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Thread: Radius and Interval of Convergence

  1. #1
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    Radius and Interval of Convergence

    Very confused about the following:

    Find the radius of convergence and interval of convergence of the series.

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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by abel2 View Post
    Very confused about the following:

    Find the radius of convergence and interval of convergence of the series.

    What have you tried? For this just use the ratio test. The interval of convergence is given by

    $\displaystyle \displaystyle \lim_{n \to \infty} \left| \frac {a_{n + 1}}{a_n} \right| < 1$

    where $\displaystyle \displaystyle a_n = \frac {x^{n + 8}}{n + 6}$

    (Be sure to check the endpoints).

    The radius of convergence, is 1/2 the length of the interval of convergence.

    Can you continue?
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  3. #3
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    Ok so using the ratio test where $\displaystyle \rho = lim_{n\rightarrow\infty}abs((a_{n+1})/(a_{n})) < 1 $

    Then we have $\displaystyle lim_{n\rightarrow\infty} abs(\frac{\frac{(-1)^(n+1)*x^(n+9)}{(n+7)}}{\frac{(-1)^(n)*x^(n+8)}{(n+6)}}$

    Simplifying, we get $\displaystyle \frac{(-1)^(n+1)*x^(n+9)*(n+6)}{(n+7)*(-1)^(n)*x^(n+8)}$

    Then I get $\displaystyle =(-1)xlim_{n\rightarrow\infty}1/n$

    $\displaystyle \rightarrow = lim_{n \rightarrow \infty} = 0 $
    $\displaystyle \rho < 1$ and therefore converges.
    Still not sure how to solve for the center point and find the two endpoints. I'm left with x = -1

    I apologize in advance for my crappy latex. The higher parentheses are superscripts. And I'm not sure about the absolute value cmd.
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  4. #4
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    Well that can't be right. 1/n = infinity. But I am sure this equation converges. Somehow I think I cancelled out the n's incorrectly. Should have ended up with a L'Hospital's method and gotten 1/1 is my suspicion. But if that is the case, wouldn't I just solve for x giving 0? But then I have no idea how to get the endpoints unless they are -1<x<1 where x=0. With radius of 1.
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  5. #5
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    $\displaystyle \displaystyle \lim_{n \to \infty} \left|\frac{x^{n+9}}{n+7} \cdot \frac{n+6}{x^{n+8}}\right| < 1$

    $\displaystyle \displaystyle |x| \cdot \lim_{n \to \infty} \frac{n+6}{n+7} < 1$

    $\displaystyle \displaystyle |x| \cdot 1 < 1$

    $\displaystyle -1 < x < 1$ ... radius of convergence is r = 1


    checking the endpoints ...

    when $\displaystyle x = 1$

    $\displaystyle \displaystyle \sum_{n=2}^{\infty} \frac{(-1)^n \cdot 1}{n+6}$ ... alternating series that converges

    when $\displaystyle x = -1$

    $\displaystyle \displaystyle \sum_{n=2}^{\infty} \frac{(-1)^n \cdot (-1)^{n+8}}{n+6} = \sum_{n=2}^{\infty} \frac{(-1)^{2n+8}}{n+6} =\sum_{n=2}^{\infty} \frac{1}{n+6}$ ... series diverges


    interval of convergence is $\displaystyle -1 < x \le 1$
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