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Math Help - Proof involving Abel's Theorem

  1. #1
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    Proof involving Abel's Theorem

    Let lim n-> ∞ a_n = L. Then, let f(x) = ∑ from 0 to ∞ of (a_n)(x^n). Show that the lim x-> 1 (1-x)f(x) = L.

    This one is pretty far over my head. I know at some point you're supposed to use Abel/SBP, but here is what I have so far.

    Let |a_n| go to |L|.

    Then, using the ratio test, let |a_n|^(1/n) go to |L|^(1/∞) = |L|^(0) = 1.

    Then, from the Ratio test, we can see that the series will converge for |x| < 1.

    Take ∑ from 0 to ∞ of (a_n)(x^n). Then, multiply through. So, we obtain, (1-x)∑ from 0 to ∞ of (a_n)(x^n) = ∑ from 0 to ∞ of (x^n - x^(n+t)). Taking b_n to equal (x^n - x^(n+t)) we can get ∑ (a_n)(b_n)..

    This is where I get stuck. I'm not really where to take it from here.
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  2. #2
    MHF Contributor chisigma's Avatar
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    Is...

    f(x)= \sum_{n=0}^{\infty} a_{n}\cdot x^{n} (1)

    ... so that...

    (1-x)\cdot f(x)= a_{0} + (a_{1}-a_{0})\cdot x + (a_{2} - a_{1}) \cdot x^{2} + \dots + (a_{n} - a_{n-1})\cdot x^{n} + \dots (2)

    It is evident that (2) is a 'telescopic series' so that is...

    \lim_{x \rightarrow 1} (1-x)\cdot f(x) = \lim_{n \rightarrow \infty} a_{n} = L (3)

    Kind regards

    \chi \sigma
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