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Math Help - Radius of convergence

  1. #1
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    Radius of convergence

    summation from n = 0 -> infinity of:

    \sum\frac{n}{2^n}x^n

    how do you find it? completely lost.

    I tried divergence and convergence stuff but it doesnt seem to net me the correct answer.
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  2. #2
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    use the ratio test to find the radius of convergence ...

    \lim_{n \to \infty} \left|\frac{(n+1)x^{n+1}}{2^{n+1}} \cdot \frac{2^n}{n x^n} \right| < 1

    \lim_{n \to \infty} \left|\frac{(n+1)x}{2n}\right| < 1

    \left|\frac{x}{2}\right| \lim_{n \to \infty} \frac{n+1}{n} < 1

    \left|\frac{x}{2}\right| \cdot 1 < 1

    -1 < \frac{x}{2} < 1

    -2 < x < 2
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  3. #3
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    Quote Originally Posted by skeeter View Post
    use the ratio test to find the radius of convergence ...

    \lim_{n \to \infty} \left|\frac{(n+1)x^{n+1}}{2^{n+1}} \cdot \frac{2^n}{n x^n} \right| < 1

    \lim_{n \to \infty} \left|\frac{(n+1)x}{2n}\right| < 1

    \left|\frac{x}{2}\right| \lim_{n \to \infty} \frac{n+1}{n} < 1

    \left|\frac{x}{2}\right| \cdot 1 < 1

    -1 < \frac{x}{2} < 1

    -2 < x < 2
    ahhhhhh I get it. I wasnt sure how to deal with the x and the n together like that.

    Thanks alot, ill try one another now. oh quck question, why you set limit to < 1.?
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  4. #4
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    Quote Originally Posted by p00ndawg View Post
    ahhhhhh I get it. I wasnt sure how to deal with the x and the n together like that.

    Thanks alot, ill try one another now. oh quck question, why you set limit to < 1.?
    The ratio test for convergence says that for the series to converge,

    \lim_{n \rightarrow \infty} \left|\frac{a_{n+1}}{a_n}\right| must be < 1
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  5. #5
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    Quote Originally Posted by mollymcf2009 View Post
    The ratio test for convergence says that for the series to converge,

    \lim_{n \rightarrow \infty} \left|\frac{a_{n+1}}{a_n}\right| must be < 1

    oh ok thats what i thought, just wanted to make sure.
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