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Math Help - asd

  1. #1
    Junior Member
    Joined
    Aug 2007
    Posts
    45

    asd

    EDIT: sorry about the title of this thread guys.. it was a mistake. if it can be changed please let me know

    how can i determine the radius of convergence of the following power series:

    \sum^{\infty}_{n=1}(log n)^2(z - i)^n

    \sum^{\infty}_{n=1}i^nz^n/n^3

    \sum^{\infty}_{n=1}n!(z+e)^n

    \sum^{\infty}_{n=1}n^2z^n/4^n + 3ni

    \sum^{\infty}_{n=0}z^5n

    Im guessing these have to be done using the ratio test but im not too sure.
    any hints please? i will try them out with any clues available thanks.
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  2. #2
    Newbie
    Joined
    Oct 2007
    Posts
    22
    you can use lim n->oo ||a_n|| / ||a_(n+1)|| if the limit exists. otherwise you'll have to cauchy-hadamard or resort to other means.

    for your first question for example, i would use cauchy hadamard.
    the thing on the bottom of the denominator for cauchy hadamard is
    limsup nth root of (log n) = limsup (log n)^(2/n) = limsup [(log n)^(1/n)]^2
    lim [n^(1/n)]^2] = 1^2 = 1
    so radius of convergence ρ = 1/1 = 1 where it's centred on the point (0,i)
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