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Sketch on the complex plane the region where the following two power series both

  1. Jul 17, 2011 #1
    1. The problem statement, all variables and given/known data

    sketch on the complex plane the region where the following two power series both converge

    1) sigma from n=0 to infinity [(z-1)^n]/[n^2]

    2) sigma from n=0 to infinity [((n!)^2)((z+4i)^n)]/[2n]!

    3. The attempt at a solution

    R=lim as n tends to infinity |(a(subscript n))/(a(subscript n+1))|

    1) R=lim n tends to infinity [(z-1)^n][[n+1]^2]/[n^2][[z-1]^(n+1)]
    =((n+1)^2)/(n^2)(z-1)
    2) R=lim as n tends to infinity [((n!)^2)((z+4i)^n)/(2n)!][(2(n+1))!/(((n+1)!)^2)((z+4i)^(n+1))

    I don't know how to proceed from here and I think I may have made a mistake somewhere
     
  2. jcsd
  3. Jul 17, 2011 #2

    HallsofIvy

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    Re: sketch on the complex plane the region where the following two power series both.

    [tex]\frac{(n+1)^2}{n^2}= \frac{n^2+ 2n+ 1}{n^2}= 1+ \frac{2}{n}+ \frac{1}{n^2}[/tex]
    [itex]\frac{(n!)^2}{((n+1)!)^2}\frac{((2n)!)^2}{((2n+1)!)^2}[/itex][itex]= \left(\frac{1}{n+1}\right)^2[/itex][itex]\left(\frac{1}{2n+1}\right)^2[/itex]
     
    Last edited: Jul 17, 2011
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