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Differentiability of a Series of Functions

  1. Nov 18, 2012 #1
    I'm working on a problem where I need to show that the series of functions, f(x) = Ʃ (xn)/n2, where n≥1, converges to some f(x), and that f(x) is continuous, differentiable, and integrable on [-1,1].

    I know how to show that f(x) is continuous, since each fn(x) is continuous, and I fn(x) converges uniformly. Because each fn(x) is also integrable, I can also show f(x) is integrable.

    The trouble I'm having is proving that f(x) is differentiable. I need to show that the series of derivatives converges uniformly. However, I don't think I can use the Weierstrass M-Test in this scenario. Any ideas?
     
  2. jcsd
  3. Nov 20, 2012 #2

    Erland

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    The differentiated series converges on the open interval (-1,1). Hence the original series is differentiable there and the derivative is the differentiated series. The differentated series diverges at 1 and converges conditionally at -1. Although I cannot prove it, this makes me suspect that the original series is not differentiable at 1 and -1.
     
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