Show that the series $\sum\limits_{k=1}^{\infty}\dfrac{x^k}{k}$ does not converge uniformly on $(-1,1)$. (Hint: Show that the sequence of partial sums is not Cauchy in the sup-norm.)
Proof: Let $s_n = \sum\limits_{k=1}^{n}\dfrac{x^k}{k}$. We need to show that there exists $\epsilon>0$ such that for all $N\in\mathbb{N}$ and for $n \geq N$ implies $\|s_n-s_N\|_{\sup} \geq \epsilon$. As a matter of fact, I claim that this must be true for any $\epsilon >0$ we choose. So let $\epsilon >0$. But, \begin{align} \|s_n-s_N\|_{\sup} &= \|\sum\limits_{k=1}^{n}\dfrac{x^k}{k} - \sum\limits_{k=1}^{N}\dfrac{x^k}{k}\|_{\sup} \\ &= \|\sum\limits_{k=N+1}^{n}\dfrac{x^k}{k}\|_{\sup} \\ &= \sum\limits_{k=N+1}^{n}\dfrac{1}{k} \end{align} which is a harmonic series and will diverge to infinity as $n\to\infty$. $\blacksquare$