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Math Help - Uniform convergence

  1. #1
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    Unhappy Uniform convergence

    Hello.
    I have proved that the series $\sum_{n=2}^{\infty}\frac{\left (\mathrm{ln}\left ( n \right ) \right )^{p}}{n}$ is convergent if p<-1 and divergent if p\geq -1.

    So the another problem is about to show the series $\sum_{n=2}^{\infty}\frac{\left (\mathrm{ln}\left ( n \right ) \right )^{x}}{n}$ is uniform convergent as $x \in ]-\infty;p]$ if p<-1.

    I am not sure what to do. If it's true that I should use the Weierstrass Test, could you please tell me what do to? Even I have read about it I feel kinda stuck when I try to answer this problem.
    (I am sorry if I have posted it in the wrong place.)
    Last edited by MathsforNewbs; May 18th 2013 at 09:25 PM.
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  2. #2
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    Re: Uniform convergence

    Hey MathsforNewbs.

    Can you use the fact that the series is continuous and that the point-wise convergence in addition to the continuity criteria of the sum imply uniform convergence?

    I'm not sure if you can use this so I'm just throwing an idea your way.
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  3. #3
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    Re: Uniform convergence

    Hello Chiro.

    I am not sure if I understood you correctly. Did you mean I had to find like $\lim_{n\rightarrow \infty}|f_{n}(x)-f(x)|=0$?

    What I need is just a "hint" where to start. And I actually not good at understanding if it's all written in words.
    Last edited by MathsforNewbs; May 18th 2013 at 10:17 PM.
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  4. #4
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    Re: Uniform convergence

    Basically I was getting at whether uniform convergence comes "for free" given other kinds of convergence and other attributes (like continuity in the function against the variable x).
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  5. #5
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    Re: Uniform convergence

    Quote Originally Posted by MathsforNewbs View Post
    Hello.
    I have proved that the series $\sum_{n=2}^{\infty}\frac{\left (\mathrm{ln}\left ( n \right ) \right )^{p}}{n}$ is convergent if p<-1 and divergent if p\geq -1.

    So the another problem is about to show the series $\sum_{n=2}^{\infty}\frac{\left (\mathrm{ln}\left ( n \right ) \right )^{x}}{n}$ is uniform convergent as $x \in ]-\infty;p]$ if p<-1.

    Over many years, I found if a property can be shown from a basic comparison then students have a better grasp of it.

    If n\ge 2~\&~\alpha>0 then it is good exercise to show that {\left( {{\alpha ^\alpha }n} \right)^{ - 1}} \leqslant {\left( {\log (n)} \right)^{ - \alpha }}.

    That tells us that \sum\limits_{n = 2}^\infty  {{{\left( {{\alpha ^\alpha }n} \right)}^{ - 1}}}  \leqslant \sum\limits_{n - 2}^\infty  {{{\left( {\log (n)} \right)}^{ - \alpha }}} .

    Thus \forall\alpha>0 the series {\sum\limits_{n - 2}^\infty  {\left( {\frac{1}{{\log (n)}}} \right)} ^\alpha } diverges.

    Using that as a lemma, you can prove your question.
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  6. #6
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    Re: Uniform convergence

    I am sorry to say it, to be honest, I don't understand anything of it. I mean I don't think it has something with the Weierstrass to do.
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  7. #7
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    Re: Uniform convergence

    Quote Originally Posted by MathsforNewbs View Post
    I am sorry to say it, to be honest, I don't understand anything of it. I mean I don't think it has something with the Weierstrass to do.
    Well, do you know or are allowed to use the integral test?
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  8. #8
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    Re: Uniform convergence

    Quote Originally Posted by Plato View Post
    Well, do you know or are allowed to use the integral test?
    Yes I know it and am allowed to use anyting (what I have learnt, of course) - just in order to answer it. I mentioned something about proving, on #1;
    Quote Originally Posted by MathsforNewbs View Post
    I have proved that the series $\sum_{n=2}^{\infty}\frac{\left (\mathrm{ln}\left ( n \right ) \right )^{p}}{n}$ is convergent if p<-1 and divergent if p\geq -1.
    by using the Integral Test.
    Last edited by MathsforNewbs; May 19th 2013 at 09:48 AM.
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  9. #9
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    Re: Uniform convergence

    Anyone? I know well that $\mathrm{ln}(n)^{x} \leq \mathrm{ln}(n){p}$ where $x\leq\ p<-1$. I am not sure how to use it with the Weierstrass convergence. I claimed it to be
    \left |\frac{\mathrm{ln}(n)^x}{n}  \right |\leq \frac{\mathrm{ln}(n)^{p}}{n}. Is it correct way?
    Last edited by MathsforNewbs; May 20th 2013 at 07:52 PM. Reason: LaTeX didn't show it correctly
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