For which positive real numbers a does the series converge

[.d9n.]

Homework Statement

For which positive real numbers a does the series

Ʃo→∞ a^log(n)

converge.

Here logarithms are to the base e

Homework 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

The Attempt at a Solution

 

[.d9n.]

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.]

thanks i'll have a look at that

 

[SammyS]

Homework Statement

For which positive real numbers a does the series

Ʃo→∞ a^log(n)

converge.

Here logarithms are to the base e

Homework 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

The Attempt at a Solution

Hello .d9n. . Welcome to PF !

Write a as e^log(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.]

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.]

is it
sum 1 to infinity of e^log(a^n)= sum 1 to infinity of a^n?
or is that just rubbish?

 

[SammyS]

Perhaps I should have said properties of exponents and logarithms.

\displaystyle \left(b^k\right)^t=b^{kt}=\left(b^t\right)^k

 

[Dick]

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.

 

[.d9n.]

Thanks so i am left with

sum 1 to infinity n^(log (a))

correct?
In regards to it converging I am assuming a must be between 1 and 9 for this to converge as log1=0 and log9=0.954..., bigger than 9 then it won't converge but increase exponentially?

 

[Dick]

Thanks so i am left with

sum 1 to infinity n^(log (a))

correct?
In regards to it converging I am assuming a must be between 1 and 9 for this to converge as log1=0 and log9=0.954..., bigger than 9 then it won't 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'.

 

[SammyS]

\displaystyle\sum_{n=1}^\infty\ n^{p} converges for what values of p ?

 

[.d9n.]

so ∑n=1,∞ n^p converges when -1<p<1 with p≠0, is that correct?

 

[.d9n.]

so therefore a=-9,-8,-7,-6,-5,-4,-3,-2,2,3,4,5,6,7,8,9?

 

[Dick]

so ∑n=1,∞ n^p converges when -1<p<1 with p≠0, is that correct?

No, that doesn't sound correct.

 

[.d9n.]

is it nearly correct?
i don't think it will converge if bigger than 1 or -1 as it will increase exponentially, so it must be between them?

 

[SammyS]

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?

\displaystyle\sum_{n=1}^\infty\ \frac{1}{\ n^r}

 

[.d9n.]

does it converge if p<-1

 

[.d9n.]

so it converges when r>1?

 

[.d9n.]

if r>1 which would make -p=r, what values for a would make log(a) negative?

 

[Dick]

does it converge if p<-1

That sounds good. So you want log(a)<(-1), right? Solve that for a. What's the inverse function to a log? And I don't think they mean log to the base 10.

 

[.d9n.]

so -1<a<1 a=/0,
say y=log(a) then the inverse is a=e^y ?

 

[Dick]

so -1<a<1 a=/0,
say y=log(a) then the inverse is a=e^y ?

No again to the first line. But yes to the second. e^(log(a))=a. Take log(a)<(-1) and exponentiate both sides of the inequality.

 

[.d9n.]

so a<e^-1= a<0.367879441?

 

[Dick]

so a<e^-1= a<0.367879441?

Now you've got it. And you also need 0<a. Otherwise the log (or your original series) isn't even defined.

 

[.d9n.]

so sum n=1 to infinity of a^log(n) converges when 0<a<0.3678...

 

[Dick]

so sum n=1 to infinity of n^log(a) converges when 0<a<0.3678...

Yes, but I'd write e^(-1) or 1/e instead of 0.3678... There's no need to replace an exact number with an approximation.

 

[.d9n.]

ok, brilliant. thanks for all your help, much appreciated