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Thread: Uniform convergence....

  1. #1
    MHF Contributor Also sprach Zarathustra's Avatar
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    Uniform convergence....

    Prove that $\displaystyle \Sigma_{n=1}^\infty \frac{x^2}{(x^2+1)^n}$ uniformly convergence at $\displaystyle [a,\infty)$ for $\displaystyle a>0$, and not uniformly convergence at $\displaystyle [-a,a]$ for $\displaystyle a>0$.


    Thank you so much!
    Last edited by Also sprach Zarathustra; Aug 2nd 2010 at 05:53 AM. Reason: k=n ( thank you Prove It)
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  2. #2
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    There isn't any $\displaystyle n$ in the expression...
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  3. #3
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    To prove uniform convergence on $\displaystyle [a, \infty)$ you could use the Weierstrass M-test - Wikipedia, the free encyclopedia.

    You can use the estimation
    $\displaystyle \displaystyle\frac{x^2}{(x^2+1)^n}\leq \frac{x^2}{{n\choose 2} x^4}=\frac{1}{x^2} \frac{1}{{n\choose 2}}\leq \frac{1}{a^2} \frac{1}{{n\choose 2}} $ for n>1

    Hope that helps.
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