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Math Help - Convergence

  1. #1
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    Convergence

    Find the interval of convergence for f(x) = x-(x^2/2)+(x^3/3)-(x^4/4)+.....
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  2. #2
    Moo
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    Hello,

    Quote Originally Posted by jules027 View Post
    Find the interval of convergence for f(x) = x-(x^2/2)+(x^3/3)-(x^4/4)+.....
    f(x)=-\sum_{n=1}^\infty \frac{(-1)^n}{n} x^n

    According to the alternate series test :
    - \frac{|x|^n}{n}=a_n should be decreasing
    - \lim_{n \to \infty} \frac{|x|^n}{n}=0

    The first one... We want : \frac{a_{n+1}}{a_n}<1

    \frac{|x|^{n+1}}{n+1} \cdot \frac{n}{|x|^n} <1

    |x| \cdot \frac{n}{n+1} <1

    |x|<\frac{n+1}{n}

    But \frac{n+1}{n} has a minimum in 1, when n \to \infty.

    Hence |x| has to be <1

    And this satisfies the second point.
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    Hello,



    f(x)=-\sum_{n=1}^\infty \frac{(-1)^n}{n} x^n

    According to the alternate series test :
    - \frac{|x|^n}{n}=a_n should be decreasing
    - \lim_{n \to \infty} \frac{|x|^n}{n}=0

    The second one is always true, whatever x is.

    Now, the first one... We want : \frac{a_{n+1}}{a_n}<1

    \frac{|x|^{n+1}}{n+1} \cdot \frac{n}{|x|^n} <1

    |x| \cdot \frac{n}{n+1} <1

    |x|<\frac{n+1}{n}

    But \frac{n+1}{n} has a minimum in 1, when n \to \infty.

    Hence |x| has to be <1
    Going off of Moo's work we knwo so far that this is convergent on

    (-1,1)


    so we must test the endpoitns....putting int netgative one yields

    \sum_{n=1}^{\infty}\frac{(-1)^n\cdot(-1)^n}{n}=\sum_{n=1}^{\infty}\frac{1}{n}

    which diverges by either the integral or the p series test

    \int_1^{\infty}\frac{dx}{x}=\infty


    inputting 1 we get

    \sum_{n=1}^{\infty}\frac{(-1)^n1^n}{n}=\sum_{n=1}^{\infty}\frac{(-1)^n}{n}

    and since \forall{n}>0,a_n>a_{n+1}

    and \lim_{n\to\infty}a_n=0

    this converges

    so the interval of convergence is

    (-1,1]
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  4. #4
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    Thank you so much guys!
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