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Thread: Weierstrass M-test

  1. #1
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    Weierstrass M-test

    So thanks to tph we know that $\displaystyle \sum_{k=0}^{\infty}\frac{k2^k}{3^k}x^{2k}$ converges on $\displaystyle (-\sqrt{\frac{3}{2}}, \sqrt{\frac{3}{2}})$.

    To use the Weierstrass M-test, how do we find a bound for that sum.

    Thanks
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  2. #2
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    Quote Originally Posted by tbyou87 View Post
    So thanks to tph we know that $\displaystyle \sum_{k=0}^{\infty}\frac{k2^k}{3^k}x^{2k}$ converges on $\displaystyle (-\sqrt{\frac{3}{2}}, \sqrt{\frac{3}{2}})$.

    To use the Weierstrass M-test, how do we find a bound for that sum.

    Thanks
    What are you trying to show? We can show that if $\displaystyle 0<r<R$ where $\displaystyle R=\sqrt{3/2}$ then $\displaystyle \sum_{k=0}^{\infty}\frac{k2^k}{3^k}x^{2k}$ converges uniformly on $\displaystyle [-r,r]$ by choosing $\displaystyle M_k = \frac{k2^k}{3^k}r^{2k}$.
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