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Math Help - INT of convergence problem

  1. #1
    Junior Member
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    INT of convergence problem

    (-1)^n(x+2)^n
    ______________
    sqrt(n)*3^n

    i use the ratio test abs(bn+1/bn), then plug and chug and I get:

    sqrt(n)|x+2|
    ___________
    3*sqrt(n+1)

    =
    |x+2|
    ________
    3

    the problem is, i try maple and it gets absolutely no where near that...rather, it gets infinity.???
    I need to be able to find the int of convergence. thanks
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  2. #2
    MHF Contributor
    Prove It's Avatar
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    Quote Originally Posted by orendacl View Post
    (-1)^n(x+2)^n
    ______________
    sqrt(n)*3^n

    i use the ratio test abs(bn+1/bn), then plug and chug and I get:

    sqrt(n)|x+2|
    ___________
    3*sqrt(n+1)

    =
    |x+2|
    ________
    3

    the problem is, i try maple and it gets absolutely no where near that...rather, it gets infinity.???
    I need to be able to find the int of convergence. thanks
    I take it that this is actually

    \sum_{n = 0}^{\infty}\frac{(-1)^n(x + 2)^n}{3^n\sqrt{n}}.


    The series converges where

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

    \lim_{n \to \infty}\left|\frac{\frac{(-1)^{n + 1}(x + 2)^{n + 1}}{3^{n + 1}\sqrt{n + 1}}}{\frac{(-1)^n(x + 2)^n}{3^n\sqrt{n}}}\right| < 1

    \lim_{n \to \infty}\left|\frac{\frac{(x + 2)^{n + 1}}{3^{n + 1}\sqrt{n +  1}}}{\frac{(x + 2)^n}{3^n\sqrt{n}}}\right| < 1

    \lim_{n \to \infty}\left|\frac{3^n(x + 2)^{n + 1}\sqrt{n}}{3^{n + 1}(x + 2)^n\sqrt{n + 1}}\right| < 1

    \lim_{n \to \infty}\left|\frac{(x + 2)\sqrt{n}}{3\sqrt{n + 1}}\right| < 1

    \left|\frac{x + 2}{3}\right|\lim_{n \to \infty}\frac{\sqrt{n}}{\sqrt{n + 1}} < 1

    \left|\frac{x + 2}{3}\right|\lim_{n \to \infty}\sqrt{\frac{n}{n + 1}} < 1

    \left|\frac{x + 2}{3}\right|\lim_{n \to \infty}\sqrt{1 - \frac{1}{n + 1}} < 1

    \left|\frac{x + 2}{3}\right|(1) < 1

    \left|\frac{x + 2}{3}\right| < 1

    -1 < \frac{x + 2}{3} < 1

    -3 < x + 2 < 3

    -1 < x < 5.


    So the interval of convergence is

    -1 < x < 5.


    You will need to check the endpoints though, as the ratio test tells us nothing when the limit = 1...
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  3. #3
    MHF Contributor alexmahone's Avatar
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    The interval of convergence is (-5, 1).

    Quote Originally Posted by Prove It View Post
    -3 < x + 2 < 3

    -1 < x < 5
    Say what?
    Last edited by mr fantastic; July 22nd 2010 at 11:25 PM. Reason: Merged posts.
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  4. #4
    Junior Member
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    reply regarding int of conv.

    I agree with member 'prove it'.
    I do get ]x+2]/3 which is then (-1, 5) for int of convergence
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  5. #5
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by alexmahone View Post
    The interval of convergence is (-5, 1).

    I get what you get as the interval of convergence.

    -3 < x+2 <3

    Now subtract 2 form both sides and you get

    -5 < x < -1

    Here is a different approach to this problem i was taught in my DE book.

     \displaystyle{\lim_{n \to \infty} \frac{\frac{1}{3^{n+1}\sqrt{n+1}}}{\frac{1}{3^n\sq  rt{n}}}}

    \lim_{n \to \infty}\frac{3^n\sqrt{n}}{3^{n+1}\sqrt{n+1}} = \frac{1}{3}

    Now take L= \frac{1}{3}

    and p = \frac{1}{L}

    and is p is equal to radius of convergence which is 3

    so now

    -3 < x +2 < 3

    -5 < x < 1

    when x = -5, the series is

    \frac{1}{\sqrt{n}} which diverges by the p test

    when x = 1, the series is

    \frac{(-1)^n}{\sqrt{n}} which converges through the alternating series test

    so

    -5 < x \leq 1
    Last edited by 11rdc11; May 16th 2010 at 04:37 PM.
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