The fourier series
$$f(t)=\sum_{n\in\mathbb N\\n\text{ odd}}\frac1n\,\sin(nt)$$
converges to a square wave. Square waves are discontinuous functions. I'm wondering if there's a continuous function that "sounds the same as" a square wave, in the sense that it has components with the same amplitude and frequency as the series above, but with different phases.
Do there exist $a_n\in\mathbb R$ such that
$$\sum_{n\in\mathbb N\\n\text{ odd}}\frac1n\,\sin(nt+a_n)$$
converges (pointwise everywhere) to a continuous function?
If we have complete freedom in picking the phase shifts $a_n$, the answer is clearly affirmative: for some sequence $S=\{a_n\}_{n\geq 0}$, $$ f_S(x) = \sum_{n\geq 0}\frac{\sin((2n+1)x+a_n)}{(2n+1)} $$ is a continuous function (proof postponed). On the other hand such continuous function is differentiable at almost no point, due to the rate of decay of the coefficients of its Fourier series, so it is a sort of Weierstrass function. And such wave does not sound as the square wave: any discontinuous signal (or even differentiable, but with large values attained by its derivative) is perceived as painful by our ear, due to the rapid changes of pressure on the eardrum (try the samples of the square wave and sawtooth wave on Wikipedia. As a folklore note, I believe the sample of the triangle wave was used in the intro of Mogwai's song sine wave).
By just considering the constant sequences, $f_S(x)$ ranges between the rectangle wave and the real part of $\text{arctanh}(e^{ix})$, i.e. $\log\left|\tan(x/2)\right|$: both functions are continous over $\mathbb{R}\setminus \pi\mathbb{Z}$, the former is bounded, the latter is not. In general
$$ (\mathcal{L} f_S)(s) = \sum_{n\geq 0}\frac{(2n+1)\cos(a_n)+s \sin(a_n)}{(2n+1)((2n+1)^2+s^2)}$$ so if we pick $a_n=n^2$ we may exploit the fact that $e^{in^2}$ is sufficiently randomly distributed on the unit circle. By Weyl's inequality both $\sum_{n=0}^{N}\sin(n^2)$ and $\sum_{n=0}^{N}\cos(n^2)$ are $\ll \sqrt{N}\log^2 N$, hence by applying summation by parts in the series defining $\mathcal{L} f_S$, then $\mathcal{L}^{-1}$, we find that $f_{\{n^2\}}(x)$ is continuous.
Here it is an approximated depiction of such Weierstrass-like function:
Not a rectangle wave at all.