I was wondering if someone could check over my proof of the question in the title. I've seen it done from the definition of continuity but I tried to use the following theorem:
Theorem: If $D$ is the common domain of a sequence of continuous functions $f_n(x)$ that converge uniformly to $f(x)$ as $n \rightarrow \infty$ then $f$ is continuous.
Prove, $f(x) = \sum_{k=1}^\infty \frac{\sin(kx)}{k(k+1)}$ is continuous
Let $f_n(x) = \sum_{k=1}^n\frac{\sin(kx)}{k(k+1)}$ and let $f(x) = \sum_{k=1}^\infty \frac{\sin(kx)}{k(k+1)}$
Suppose $\epsilon > 0$ then let $N > \frac{1}{\epsilon}$, so we have $\forall n\geq N$
$|f_n(x) - f(x)| = |\sum_{k=1}^n\frac{\sin(kx)}{k(k+1)} - \sum_{k=1}^\infty \frac{\sin(kx)}{k(k+1)}| = |\sum_{k=n+1}^\infty\frac{\sin(kx)}{k(k+1)}| \leq \sum_{k=n+1}^\infty\frac{|\sin(kx)|}{|k(k+1)|} \leq \sum_{k=n+1}^\infty\frac{1}{k(k+1)} = \sum_{k=n+1}^\infty\frac{1}{k} - \frac{1}{k+1} = \frac{1}{n+1} < \frac{1}{n} < \epsilon$.
Thus, $\ f_n(x)$ converges uniformly to $\sum_{k=1}^\infty \frac{\sin(kx)}{k(k+1)}$ and by the previous theorem $f(x)$ is continuous.