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Math Help - Finding interval/radius of convergence of power series

  1. #1
    Senior Member Pinkk's Avatar
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    Finding interval/radius of convergence of power series

    I'm fairly certain I have the correct answer, but I just want to verify.

    \sum_{k=1}^{\infty}\frac{k^{2}x^{k}}{2\cdot 4\cdot 6\cdot \cdot \cdot (2k)}

    So I used the ratio test:

    \lim_{k\to \infty}|\frac{a_{k+1}}{a_{k}}| = \lim_{k\to \infty}\frac{(k+1)^{2}|x|^{k+1}}{4\cdot 6\cdot 8\cdot \cdot \cdot (2k+2)}\cdot \frac{2\cdot 4\cdot 6\cdot \cdot \cdot (2k)}{k^{2}|x|^{k}}=2|x|\lim_{k\to \infty}\frac{k^{2}+2k+1}{2k^{3}+2k^{2}}<1

    Well, the limit goes to zero, so I end up with:

    2|x|\cdot 0 < 1

    0<1

    So that means the series is convergent for all values of x, so the radius of convergence is \infty and the interval of convergence is (-\infty,\infty).
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  2. #2
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    Absolutely Correct.
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  3. #3
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    Quote Originally Posted by Pinkk View Post
    I'm fairly certain I have the correct answer, but I just want to verify.

    \sum_{k=1}^{\infty}\frac{k^{2}x^{k}}{2\cdot 4\cdot 6\cdot \cdot \cdot (2k)}

    So I used the ratio test:

    \lim_{k\to \infty}|\frac{a_{k+1}}{a_{k}}| = \lim_{k\to \infty}\frac{(k+1)^{2}|x|^{k+1}}{4\cdot 6\cdot 8\cdot \cdot \cdot (2k+2)}\cdot \frac{2\cdot 4\cdot 6\cdot \cdot \cdot (2k)}{k^{2}|x|^{k}}=2|x|\lim_{k\to \infty}\frac{k^{2}+2k+1}{2k^{3}+2k^{2}}<1

    Well, the limit goes to zero, so I end up with:

    2|x|\cdot 0 < 1

    0<1

    So that means the series is convergent for all values of x, so the radius of convergence is \infty and the interval of convergence is (-\infty,\infty).
    Just a note about your ratio test.

    Note that

    \frac{k^2 + 2k + 1}{2k^3 + 2k^2} = \frac{(k + 1)^2}{2k(k + 1)}

     = \frac{k + 1}{2k}

     = \frac{k}{2k} + \frac{1}{2k}

     = \frac{1}{2} + \frac{1}{2k}


    As k \to \infty, the limit tends to \frac{1}{2}. So you're right about the ratio being less than 1 and so the series is convergent. Just looks better :P
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  4. #4
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    Hello, Pinkk!

    You're absolutely correct . . . nice work!


    \sum_{k=1}^{\infty}\frac{k^{2}x^{k}}{2\cdot 4\cdot 6\cdot \cdot \cdot (2k)}
    I would do it like this . . .


    Note that: . 2\cdot4\cdot6\hdots (2k) \;=\;2^k(1\cdot2\cdot3\hdots k) \;=\;2^k\,k!

    . . So we have: . \sum^{\infty}_{k=1}\frac{k^2x^k}{2^k\,k!}


    Ratio: . \left|\frac{(k+1)x^{k+1}}{2^{k+1}(k+1)!}\cdot\frac  {2^k\,k!}{k^2x^l}\right| \;\;=\;\;\left|\frac{2^o}{2^{k+1}}\cdot\frac{x^{k+  1}}{x^k}\cdot\frac{k!}{(k+1)!} \cdot\frac{k+1)^2}{k^2}\right|

    . . . = \;\;\frac{1}{2}\cdot|x|\cdot\frac{1}{k+1}\cdot\fra  c{(k+1)^2}{k^2} \;\;=\;\;\frac{1}{2}\cdot|x|\cdot\frac{k+1}{k^2}\;  \;=\;\;\frac{1}{2}\left(\frac{1}{k}+\frac{1}{k^2}\  right)|x|


    And: . \lim_{k\to\infty}\left[\frac{1}{2}\left(\frac{1}{k} + \frac{1}{k^2}\right)|x|\right] \;=\;0 \quad\hdots\;\text{ etc.}

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  5. #5
    Senior Member Pinkk's Avatar
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    @Prove It

    Isn't it actually:

    \lim_{k\to \infty}\frac{k+1}{k^{2}}=0

    If the limit somehow equaled \frac{1}{2}, then the radius of convergence would not be \infty.
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  6. #6
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    Quote Originally Posted by Pinkk View Post
    @Prove It

    Isn't it actually:

    \lim_{k\to \infty}\frac{k+1}{k^{2}}=0

    If the limit somehow equaled \frac{1}{2}, then the radius of convergence would not be \infty.
    Oops, you are right. It should have been \frac{k + 1}{2k^2}.

    So you're right, that limit is not \frac{1}{2}.

    Still tends to a value less than 1 though.

    \frac{k + 1}{2k^2} = \frac{k}{2k^2} + \frac{1}{2k^2}

     = \frac{1}{2k} + \frac{1}{2k^2}.


    So as k \to \infty the limit tends to 0.


    Still converges though.
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