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Math Help - unif convergence of series

  1. #1
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    unif convergence of series

    Hello all,I need a bit of help to proove that
    \sum_{n=1}^{\infty} \dfrac{x}{(1+{x})^n} uniformally converges , x in [1,2]

    , n \in \mathbb{N}=1,2,3....
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Kiki View Post
    Hello all,I need a bit of help to proove that
    \sum_{n=1}^{\infty} \dfrac{x}{(1+{x})^n} uniformally converges , x in [1,2]

    , n \in \mathbb{N}=1,2,3....
    Merely note \forall x\in [1,2] that \frac{x}{(1+x)^n}\leqslant \frac{1}{2^n} and apply Weierstrass.
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  3. #3
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    Quote Originally Posted by Drexel28 View Post
    Merely note \forall x\in [1,2] that \frac{x}{(1+x)^n}\leqslant \frac{1}{2^n} and apply Weierstrass.
    Sweet! Thank you vm
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Kiki View Post
    Sweet! Thank you vm
    You better prove that. I could have said that your function is less than \frac{1}{\pi^{\pi^n}} and you still would have believed me.
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  5. #5
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    Quote Originally Posted by Drexel28 View Post
    You better prove that. I could have said that your function is less than \frac{1}{\pi^{\pi^n}} and you still would have believed me.
    lol man you are right, when I first saw your answer I did mistakenly thought of it as a correct one cause I thought that the numerator was "1" , but instead it was "x" . (I suppose the same thing did happen to you too?)

    So I think that its not correct, because for n=1 (and lets say x=2)

    As far as mentioning \frac{1}{\pi^{\pi^n}} do we get this from a Taylor Series?
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  6. #6
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    I think I ve found it.

    \frac{x}{(1+x)^n}\leqslant \frac{x^n}{(1+x)^n} = (\frac{x}{1+x})^n \leqslant {(\frac{2}{3})^n}

    Is this ok?
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Kiki View Post
    I think I ve found it.

    \frac{x}{(1+x)^n}\leqslant \frac{x^n}{(1+x)^n} = (\frac{x}{1+x})^n \leqslant {(\frac{2}{3})^n}

    Is this ok?
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