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Math Help - Uniform COnvergence

  1. #1
    Super Member Showcase_22's Avatar
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    Uniform COnvergence

    For f(x)=\sum_{n=1}^{\infty} \frac{1}{n(1+nx^2}, on what intervals of x in the form (a,b) does the series converge uniformly?

    My answer:

    \left| \frac{1}{n(1+nx^2)} \right| \leq \left| \frac{1+nx^2}{n(1+nx^2)} \right|= \left| \frac{1}{n} \right|

    I would now choose my  M_n=\frac{1}{n} . However,  \sum_{n=1}^{\infty}\frac{1}{n} does not converge!

    I can't see another way of doing it so that I get it to be less than a convergent sequence. =S
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  2. #2
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    Quote Originally Posted by Showcase_22 View Post
    For f(x)=\sum_{n=1}^{\infty} \frac{1}{n(1+nx^2}, on what intervals of x in the form (a,b) does the series converge uniformly?

    My answer:

    \left| \frac{1}{n(1+nx^2)} \right| \leq \left| \frac{1+nx^2}{n(1+nx^2)} \right|= \left| \frac{1}{n} \right|

    I would now choose my  M_n=\frac{1}{n} . However,  \sum_{n=1}^{\infty}\frac{1}{n} does not converge!

    I can't see another way of doing it so that I get it to be less than a convergent sequence. =S

    \left |\frac{1}{n(1+nx^2)} \right|=\frac{1}{n+n^2x^2}\leq \frac{1}{n^2x^2}

    As the series \frac{1}{x^2}\sum\limits_{n=1}^\infty\frac{1}{n^2} converges no matter what is x \neq 0 you're done.

    Tonio
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