I've been tasked to show the following series is uniformly convergent in $(0,1]$
$$\sum_{n=1}^{\infty}(x\ln(x))^n$$
I've proven that $\forall x\in(0,1] : (|x \ln(x)|<1)$, which makes the original a series a geometric series. Which means that:
$$\sum_{n=1}^{\infty}(x \ln(x))^n=\frac1{1 - x \ln(x)}$$
But how do I prove that the series is uniformly convergent?
Thanks!
Are you familiar with the Weierstrass $M$-test? With the observations you've provided, the uniform convergence of this series is immediate; since $x \ln{x} \to 0$ as $x \to 0$ and $x \ln{x} = 0$ when $x = 1$, $x\ln{x}$ is bounded over $(0, 1]$. Since you've already shown that $|x\ln{x}| < 1$ for all $x \in (0, 1]$, this implies $$\sum_{n=0}^\infty (x \ln{x})^n$$ is dominated by the series $$\sum_{n=0}^\infty a^n$$ for some $0 < a < 1$, and therefore converges uniformly by the Weierstrass $M$-test.
If not, that's okay; you're mostly there. As @Neckverse Herdman suggested, try comparing your situation to the convergence of a geometric series on the same interval. If you're familiar with the proofs related to series of that sort, then the transition to this one should be natural.