Uniform convergence of $\sum_{n=1}^{\infty}(x\cdot\ln x)^n$ for $x\in(0,1]$

Question

How can I prove or disprove uniform convergence $$ \sum_{n=1}^{\infty}(x\cdot\ln x)^n$$ for $x\in(0,1]$?

Answer 1

For every $x \in (0,1]$ you have that $|x\ln(x)|\leq e^{-1}$, thus $$ \left| {\left( {x\ln x} \right)^n } \right| = \left| {\left( {x\ln x} \right)} \right|^n \le e^{ - n} $$ for every integer $n\geq 1$. Since the numerical series $$ \sum\limits_{n = 1}^\infty {e^{ - n} } $$ by Weierstrass M-Test you get the uniform convergence of your series.

Answer 2

Notice that

$$\lvert x\ln x\rvert \leq x^2,$$

and so $x\ln x\to 0$ as $x\to0^+$ This means that $x\ln x$ is bounded for $x\in(0,1]$ (this is easy to see as we could extend it to a continuous function on $[0,1]$ as the limit as $x\to0^+$ exists, and all continuous functions on compact intervals are bounded), and so there is some $\alpha>0$ such that

$$\lvert x\ln x\rvert\leq \alpha.$$

Using some basic calculus is it also easy to see that $\alpha<1$. Thus, setting $M_n=\alpha^n$, we have that

$$\forall n\in\mathbb{Z}^+ :\forall x\in(0,1] : \lvert x\ln x\rvert^n\leq M_n, \quad \text{and} \quad \sum_{j=1}^\infty M_j<\infty,$$

where the series converges as it is a convergent geometric series (as $\alpha\in(0,1)$). It follows by the Weierstrass M-test that the series

$$\sum_{j=1}^\infty (x\ln x)^j$$

converges absolutely and uniformly on $(0,1]$.