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Radius of Convergent

  1. Dec 8, 2007 #1
    [Solved] Radius of Convergent

    1. The problem statement, all variables and given/known data

    Find the radius of convergent for [tex] \sum_{n=1}^\infty (1-2^n)(ln(n))x^n [/tex]

    2. Relevant equations

    [tex]\frac {1}{R} = L = \lim \frac{a_{n+1}}{a_n} [/tex]

    3. The attempt at a solution

    [tex] lim \frac {(1-2^{n+1})(ln(n+1)}{(1-2^n)(ln(n))} = L [/tex]

    [tex] lim \frac {(1-2^n)(ln(n))}{(1-2^{n+1})(ln(n+1))} = R [/tex]

    I'm dizzy looking at this but how can I find:

    [tex] \lim_{n\rightarrow\infty} R [/tex]
    Last edited: Dec 8, 2007
  2. jcsd
  3. Dec 8, 2007 #2
    my idea would be to take the quotient of the logs as 1 because the derivatives both go as 1/x and so both functions should behave the same in infinity.
    The ones up front can be thrown away don´t make any contribution at infinity so you´re left with 2^n/2^n+1 = 1/2 = R :)

    No make it a bit more rigorous if you like :)
  4. Dec 8, 2007 #3
    [tex] lim\frac{1-2^n}{1-2^{n+1}} * \frac{ln(n)}{ln(n+1)} = \frac{2^n}{2^{n+1}} = 2^{n-(n+1)} = 2^{-1} = \frac{1}{2} [/tex]

    Indeed this is right, I've checked the results and it is a hit!

    But I don't underssand why you can just skip the log parts?

    Ahh, now when I think about I see that [tex] lim \frac{ln(n)}{ln(n+1)} = 1 [/tex] so that cancels out!

    I usually get stuck with the obvious.

    Thanks mr. Brown!
  5. Dec 8, 2007 #4
    you could use l´hospital on the logs to make it rigorous be the derivative quotient would be
    n^-1/(n+1)^(-1)=1+1/n goes to n as n goes to infinity :)
  6. Dec 8, 2007 #5
    Yeah I know but this lim is just so much for the eye to solve :)
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