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Math Help - radius of convergence (power series)

  1. #1
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    radius of convergence (power series)

    Hey guys,

    if |a_k|<|b_k| for large k, show that if (b_k)(x^k) converges on an open interval I, then (a_k)(x^k) also converges on I.

    Please help.
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  2. #2
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    Quote Originally Posted by thahachaina View Post
    Hey guys,

    if |a_k|<|b_k| for large k, show that if (b_k)(x^k) converges on an open interval I, then (a_k)(x^k) also converges on I.

    Please help.
    Given a sequence \{ x_n\} the series \sum_{n=1}^{\infty}x_n converges if and only if \sum_{n=N}^{\infty} x_n where N>1.
    Therefore, \sum_{k=1}^{\infty}a_kx^k converges if and only if  \sum_{k=K}^{\infty}a_kx^k where K>1 is chosen so large that |a_k| < |b_k|.

    Remember that an interval of converges of a series (centered as 0) is an interval centered at 0. If we look at the open interval then the series also converges absolutely. Now if x\in I then x is on the open interval of convergence of \sum_{k=K}^{\infty}b_kx^k and so \sum_{k=K}^{\infty}|b_kx^k| converges too. But we have that |a_kx^k| \leq |b_kx^k| and so \sum_{k=K}^{\infty}a_kx^k converges too.
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