# Uniform convergence

Printable View

• May 18th 2013, 09:21 PM
MathsforNewbs
Uniform convergence
Hello.
I have proved that the series $\displaystyle$\sum_{n=2}^{\infty}\frac{\left (\mathrm{ln}\left ( n \right ) \right )^{p}}{n}$$is convergent if \displaystyle p<-1 and divergent if \displaystyle p\geq -1. So the another problem is about to show the series \displaystyle \sum_{n=2}^{\infty}\frac{\left (\mathrm{ln}\left ( n \right ) \right )^{x}}{n}$$ is uniform convergent as $\displaystyle$x \in ]-\infty;p]$$if \displaystyle p<-1. I am not sure what to do. If it's true that I should use the Weierstrass Test, could you please tell me what do to? Even I have read about it I feel kinda stuck when I try to answer this problem. (I am sorry if I have posted it in the wrong place.) • May 18th 2013, 09:45 PM chiro Re: Uniform convergence Hey MathsforNewbs. Can you use the fact that the series is continuous and that the point-wise convergence in addition to the continuity criteria of the sum imply uniform convergence? I'm not sure if you can use this so I'm just throwing an idea your way. • May 18th 2013, 10:14 PM MathsforNewbs Re: Uniform convergence Hello Chiro. I am not sure if I understood you correctly. Did you mean I had to find like \displaystyle \lim_{n\rightarrow \infty}|f_{n}(x)-f(x)|=0$$?

What I need is just a "hint" where to start. And I actually not good at understanding if it's all written in words.
• May 18th 2013, 10:38 PM
chiro
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).
• May 19th 2013, 04:38 AM
Plato
Re: Uniform convergence
Quote:

Originally Posted by MathsforNewbs
Hello.
I have proved that the series $\displaystyle$\sum_{n=2}^{\infty}\frac{\left (\mathrm{ln}\left ( n \right ) \right )^{p}}{n}$$is convergent if \displaystyle p<-1 and divergent if \displaystyle p\geq -1. So the another problem is about to show the series \displaystyle \sum_{n=2}^{\infty}\frac{\left (\mathrm{ln}\left ( n \right ) \right )^{x}}{n}$$ is uniform convergent as $\displaystyle$x \in ]-\infty;p]$$if \displaystyle p<-1. Over many years, I found if a property can be shown from a basic comparison then students have a better grasp of it. If \displaystyle n\ge 2~\&~\alpha>0 then it is good exercise to show that \displaystyle {\left( {{\alpha ^\alpha }n} \right)^{ - 1}} \leqslant {\left( {\log (n)} \right)^{ - \alpha }}. That tells us that \displaystyle \sum\limits_{n = 2}^\infty {{{\left( {{\alpha ^\alpha }n} \right)}^{ - 1}}} \leqslant \sum\limits_{n - 2}^\infty {{{\left( {\log (n)} \right)}^{ - \alpha }}} . Thus \displaystyle \forall\alpha>0 the series \displaystyle {\sum\limits_{n - 2}^\infty {\left( {\frac{1}{{\log (n)}}} \right)} ^\alpha } diverges. Using that as a lemma, you can prove your question. • May 19th 2013, 08:50 AM MathsforNewbs Re: Uniform convergence I am sorry to say it, to be honest, I don't understand anything of it. I mean I don't think it has something with the Weierstrass to do. • May 19th 2013, 09:03 AM Plato Re: Uniform convergence Quote: Originally Posted by MathsforNewbs I am sorry to say it, to be honest, I don't understand anything of it. I mean I don't think it has something with the Weierstrass to do. Well, do you know or are allowed to use the integral test? • May 19th 2013, 09:43 AM MathsforNewbs Re: Uniform convergence Quote: Originally Posted by Plato Well, do you know or are allowed to use the integral test? Yes I know it and am allowed to use anyting (what I have learnt, of course) - just in order to answer it. I mentioned something about proving, on #1; Quote: Originally Posted by MathsforNewbs I have proved that the series \displaystyle \sum_{n=2}^{\infty}\frac{\left (\mathrm{ln}\left ( n \right ) \right )^{p}}{n}$$ is convergent if $\displaystyle p<-1$ and divergent if $\displaystyle p\geq -1$.

by using the Integral Test. (Happy)
• May 20th 2013, 07:49 PM
MathsforNewbs
Re: Uniform convergence
Anyone? I know well that $\displaystyle$\mathrm{ln}(n)^{x} \leq \mathrm{ln}(n){p}$$where \displaystyle x\leq\ p<-1$$. I am not sure how to use it with the Weierstrass convergence. I claimed it to be
$\displaystyle \left |\frac{\mathrm{ln}(n)^x}{n} \right |\leq \frac{\mathrm{ln}(n)^{p}}{n}$. Is it correct way?