Uniform Convergence of $Y(x)$ in $(0,1]$

[Also sprach Zarathustra]

Hello!A little problem:With the given series,$$Y(x)= \sum_{n=1}^{\infty}(-1)^n\frac{x^n\ln^nx}{n!} $$ ,why $Y(x)$ is Uniformly converges for all $x\in(0,1]$ ?Ok, I know that $Y(x)$ is u.c by M-test:

$$\max{|x\ln{x}|}=\frac{1}{e}$$

And,

$$ \sum_{n=0}^{\infty}\frac{(\frac{1}{e})^n}{n!} $$

Is converges! But why only in $(0,1]$ ?
Thank you!

 

[tarantella]

We need some clarifications. First, we talk about uniform convergence on a set, not only at one point (in this case, it's pointwise convergence). Where do you take the $\max$?

 

[Also sprach Zarathustra]

Where do you take the $\max$?

On positive $\mathbb{R}$ .

 

[tarantella]

In this case, the $\max$ is infinite. But if you take it on $(0,1]$, you will notice that the series is normally convergent on $(0,1]$.

 

[Also sprach Zarathustra]

In this case, the $\max$ is infinite. But if you take it on $(0,1]$, you will notice that the series is normally convergent on $(0,1]$.

Normally convergence is not enough for this problem that I have, I need integrate that sum,- I must first prove the U.C. on (0,1].

I'll rephrase my question: is that the only interval,(0,1], that Y(x) U.N in ?

 

[tarantella]

You have the normal (hence uniform) convergence on each interval of the form $(0,a]$, $a>0$.