n=1}^\infty\ n^{p}[/itex] converges for what values of p ?
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#12
Feb1-12, 05:33 AM
P: 61
so ∑n=1,∞ n^p converges when -1<p<1 with p≠0, is that correct?
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#13
Feb1-12, 05:45 AM
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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#14
Feb1-12, 08:54 AM
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Quote Quote by .d9n. View Post
so ∑n=1,∞ n^p converges when -1<p<1 with p≠0, is that correct?
No, that doesn't sound correct.
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#15
Feb1-12, 09:02 AM
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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#16
Feb1-12, 09:07 AM
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Quote Quote by .d9n. View Post
so ∑n=1,∞ n^p converges when -1<p<1 with p≠0, is that correct?
No.

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

For what value of r, will the following converge?

[itex]\displaystyle\sum_{n=1}^\infty\ \frac{1}{\ n^r}[/itex]
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#17
Feb1-12, 09:12 AM
P: 61
does it converge if p<-1
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#18
Feb1-12, 09:13 AM
P: 61
so it converges when r>1?


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