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Radius of convergence

  1. Dec 9, 2007 #1
    Radius of convergence, interval of convergence

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

    Find the radius of convergence and the interval of convergence of the following series.

    a) [tex]\sum_{n=0}^\infty \frac{x^n}{(n^2)+1}[/tex]

    c) [tex]\sum_{n=2}^\infty \frac{x^n}{ln(n)}[/tex]

    e) [tex]\sum_{n=1}^\infty \frac{n!x^n}{n^2}[/tex]

    f) [tex]\sum_{n=1}^\infty \frac{n!2^nx^n}{n^n}[/tex]

    2. Relevant equations

    I think I use the Ratio test:

    [tex]\lim_{n\rightarrow\infty} \frac{a_{n+1}}{a_n} = L < 1[/tex]

    3. The attempt at a solution

    a) Let [tex]{a_n} = \frac{x^n}{(n^2)+1}[/tex]
    So:

    [tex]\lim_{n\rightarrow\infty} \frac{a_{n+1}}{a_n} = \lim_{n\rightarrow\infty} \frac{x^(n+1)}{((n+1)^2)+1} * \frac{n^2+1}{x^n} = \lim_{n\rightarrow\infty} \frac{x*n^2+1}{n^2+2n+2} = \lim_{n\rightarrow\infty} x[/tex]
    Radius: 1, Interval of convergence (-1, 1)

    c) Let [tex]{a_n} = \frac{x^n}{ln(n)}[/tex]
    So:

    [tex]\lim_{n\rightarrow\infty} \frac{a_{n+1}}{a_n} = \lim_{n\rightarrow\infty} \frac{x^(n+1)}{ln(n+1)} * \frac{ln(n)}{x^n} = \lim_{n\rightarrow\infty} \frac{x*ln(n)}{ln(n+1)} = \lim_{n\rightarrow\infty} x[/tex]
    Radius: 1, Interval of convergence (-1, 1)

    e) Let [tex]{a_n} = \frac{n!x^n}{n^2}[/tex]
    So:

    [tex]\lim_{n\rightarrow\infty} \frac{a_{n+1}}{a_n} = \lim_{n\rightarrow\infty} \frac{(n+1)!x^(n+1)}{(n+1)^2} * \frac{n^2}{n!x^n} = \lim_{n\rightarrow\infty} \frac{n^2x}{n+1} = \lim_{n\rightarrow\infty} x[/tex]
    Radius: 1, Interval of convergence (-1, 1)

    f) Let [tex]{a_n} = \frac{n!2^nx^n}{n^n}[/tex]
    So:

    [tex]\lim_{n\rightarrow\infty} \frac{a_{n+1}}{a_n} = \lim_{n\rightarrow\infty} \frac{n+1!2^(n+1)x^(n+1)}{(n+1)^(n+1)} * \frac{n^n}{n!2^nx^n} = \lim_{n\rightarrow\infty} 2x * (\frac{n}{n+1})^n = \lim_{n\rightarrow\infty} 2x[/tex]
    Radius: 1/2, Interval of convergence (-0.5, 0.5)
     
    Last edited: Dec 9, 2007
  2. jcsd
  3. Dec 9, 2007 #2

    dynamicsolo

    User Avatar
    Homework Helper

    The radii for three of these seem alright; you may wish to check your ratio in part (e). A number of these series converge for at least one of the endpoints of their intervals of convergence.
     
  4. Dec 9, 2007 #3
    I thought it seemed weird that three of the radii were 1, thanks for checking them over :)

    May I inquire which part of (e) you think I might have stumbled? Is it from:
    [tex]\lim_{n\rightarrow\infty} \frac{(n+1)!x^(n+1)}{(n+1)^2} * \frac{n^2}{n!x^n} = \lim_{n\rightarrow\infty} \frac{n^2x}{n+1}[/tex]

    or

    [tex]\lim_{n\rightarrow\infty} \frac{n^2x}{n+1} = \lim_{n\rightarrow\infty} x[/tex]

    Will check the endpoints in a bit, thanks again.
     
  5. Dec 9, 2007 #4

    cristo

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    Staff Emeritus
    Science Advisor

    Just a note on LaTex: if you want exponents to show up you need to enclose them in curly brackets; i.e. x^{n+1}.
     
  6. Dec 9, 2007 #5
    Oh thanks. I'm still working my way around with LaTeX.
     
  7. Dec 9, 2007 #6

    dynamicsolo

    User Avatar
    Homework Helper

    This is not an equation:

    [tex]\lim_{n\rightarrow\infty} \frac{n^2x}{n+1} = \lim_{n\rightarrow\infty} x[/tex]

    The left-hand side is correct. I believe you should find the radius of convergence for this series to be R = 0. (In fact, the original general term for this series should make one suspicious...)
     
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