# Uniform convergence

• May 18th 2013, 10:21 PM
MathsforNewbs
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.)
• May 18th 2013, 10:45 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, 11:14 PM
MathsforNewbs
Re: Uniform convergence
Hello Chiro.

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

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, 11:38 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, 05:38 AM
Plato
Re: Uniform convergence
Quote:

Originally Posted by MathsforNewbs
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.
• May 19th 2013, 09:50 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, 10:03 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, 10:43 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 $\sum_{n=2}^{\infty}\frac{\left (\mathrm{ln}\left ( n \right ) \right )^{p}}{n}$ is convergent if $p<-1$ and divergent if $p\geq -1$.

by using the Integral Test. (Happy)
• May 20th 2013, 08:49 PM
MathsforNewbs
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?