We can see that $f$ isn't differentiable at $0$ [and generally at $2m\pi$ by periodicity] by estimating the difference quotient
$$\frac{f(x_k) - f(0)}{x_k}$$
for a suitable sequence $(x_k)$ converging to $0$. Let's choose $x_k = \frac{1}{k}$. Then we have
$$f(x_k) \geqslant \sum_{n = 1}^k \frac{1}{n^2}\sin \frac{n}{k} - \sum_{n = k+1}^{\infty} \frac{1}{n^2} > \frac{2}{\pi k} \sum_{n = 1}^k \frac{1}{n} - \frac{1}{k} > \biggl(\frac{2}{\pi}\log k - 1\biggr)x_k$$
since $\sin x > \frac{2}{\pi} x$ for $0 < x < \pi/2$, $\sum_{n = 1}^k \frac{1}{n} > \log k$, and $\sum_{n = k+1}^{\infty} \frac{1}{n^2} < \sum_{n = k+1}^{\infty} \frac{1}{n(n-1)} = \frac{1}{k}$. So we directly see that
$$\lim_{k\to\infty} \frac{f(x_k)}{x_k} = +\infty.$$
In the sine series
$$f(x)=\sum_{n\geq1}\frac{\sin n x}{\sqrt{n}}$$
The coefficients $c_n:=\frac{1}{\sqrt{n}}$ form a decreasing sequence converging to $0$, and
$$
\sum_{n\geq1}\frac{c_n}{n}=\sum_{n\geq1}\frac{1}{n^{3/2}}<\infty$$
Therefore
- $f\in L_1(\mathbb{T})$, and
- $S_n(x)=\sum^n_{k=1}\frac{\sin kx}{\sqrt{k}}$ converges in $L_1(\mathbb{T})$ to $f(x)$.
Here, I am using the following result for special type of trigonometric series(See F. Jones, Lebesgue Integration on Euclidean space, pp. 437 for example).
Theorem: Suppose $c_n\searrow0$.
$f(x)=\sum_{n\geq1} c_n\sin nx$ is in $L_1(\mathbb{T})$ iff $\sum_n\frac{c_n}{n}<\infty$. In either case, $S_n(x)=\sum^n_{k=1} c_k\sin kx$ converges to $f$ in $L_1$.
Ideas of the proof of Theorem: In one direction, namely that $f\in L_1(\mathbb{T})$ implies the convergence of $\sum_n\frac{c_n}{n}$ follows from a more general result that can be found here. One may also construct a specific proof for the case of sine series of the type considered in the statement of the Theorem (See reference).
On the other direction, the reference above shows that convergence of $\sum_n\frac{c_n}{n}$ implies that $S_n(x)$ converges in $L_1(\mathbb{T})$. This is done by
- Showing that $c_n\log n=o(1)$ as $n\rightarrow\infty$
- The sine kernel $K_n(x)=\sin x+\ldots \sin nx$ satisfies $\|K_n\|_{L_1(\mathbb{T})}=O(\log (n+1))$.
- Using summation by parts, one shows that
$$\|f-S_n\|_{L_1(\mathbb{T})}\leq \kappa\sum_{n>N}\frac{c_n}{n}+2\kappa c_{N+1}\log(N+1)$$
for some constant $\kappa>0$.
Edit:
F. Jone's gives no references for this result, but I suppose this is well known to every Harmonic Analyst. Hopefully someone who reads this answer can provide more references. I ignore if Zigmund's book contain these sort of results.
Best Answer
To inspect the discontinuity of the summation, let's calculate the sum. By the Abel's theorem,
$$ f(x) := \sum_{n=1}^{\infty} \frac{\sin nx}{n} = \lim_{s\to 0^{+}} \sum_{n=1}^{\infty} \frac{\sin nx}{n} e^{-ns}. $$
By utilizing Taylor expansion of the logarithm,
\begin{align*} \sum_{n=1}^{\infty} \frac{\sin nx}{n} e^{-ns} &= \Im \sum_{n=1}^{\infty} \frac{e^{n(ix-s)}}{n} = - \Im \log (1 - e^{ix-s}) \\ &= -\Im \log (1 - e^{-s}\cos x - ie^{-s}\sin x) \\ &= \arctan \left(\frac{e^{-s}\sin x}{1 - e^{-s}\cos x}\right). \end{align*}
Thus taking $s \to 0^{+},$
$$ f(x) = \arctan \left(\frac{\sin x}{1 - \cos x}\right) = \arctan \left(\cot \frac{x}{2}\right) = \arctan \left(\tan \frac{\pi-x}{2}\right). $$
Therefore
$$ f(x) = \begin{cases} \frac{\pi - x}{2} & x \in (0, 2\pi),\\ 0 & x = 0, \\ f(x+2\pi), & x \in \Bbb{R}. \end{cases} $$
This shows a clear-cut jump discontinuity at each $x \in 2\pi \Bbb{Z}$.