Study of functions and geometric series I have the following function: $$f(x):= \sum_{n=1}^\infty \frac{1}{n}\sin\left(\dfrac x n\right)$$
I want to know two things:


*

*If it's continuous on a certain interval

*If it's differentiable
To verify if it's continuous I can use Abel's theorem that says that if there exists an $x_0$ such that $f(x):=\sum_{n=1}^\infty \dfrac{1}{n}\sin\left(\dfrac{x_0} n \right)$ converges then it is continuous on $[-x_0,x_0]$.
We can say that $f(x)\leq \sum_{n=1}^\infty \dfrac T n$ with $T\in \mathbb{R}/(-1;1)$
Now I don't where to go and I have no clue how to show that it's differentiable.
 A: 
PRIMER: Uniform Convergence and Differentiability
Let $\phi_n(x)$ be a sequence of differentiable functions for $x\in [a,b]$ such that $\lim_{n\to \infty}\phi_n(x_0)$ converges for some $x_0\in [a,b]$.  If $\phi_n'(x)$ converges uniformly on $[a,b]$, then $\phi)n(x)$ converges to some function, say $\phi(x)$, and $\lim_{n\to \infty}\phi_n'(x)=\phi'(x)$ for $x\in [a,b]$.

Let $f(x)$ be the function with series representation given by
$$f(x)=\sum_{n=1}^\infty \frac{1}{n}\sin(x/n) \tag 1$$
The series $S(x)$, formed by differentiating $(1)$ term-by-term, is given by
$$S(x)=\sum_{n=1}^\infty \frac{1}{n^2}\cos(x/n) \tag 2$$
Note that since $|\cos(x/n)|\le 1$, $|S(x)|\le \sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$.  Hence, the Weierstrass M-Test guarantees that the series in $(2)$ converges uniformly for all $x$ on any closed interval.  Inasmuch as the series in $(1)$ converges for all $x$ (actually, we only need convergence at a single point), then we have
$$f'(x)=\sum_{n=1}\frac{1}{n^2}\cos(x/n)$$
for all $x$.  And since differentiability implies continuity, we are done!
