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Thread: Radius of Convergence of Power Series

  1. #1
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    Question Radius of Convergence of Power Series

    Hello,

    I need a little bit help with the following power series. The exercise is to calculate the radius of convergence.

    $\displaystyle \sum\limits_{n=1}^{\infty} (\mathrm{ln}(n))^{n} x^{n} $

    I know the criterion of Euler and the stronger criterion of Hadamard.

    Euler:

    $\displaystyle r= \frac{\mid a_{n} \mid}{ \mid a_{n+1} \mid }$

    Hadamard:

    $\displaystyle r= \frac{1}{\mathrm{limsup} ~ \sqrt[n]{\mid a_{n} \mid} } $

    and of course $\displaystyle a_{n}$ is

    $\displaystyle a_{n} = (\mathrm{ln(n)})^{n}$

    My problem now, is that I don't know how to calculate this expressions with this logarithm

    Thanks for help
    Last edited by Besserwisser; Jan 26th 2010 at 01:25 AM.
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  2. #2
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    Quote Originally Posted by Besserwisser View Post
    Hello,

    I need a little bit help with the following power series. The exercise is to calculate the radius of convergence.

    $\displaystyle \sum\limits_{n=1}^{\infty} (\mathrm{ln})^{n} x^{n} $

    I know the criterion of Euler and the stronger criterion of Hadamard.

    Euler:

    $\displaystyle r= \frac{\mid a_{n} \mid}{ \mid a_{n+1} \mid }$

    Hadamard:

    $\displaystyle r= \frac{1}{\mathrm{limsup} ~ \sqrt[n]{\mid a_{n} \mid} } $

    and of course $\displaystyle a_{n}$ is

    $\displaystyle a_{n} = \mathrm{ln}^{n}$

    My problem now, is that I don't know how to calculate this expressions with this logarithm

    Thanks for help
    Is it $\displaystyle \sum_{n = 1}^{\infty}(\ln{x})^nx^n$?
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  3. #3
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    Quote Originally Posted by Prove It View Post
    Is it $\displaystyle \sum_{n = 1}^{\infty}(\ln{x})^nx^n$?
    I made typing errors. Sorry. The Power Seris is

    $\displaystyle \sum_{n = 1}^{\infty}(\mathrm{ln}(n))^{n} x^{n}$

    and

    $\displaystyle a_{n} = (\mathrm{ln}(n))^{n} $
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  4. #4
    MHF Contributor chisigma's Avatar
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    What is the neccesary condition for the convergence of an 'infinite sum'?...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  5. #5
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    Quote Originally Posted by Prove It View Post
    Is it $\displaystyle \sum_{n = 1}^{\infty}(\ln{x})^nx^n$?
    But that would not be a power series.
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Besserwisser View Post
    Hello,

    I need a little bit help with the following power series. The exercise is to calculate the radius of convergence.

    $\displaystyle \sum\limits_{n=1}^{\infty} (\mathrm{ln}(n))^{n} x^{n} $

    I know the criterion of Euler and the stronger criterion of Hadamard.

    Euler:

    $\displaystyle r= \frac{\mid a_{n} \mid}{ \mid a_{n+1} \mid }$

    Hadamard:

    $\displaystyle r= \frac{1}{\mathrm{limsup} ~ \sqrt[n]{\mid a_{n} \mid} } $

    and of course $\displaystyle a_{n}$ is

    $\displaystyle a_{n} = (\mathrm{ln(n)})^{n}$

    My problem now, is that I don't know how to calculate this expressions with this logarithm

    Thanks for help
    For it to converge you clearly need $\displaystyle \ln^n(n)\cdot x^n=\left(x\ln(n)\right)^n\to0$. For this to happen you need only prove that $\displaystyle x\ln(n)$ goes to zero for some $\displaystyle x$..but...there is a problem with that.
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