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Math Help - Derivative of an infinite series of functions

  1. #1
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    Derivative of an infinite series of functions

    Let f(z)=\sum^\infty_{n=1}\frac{(-1)^{n-1}z^n}{n}, -1<z<1. Prove that f'(z)=\sum^\infty_{n=1}(-1)^{n-1}z^{n-1} on (-1,1).

    I know that I need to prove uniform convergence on (-1,1). so, Let h_k(x)=\frac{(-1)^{n-1}}{n} and g_k(x)=z^n for all n\in\mathbb{N}. I know that h_k(z) converges to 0 and that g_k(z) is always between (-1,1). So using the Dirichlet's test for uniform convergence i should be able to prove f(z) is uniformly convergent and therefor (\sum^\infty_{n=1}f_n(z))'=\sum^\infty_{n=1}f'_n(z  ). Will this work? and how do i write a formal proof if it is?
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  2. #2
    Super Member
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    Apr 2009
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    Re: Derivative of an infinite series of functions

    Uniform convergence of a sequence doesn't guarantee even pointwise convergence of the sequence of derivatives, on the other hand you have the following:

    If f_k,f : (-1,1) \to \mathbb{R} and f_k(x_0)\to f(x_0) for one x_0 \in (-1,1) and f'\to g uniformly on compact subintervals of (-1,1) then f'=g.

    In your case take x_0=0 and since the sequence (series) of derivatives is a geometric series the result is immediate.
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