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Math Help - converging "normally"

  1. #1
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    converging "normally"

    for this problem i think i need to use the abel's lemme which states that
    for r,r_0 in the reals such that 0<r<r_0, if there exists real number M
    such that M>0 and (abs(a_n))(r_0)^n≤M for all natural numbers, then
    the series Σa_nz^n for n≥0 converges normally for abs (z)<r.
    converging normally means that if (u_n(z))_n is a sequence of complex
    valued functiosn on E, then the series Σu_n converges normally on E
    given that the series Σllull converges (llull is the sup-norm of u,
    defined by llull=sup(z in E) abs (u(z))

    problem: if ρ is the radius of convergence of the series Σa_nz^n
    (n≥0), and r is a real number less than ρ, prove that the
    latter series converges normally for abs (z)<r (btw z is a modulus defined on each complex number by abs (z) = sqrt (x^2+y^2) and with property abs (z_1*z_2)=abs(z_1)abs(z_2) and abs (z_1+z_2)<= sqrt (z_1)+sqrt (z_2)
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  2. #2
    Member
    Joined
    Apr 2008
    From
    Seoul, South Korea
    Posts
    128
    i thought more about this question and would this proof work?

    pick r_0 such that r<r_0<p (rho)
    since r_0<p and radius of convergence is p, the series converges for z=r_0.
    => the series converges normally for abs (z)<r.

    how would i go about proving the divergence for abs (z)> p?
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