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Math Help - Radius of convergence

  1. #1
    MHF Contributor Mathstud28's Avatar
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    Radius of convergence

    could someone give me an unsolved radius of convergence problem and I will work it and they can critique me?
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  2. #2
    Flow Master
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    Quote Originally Posted by Mathstud28 View Post
    could someone give me an unsolved radius of convergence problem and I will work it and they can critique me?
    Power Series and the Radius of Convergence
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Ok

    so follow my work here if you wouldnt mind \sum_{n=1}^{\infty}\frac{(-1)^{n+1}(x-5)^{n}}{n5^{n}}...ok so I must find all x such that \lim_{n \to{\infty}}\bigg|\frac{a_{n+1}}{a_n}\bigg|<1...so \lim_{n \to{\infty}}\bigg|\frac{(x-5)^{n+1}}{(n+1)5^{n+1}}\cdot{\frac{n5^{n}}{(x-5)^{n}}}\bigg|<1..we are left with \lim_{n \to{\infty}}\bigg|\frac{n(x-5)}{5(n+1)}\bigg|<1...now evaluating the limit we get \frac{|x-5|}{5}<1 so 0<x<10...now I must test the endpoints of the test intervals..when I input x=0 we get \sum_{n=1}^{\infty}\frac{(-1)^{n+1}(-5)^{n}}{n5^{n}}=\sum_{n=1}^{\infty}\frac{(-1)^{2n+1}5^{n}}{n5^{n}}=\sum_{n=1}^{\infty}\frac{(-1)^{2n+1}}{n} and now realizing that for integral numbers (-1)^{2x+1}=-1 since 2n+1 is a series for all odd integers...so now we have -\sum_{n=1}^{\infty}\frac{1}{n} which is obviously a divergent harmonic series...now when we put 10 in for x after all the simplification we would get \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n} and seeing that this is equal to -\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n} and using the Root test we get that \lim_{n \to {\infty}}\bigg[\frac{(-1)^{n}}{n}\bigg]^{\frac{1}{n}}=\lim_{n \to{\infty}}\frac{-1}{n^{\frac{1}{n}}}=-1 and since that is <1 it is divergent...so the series only converges on 0<x<10...is that right?
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  4. #4
    Moo
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    Hello,

    Your last sentence is strange...

    Of course -1<1, but when you study radius of convergence, you always work with absolute values
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Wow

    I'm off todeay...thats sad because I knew that...well then by using the alternating series test we can conclude that it diverges so my answer stands
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    So

    can someone tell me after my addendum if my work is correct?
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