TITLE: Differentiability of the sum of $\sum_{n=1}^{\infty}\frac{\sin(nx)}{n^2}$
QUESTION [5 upvotes]: Consider the series $$\sum_{n=1}^{\infty}\frac{\sin(nx)}{n^2}$$
This series converges uniformly to a continuous function, by Weierstrass's test, and it is the Fourier series of its sum, I'll call $f(x)$.
Is $f$ continuously differentiable in $[-\pi, \pi]$?
The common criterion to determine differentiability of the sum of uniformly convergent series is by testing uniform convergence of the series of derivatives. But I don't see how it's applicable here.
This question:
The Fourier series $\sum_{n=1}^\infty (1/n)\cos nx$ calculates the sum of the derivatives.
But is there an argument to say that $f$ is not differentiable (based on the answer there), without calculating the explicit sum?

REPLY [4 votes]: By the Dirichlet's test, the series $\sum_{n=1}^{\infty} \frac{\cos(nx)}{n}$ converges uniformly on $[-\pi,\pi]\setminus(-\delta,\delta)$ for every $\delta > 0$. So it follows that $f$ is differentiable on $[-\pi,\pi]\setminus\{0\}$ and its derivative is given by the term-wise differentiation
$$ f'(x) = \sum_{n=1}^{\infty} \frac{\cos (nx)}{n}, \qquad x \in [-
\pi,\pi]\setminus\{0\}. $$
At $x = 0$, @Daniel Fischer's answer shows that $f'(0)$ does not exist. In fact, accompanied by the actual computation $f'(x) = -\frac{1}{2}\log(2 - 2\cos x)$ for $x \neq 0$ and by mean value theorem, we can prove that $f'(0) = +\infty$.
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