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Math Help - Convergence of a Series

  1. #1
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    Convergence of a Series

    Hey I've been having a bit of trouble find the interval of convergence for:

    \sum_{n = 0}^{\infty} \frac{(-1)^n (2x + 1)^n}{n 2^{2n}}

    I did the ratio test:

    \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{(-1) |2x + 1| n}{2^2 (n+1)} \right| = \frac{1}{4} \left| 2x + 1 \right| < 1

    So \frac{-5}{2} < x < \frac{3}{2}

    Now test x = -5/2:

    \sum_{n = 0}^{\infty} \frac{(-1)^n (-4)^n}{n 2^{2n}} = \sum_{n = 0}^{\infty} \frac{1}{n} \ \ \mbox{DV}

    Now test x = 3/2:

    \sum_{n = 0}^{\infty} \frac{(-1)^n (4)^n}{n 2^{2n}} = \sum_{n = 0}^{\infty} \frac{(-1)^n}{n} \ \ \mbox{CV}

    So the interval of convergence is: \boxed{\frac{-5}{2} < x \leq \frac{3}{2}}

    Does that seem right???
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  2. #2
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    Quote Originally Posted by TrevorP View Post
    Hey I've been having a bit of trouble find the interval of convergence for:

    \sum_{n = 0}^{\infty} \frac{(-1)^n (2x + 1)^n}{n 2^{2n}}
    You can kill this with the root test instead.

    \left| (-1)^n \frac{(2x+1)^n}{n2^{2n}} \right|^{1/n} = \frac{|2x+1|}{n^{1/n} 4} \to \frac{|2x+1|}{4}

    We want, |2x+1| < 4 \implies -4 < 2x + 1 < 4 \implies -\frac{5}{2} < x < \frac{3}{2}.

    If x=3/2 then \sum_{n=0}^{\infty} \frac{(-1)^n 4^n}{n4^n} which converges.

    If x=-5/2 then \sum_{n=0}^{\infty} \frac{(-1)^n (-1)^n 4^n}{n4^n} which diverges.
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  3. #3
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    Thanks, I'm not familiar with the root test but thanks for confirming my answer.
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