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Math Help - Interval of convergence

  1. #1
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    Interval of convergence

    I need some help on the following

    Determine the interval of convergence of the power series

    infinity
    sum of series (-1)^n/(n+3)4^n * (x-1)^n
    n=1

    I have

    Here An = (-1)^n/(n+3)4^n, for n=1,2....

    1. since

    |An+1/An| = 1/(n+4)4^n+1 * (n+3)4^n/1

    = ? ----> as n ----> ?

    we have R = ? , by the Ratio Test.

    Thus

    2. If x = ? , the the power series is

    ?


    Hence the interval of convergence is ( ? )
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  2. #2
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    Re: Interval of convergence

    Is this your mid-term quiz? You should be demonstrating SOME knowledge fo the subject. You seem to be just guessing.
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    Re: Interval of convergence

    your series is not real since (-1)^\frac{n}{n+3} is imaginary for n as odd
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    Re: Interval of convergence

    Quote Originally Posted by waqarhaider View Post
    your series is not real since (-1)^\frac{n}{n+3} is imaginary for n as odd
    You have miss read the question.
    It is \sum\limits_{n = 1}^\infty  {\frac{{( - 1)^n }}{{(n + 3)4^n }}(x - 1)^n } ~.

    Consider the root test.
    \frac{{\left| {x - 1} \right|}}{{\sqrt[n]{{n + 3}}(4)}} \to ?
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    Re: Interval of convergence

    I am not sure about the root test, mainly been working on the Ratio test.
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    Re: Interval of convergence

    Quote Originally Posted by Arron View Post
    I am not sure about the root test, mainly been working on the Ratio test.
    Essentially the two test are in most cases equivalent.
    They both rely upon the same conclusion
    If \lim _{n \to \infty }  {{\sqrt[n]{{\left|a_n\right| }}} }  < 1 the series converges.
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    Re: Interval of convergence

    So does that tend to 1 as n ----> infinity
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    Re: Interval of convergence

    Quote Originally Posted by Arron View Post
    So does that tend to 1 as n ----> infinity
    NO!
    \frac{{\left| {x - 1} \right|}}{{\sqrt[n]{{n + 3}}(4)}} \to \frac{{\left| {x - 1} \right|}}{4}

    Now solve {\frac{|x-1|}{4}<1.

    Be sure to test the endpoints.
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    Re: Interval of convergence

    Sorry I find this really hard. Would it be 1/4. Thanks for all your help.
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  10. #10
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    Re: Interval of convergence

    Quote Originally Posted by Arron View Post
    Sorry I find this really hard. Would it be 1/4. Thanks for all your help.
    Would what be 1/4?? Please quote the replies you are refering to when asking further help.
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  11. #11
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    Re: Interval of convergence

    By using d'Alembert test you will come to the same conclusion. We have to calculate (in general):
    \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|=r
    So in this case:
    \lim_{n\to \infty} \left|\frac{\frac{(-1)^{n+1}\cdot (x-1)^{n+1}}{(n+4)\cdot 4^{n+1}}}{\frac{(-1)^n\cdot (x-1)^n}{(n+3) /cdot 4^n}}\right |
    =\lim_{n \to \infty} \left | \frac{(-1)^{n+1}\cdot (x-1)^{n+1}\cdot (n+3)\cdot 4^n}{(n+4)\cdot 4^{n+1}\cdot (-1)^n\cdot (x-1)^n}\right|
    =\lim_{n \to \infty} \left |(-1)\cdot (x-1)\cdot \frac{1}{4}\cdot \frac{n+3}{n+4}\right |
    =\frac{|x-1|}{4}\lim_{n \to \infty} \left|\frac{n+3}{n+4}\right |=\frac{|x-1|}{4}

    Along the test the serie wil converge if 0<\frac{|x-1|}{4}<1 and diverge if \frac{|x-1|}{4}>1 so you've to calculate the x values wherefore:
    \frac{|x-1|}{4}<1\Leftrightarrow |x-1|<4
    So two different cases:
    (1) x-1<4 \Leftrightarrow ...
    (2) -x+1<4 \Leftrightarrow ...
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