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Math Help - find the radius of convergence for 1st derivative of f.

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    find the radius of convergence for 1st derivative of f.

    If f = n=0, infinite ((-1)^(n+1))(x-1)^(n))/n, find the radius of convergence for 1st derivative of f.
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    Quote Originally Posted by ewkimchi View Post
    If f = n=0, infinite ((-1)^(n+1))(x-1)^(n))/n, find the radius of convergence for 1st derivative of f.
    f(x) = \sum_{n = 0}^{\infty}\frac{(-1)^{n + 1}(x - 1)^n}{n}


    f'(x) = \sum_{n = 0}^{\infty}\frac{(-1)^{n + 1}n(x - 1)^{n - 1}}{n}

     = \sum_{n = 0}^{\infty}(-1)^{n + 1}(x - 1)^{n - 1}.


    The radius of convergence of f'(x) is given where

    \lim_{n \to \infty}\left|\frac{t_{n + 1}}{t_n}\right| < 1

    \lim_{n \to \infty}\left|\frac{(-1)^{n + 2}(x - 1)^{n}}{(-1)^{n + 1}(x - 1)^{n - 1}}\right| < 1

    \lim_{n \to \infty}\left|x - 1\right|< 1

    |x - 1| < 1

    -1 < x - 1 < 1

    0 < x < 2.

    The radius of convergence is 1, and the interval of convergence is 0 < x < 2.
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