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Math Help - interval of convergence

  1. #1
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    interval of convergence

    sum n=o to oo (2n)!(x/2)^n

    = (2n+2)!(x/2)^(n+1) / (2n)!(x/2)^n

    = (2n+2)(2n+1)(x/2)

    where do i go from there? ><
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jeph View Post
    sum n=o to oo (2n)!(x/2)^n

    = (2n+2)!(x/2)^(n+1) / (2n)!(x/2)^n

    = (2n+2)(2n+1)(x/2)

    where do i go from there? ><
    take the limit as n goes to infinity, and set whatever the result is to be < 1. then you find the x's that satisfy it and check the end points
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  3. #3
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    Quote Originally Posted by jeph View Post
    sum n=o to oo (2n)!(x/2)^n

    = (2n+2)!(x/2)^(n+1) / (2n)!(x/2)^n

    = (2n+2)(2n+1)(x/2)

    where do i go from there? ><
    Note,
     (2n)!(x/2)^n = \frac{(2n)!}{2^n}\cdot x^n

    Ratio test for x\not =0,
    \left| \frac{(2n+2)!}{2^{n+1}} \cdot \frac{2^n}{(2n)!} \cdot x\right| = \frac{(2n+2)(2n+1)}{2}\cdot |x|\to +\infty

    So, this convergense only for x=0
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  4. #4
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    so does it go

    lim n to oo [(2n+2)(2n+1)/2]x

    = |[oo/2]x|<1

    -1< [oo/2]x < 1
    -2/oo< x < 2/oo
    0< x <0

    x=0 ?
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  5. #5
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    Quote Originally Posted by jeph View Post
    so does it go

    lim n to oo [(2n+2)(2n+1)/2]x

    = |[oo/2]x|<1

    -1< [oo/2]x < 1
    -2/oo< x < 2/oo
    0< x <0

    x=0 ?
    Is there a number x such that 0<x<0? No! But the thing is that we assumed (secretly) that x\not = 0. So the only point to be tested is x=0, which of course converges.
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