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Math Help - Finding interval of convergance

  1. #1
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    Finding interval of convergance

    How do I find the interval of convergance for this?

    I did the ratio test and get .25<x<.25
    The book has .5<x<.5

    I will do the endpoint test after I see why I'm not getting one half as my interval!Finding interval of convergance-imageuploadedbytapatalk-21373835141.519519.jpg
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  2. #2
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    Re: Finding interval of convergance

    Quote Originally Posted by minneola24 View Post
    I will do the endpoint test after I see why I'm not getting one half as my interval!
    This is because you first need to represent the series as \sum_{n=0}^\infty c_nx^n for some c_n and not as \sum_{n=0}^\infty c_nx^{2n}.
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  3. #3
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    Re: Finding interval of convergance

    Ok.

    How exactly do I do that, and why?
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    Re: Finding interval of convergance

    Quote Originally Posted by minneola24 View Post
    Ok.
    How exactly do I do that, and why?
    I answered that here.

    Use the root on ~\;\;\sqrt[n]{{\left| {\frac{{{2^{2n + 3}}{x^{2n + 2}}}}{{2n + 2}}} \right|}}
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  5. #5
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    Re: Finding interval of convergance

    Hello, minneola24!

    \text{Find the interval of convergence: }\;\sum^{\infty}_{n=0}(-1)^n\frac{4^n\cdot8\cdot x^{2n+2}}{2n+2}

    |R| \;=\;\left|\frac{a_{n+1}}{a_n}\right| \;=\;\left|\frac{4^{n+1}\cdot 8\cdot x^{2n+4}}{2n+4} \cdot \frac{2n+2}{4^n\cdot8\cdot x^{2n+2}}\right|

    . . . . . . . . . . =\;\left|\frac{4^{n+1}\cdot8}{4^n\cdot8}\cdot \frac{x^{2n+4}}{x^{2n+2}}\cdot \frac{2n+2}{2n+4}\right|

    . . . . . . . . . . =\;\left|4\cdot x^2\cdot\frac{n+1}{n+2}\right|

    . . . . . . . . . . =\;\left|4\cdot x^2\cdot\frac{1+\frac{1}{n}}{1+\frac{2}{n}} \right|


    \lim_{n\to\infty}|R| \;=\;\lim_{n\to\infty}\left|4\cdot x^2\cdot\frac{1+\frac{1}{n}}{1+\frac{2}{n}} \right| \;=\;\left|4\cdot x^2\cdot\frac{1+0}{1+0}\right| \;=\;|4x^2|


    Then: . |4x^2| \:<\:1 \quad\Rightarrow\quad |x^2| \:<\:\tfrac{1}{4}

    Therefore: . |x|\:<\:\tfrac{1}{2} \quad\Rightarrow\quad \text{-}\tfrac{1}{2}\:<\:x \:<\:\tfrac{1}{2}
    Thanks from minneola24
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    Re: Finding interval of convergance

    Quote Originally Posted by minneola24 View Post
    How exactly do I do that, and why?
    I intended to use facts specifically about the radius of convergence of a power series instead of tests that apply to any series.

    \sum_{n=0}^\infty(-1)^n\frac{4^n\cdot8\cdot x^{2n+2}}{2n+2}= (set k = n + 1, i.e., n = k - 1)
    \sum_{k=1}^\infty(-1)^{k-1}\frac{4^{k-1}\cdot8\cdot x^{2k}}{2k}=
    \sum_{n=1}^\infty c_nx^n (set n = 2k, i.e., k = n/2)

    where

    c_n= \begin{cases}\displaystyle (-1)^{n/2-1}\frac{4^{n/2-1}\cdot8}{n} & n \text{ is even}\\ 0 & n \text{ is odd} \end{cases}

    Now, strictly speaking, we cannot use the formula

    r = \lim_{n\rightarrow\infty} \left| \frac{c_n}{c_{n+1}} \right|

    because some c_n are equal to 0 (but see Soroban's answer on how to use ratio test for arbitrary series), but we can use

    r = \frac{1}{\limsup_{n\rightarrow\infty}\sqrt[n]{|c_n|}}

    Since \sqrt[n]{n}\to 1 as n\to\infty, \limsup_{n\to\infty}|c_n|^{1/n}= \lim_{n\to\infty}\left(4^{n/2-1}\right)^{1/n} =4^{1/2}=2, so r = 1/2. If we did not convert the series to the form \sum_{n=1}^\infty c_nx^n and took the nth root of \frac{4^n\cdot8}{2n+2}, we would get r = 1/4.
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