# Thread: 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]$$if \displaystyle p<-1. I am not sure what to do. If it's true that I should use the Weierstrass Test, could you please tell me what do to? Even I have read about it I feel kinda stuck when I try to answer this problem. (I am sorry if I have posted it in the wrong place.) 2. ## Re: Uniform convergence Hey MathsforNewbs. Can you use the fact that the series is continuous and that the point-wise convergence in addition to the continuity criteria of the sum imply uniform convergence? I'm not sure if you can use this so I'm just throwing an idea your way. 3. ## Re: Uniform convergence Hello Chiro. I am not sure if I understood you correctly. Did you mean I had to find like \displaystyle \lim_{n\rightarrow \infty}|f_{n}(x)-f(x)|=0$$?

What I need is just a "hint" where to start. And I actually not good at understanding if it's all written in words.

4. ## 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).

5. ## Re: Uniform convergence

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]$$if \displaystyle p<-1. Over many years, I found if a property can be shown from a basic comparison then students have a better grasp of it. If \displaystyle n\ge 2~\&~\alpha>0 then it is good exercise to show that \displaystyle {\left( {{\alpha ^\alpha }n} \right)^{ - 1}} \leqslant {\left( {\log (n)} \right)^{ - \alpha }}. That tells us that \displaystyle \sum\limits_{n = 2}^\infty {{{\left( {{\alpha ^\alpha }n} \right)}^{ - 1}}} \leqslant \sum\limits_{n - 2}^\infty {{{\left( {\log (n)} \right)}^{ - \alpha }}} . Thus \displaystyle \forall\alpha>0 the series \displaystyle {\sum\limits_{n - 2}^\infty {\left( {\frac{1}{{\log (n)}}} \right)} ^\alpha } diverges. Using that as a lemma, you can prove your question. 6. ## Re: Uniform convergence I am sorry to say it, to be honest, I don't understand anything of it. I mean I don't think it has something with the Weierstrass to do. 7. ## Re: Uniform convergence Originally Posted by MathsforNewbs I am sorry to say it, to be honest, I don't understand anything of it. I mean I don't think it has something with the Weierstrass to do. Well, do you know or are allowed to use the integral test? 8. ## Re: Uniform convergence Originally Posted by Plato Well, do you know or are allowed to use the integral test? Yes I know it and am allowed to use anyting (what I have learnt, of course) - just in order to answer it. I mentioned something about proving, on #1; Originally Posted by MathsforNewbs I have proved that the series \displaystyle \sum_{n=2}^{\infty}\frac{\left (\mathrm{ln}\left ( n \right ) \right )^{p}}{n}$$ is convergent if $\displaystyle p<-1$ and divergent if $\displaystyle p\geq -1$.
by using the Integral Test.

9. ## 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?