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For which positive real numbers a does the series converge

  1. Jan 26, 2012 #1
    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
     
  2. jcsd
  3. Jan 26, 2012 #2
    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?
     
  4. Jan 26, 2012 #3
    thanks i'll have a look at that
     
  5. Jan 26, 2012 #4

    SammyS

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    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.
     
  6. Jan 26, 2012 #5
    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
     
  7. Jan 26, 2012 #6
    is it
    sum 1 to infinity of e^log(a^n)= sum 1 to infinity of a^n?
    or is that just rubbish?
     
  8. Jan 26, 2012 #7

    SammyS

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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]
     
  9. Jan 26, 2012 #8

    Dick

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    Rubbish. (e^log(a))^log(n)=e^(log(a)*log(n))=(e^log(n))^log(a). Think about that.
     
  10. Jan 31, 2012 #9
    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?
     
  11. Jan 31, 2012 #10

    Dick

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    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'.
     
  12. Jan 31, 2012 #11

    SammyS

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    [itex]\displaystyle\sum_{n=1}^\infty\ n^{p}[/itex] converges for what values of p ?
     
  13. Feb 1, 2012 #12
    so ∑n=1,∞ n^p converges when -1<p<1 with p≠0, is that correct?
     
  14. Feb 1, 2012 #13
    so therefore a=-9,-8,-7,-6,-5,-4,-3,-2,2,3,4,5,6,7,8,9?
     
  15. Feb 1, 2012 #14

    Dick

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    No, that doesn't sound correct.
     
  16. Feb 1, 2012 #15
    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?
     
  17. Feb 1, 2012 #16

    SammyS

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    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]
     
  18. Feb 1, 2012 #17
    does it converge if p<-1
     
  19. Feb 1, 2012 #18
    so it converges when r>1?
     
  20. Feb 1, 2012 #19
    if r>1 which would make -p=r, what values for a would make log(a) negative?
     
  21. Feb 1, 2012 #20

    Dick

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    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.
     
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