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Thread: 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 $\displaystyle (-1)^\frac{n}{n+3}$ 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 $\displaystyle (-1)^\frac{n}{n+3}$ is imaginary for n as odd
    You have miss read the question.
    It is $\displaystyle \sum\limits_{n = 1}^\infty {\frac{{( - 1)^n }}{{(n + 3)4^n }}(x - 1)^n } ~.$

    Consider the root test.
    $\displaystyle \frac{{\left| {x - 1} \right|}}{{\sqrt[n]{{n + 3}}(4)}} \to ?$
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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 $\displaystyle \lim _{n \to \infty } {{\sqrt[n]{{\left|a_n\right| }}} } < 1$ 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!
    $\displaystyle \frac{{\left| {x - 1} \right|}}{{\sqrt[n]{{n + 3}}(4)}} \to \frac{{\left| {x - 1} \right|}}{4}$

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

    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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    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):
    $\displaystyle \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|=r$
    So in this case:
    $\displaystyle \lim_{n\to \infty} \left|\frac{\frac{(-1)^{n+1}\cdot (x-1)^{n+1}}{(n+4)\cdot 4^{n+1}}}{\frac{(-1)^n\cdot (x-1)^n}{(n+3) /cdot 4^n}}\right |$
    $\displaystyle =\lim_{n \to \infty} \left | \frac{(-1)^{n+1}\cdot (x-1)^{n+1}\cdot (n+3)\cdot 4^n}{(n+4)\cdot 4^{n+1}\cdot (-1)^n\cdot (x-1)^n}\right|$
    $\displaystyle =\lim_{n \to \infty} \left |(-1)\cdot (x-1)\cdot \frac{1}{4}\cdot \frac{n+3}{n+4}\right |$
    $\displaystyle =\frac{|x-1|}{4}\lim_{n \to \infty} \left|\frac{n+3}{n+4}\right |=\frac{|x-1|}{4}$

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