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Thread: Radius of convergence

  1. #1
    Super Member Showcase_22's Avatar
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    Radius of convergence

    Find the radius of convergence of the power series $\displaystyle \sum a_nx^n$ for the following choices of $\displaystyle (a_n; n \geq 0).$

    $\displaystyle a_n=a^{n^2}$ for some constant $\displaystyle a \in (0, \infty).$
    So far I have this:

    $\displaystyle \sum a^{n^2}x^n=1+a^4x^2+a^9x^3+......$

    but $\displaystyle |a^{n^2}| \leq |a^n|$

    Therefore: $\displaystyle |\sum a^{n^2}x^n|=|1+a^4x^2+a^9x^3+... |\leq | \sum a^{n}x^n|=|1+(ax)^2+(ax)^3+...|$

    This will only converge if $\displaystyle a<1$. Hence the radius of convergence of $\displaystyle | \sum a^{n}x^n|=|1+(ax)^2+(ax)^3+...|$ is 1 (ie. $\displaystyle R=1$).

    So I have an upper bound, but I can't find a lower bound. Since x can be negative (i'm assuming from the question that $\displaystyle x \in \mathbb{R}$) I can't just make the original power series greater than 1!

    Any help would be great!
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  2. #2
    Moo
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    A Cute Angle Moo's Avatar
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    Hello,
    Quote Originally Posted by Showcase_22 View Post
    but $\displaystyle |a^{n^2}| \leq |a^n|$
    This is only true if a>1 !

    Then this inequality
    Therefore: $\displaystyle |\sum a^{n^2}x^n|=|1+a^4x^2+a^9x^3+... |\leq | \sum a^{n}x^n|=|1+(ax)^2+(ax)^3+...|$
    is false
    I don't know for the rest
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  3. #3
    Super Member Showcase_22's Avatar
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    ah, okay then.

    Well how about this:

    $\displaystyle a^{n^2}<1$ for the power series to converge.

    Hence:

    $\displaystyle n^2 ln(a)<0$

    $\displaystyle ln(a)<0$

    hence $\displaystyle a<1$.

    Therefore the radius of convergence is 1.

    $\displaystyle ie. \ R=1$.

    I would have to prove that the series diverges for $\displaystyle a \geq 1$ but I think I can do that.
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