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Math Help - Rational function from series?

  1. #1
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    Rational function from series?

    Given the series:

    f(z)=\sum_{n=2}^{\infty} \frac{n(n-1)}{3^n}z^{n-2}

    a) find the radius of convergence (got that, it's z>1/3)

    b) Prove that f is a rational function inside its circle of convergence

    c) find the rational number f(2)

    Well, I mess around with this for a while and get:

    f(z)=\sum_{n=2}^{\infty} \frac{n!}{3^2}(\frac{z}{3})^{n-2}\frac{1}{(n-2)!}

    Which is interesting, and sort of reminds me of \frac{1}{e} and e^z, but a) I know multiplication of series doesn't work like that and b) that wouldn't be rational anyway. Any hints?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by igopogo View Post
    Given the series:

    f(z)=\sum_{n=2}^{\infty} \frac{n(n-1)}{3^n}z^{n-2}

    a) find the radius of convergence (got that, it's z>1/3)

    b) Prove that f is a rational function inside its circle of convergence

    c) find the rational number f(2)

    Well, I mess around with this for a while and get:

    f(z)=\sum_{n=2}^{\infty} \frac{n!}{3^2}(\frac{z}{3})^{n-2}\frac{1}{(n-2)!}

    Which is interesting, and sort of reminds me of \frac{1}{e} and e^z, but a) I know multiplication of series doesn't work like that and b) that wouldn't be rational anyway. Any hints?
    Two derivatives of a well known function.
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  3. #3
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    Nope, feeling dense here. Can you push me toward a *method* of looking at a problem like this? This is study, not homework
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  4. #4
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    nvm...solved. Thanks for the direction!
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