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Let $f$ be a function defined by a series $$f\left(x\right)=\sum c_n e^{inx}.$$ Sometimes, I can prove that the series converges pointwise (when it does), using the Dirichlet test. When the convergence is uniform by the Weierstrass test (i.e., $\sum\left|c_n\right|<\infty$), I can show that $f$ is continuous. When the series of derivatives converges uniformly ($\sum\left|nc_n\right|<\infty$), I can show that $f$ is differentiable. However, very often I plot the series, and $f$ looks clearly continuous and/or $C^1$ (sometimes piecewise, sometimes globally), but I can't prove it. What are some other tools, besides these basic theorems/tests I mentioned above?

Context: I want to show that $$f\left(x\right) = \sum_{n\geqslant 2} \frac{1-\cos nx}{n^2 \log n}$$ is $C^1$. It's enough to define $$g\left(x\right) = \sum_{n\geqslant 2} \frac{\sin nx}{n \log n},$$ show that this series converges (easy: Dirichlet test), $g$ is continuous (hard), and then integrate it (easy again). When I plot the graph, this series looks like it's uniformly convergent, but I have no idea how to show this. I would by satisfied with a proof that it is continuous at $x=0$, and that it conveges uniformly on any compact set away from zero. I can't show either of these things. My only clue is: I know that $\sum \left(\sin nx\right)/n$ is the sawtooth, which is discontinuous at zero, but with limits at the right and at the left. Multiplying the Fourier coefficients by $1/\log n$ looks like "smoothing", like a convolution, so it should be continuous indeed. How do I show this?

fonini
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There are some fairly sophisticated results that relate the asymptotic behavior of Fourier coefficients to the convergence of the series. One that would be particularly useful to you in this case is the following, which can be found in Section 7.2 of Edwards' book Fourier Series: A Modern Introduction (other such results may be found there as well).

Let $\{a_n\}_{n=1}^\infty \subset(0,\infty)$ be non-increasing. Then the series $\sum_{n=1}^\infty a_n \sin(nx)$ is uniformly convergent if and only if $\lim_{n \to \infty} n a_n =0$.

This result shows that indeed your series defining $g$ converges uniformly, and so the rest of your argument goes through.

Glitch
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