Question 1: I want to show that $\sum\limits_{n=1}^\infty\frac{x^n}{n}$ converges uniformly on $[-1,0]$, so I am trying to show equivalently that $$\sup\limits_{x\in[-1,0]}\left|\sum\limits_{n=1}^\infty \frac{x^n}{n} - \sum\limits_{n=1}^m \dfrac{x^n}{n}\right|=\sup\limits_{x\in[-1,0]}\left|\sum\limits_{n=m+1}^\infty \dfrac{x^n}{n}\right|\rightarrow 0$$ as $m\rightarrow\infty$. I'm having some difficulties, perhaps there's a clever trick I'm missing. I think the supremum is when $x=-1$ after computing the sum for a couple values in $[-1,0]$ and seeing that the sums get closer to $0$ as $x\rightarrow 0$.
Question 2 I know that the series fails to converge uniformly on $[-1,1)$. This is unsurprising, as the supremum is no longer at $x=-1$ (I think the supremum does not exist, after calculating several values close to $1$). How can I make this more precise?
A possible sketch of question $2$ might look like:
From here you can use the fact that your series approaches the harmonic series as $x \to 1$, which is well known to diverge to $+\infty$. This will serve to establish that for all $K \in \Bbb{R}_{>0}$ you can find $x \in [-1,1)$ such that $\left|\sum_{n=1}^\infty \frac{x^n}{n} \right| > K$.