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Math Help - Proving a function is continuous using uniform convergence

  1. #1
    Senior Member Pinkk's Avatar
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    Proving a function is continuous using uniform convergence

    On the domain [-1,1), is \sum_{n=1}^{\infty}\frac{1}{n}x^{n} continuous?

    If we let f_{k}(x) = \sum_{n=1}^{k}\frac{1}{n}x^{n}, then it is clear that for all k\in \mathbb{N}, f_{k}(x) is continuous, so now it will suffice to show that f_{k}(x) uniformly converges to f(x) for x\in [-1,1). So for any \epsilon > 0, there exists N such that if n>N, then |f_{k}(x) - f(x)| < \epsilon.

    This is where I am running into some trouble because |f_{k}(x) - f(x)| = \sum_{n=k}^{\infty}\frac{1}{n}x^{n}. After this point I am not exactly sure where to proceed. I have a hunch that the fact that f(x) converges on that domain will come into play, but I am not sure. Thanks.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Pinkk View Post
    On the domain [-1,1), is \sum_{n=1}^{\infty}\frac{1}{n}x^{n} continuous?

    If we let f_{k}(x) = \sum_{n=1}^{k}\frac{1}{n}x^{n}, then it is clear that for all k\in \mathbb{N}, f_{k}(x) is continuous, so now it will suffice to show that f_{k}(x) uniformly converges to f(x) for x\in [-1,1). So for any \epsilon > 0, there exists N such that if n>N, then |f_{k}(x) - f(x)| < \epsilon.

    This is where I am running into some trouble because |f_{k}(x) - f(x)| = \sum_{n=k}^{\infty}\frac{1}{n}x^{n}. After this point I am not exactly sure where to proceed. I have a hunch that the fact that f(x) converges on that domain will come into play, but I am not sure. Thanks.
    I assume that you don't know that a power series is uniformly convergent on it's radius of convergence, huh?
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