Fourier series: Jump discontinuity from $-\infty$ to $\infty$-convergence?

Question

To my surprise, there isn’t much information about the Fourier series of $\tan(x)$ on the internet. The Fourier series is $$\tan(x)=2\sum^\infty_{n=1}(-1)^{n-1}\sin(2nx)$$

It is well known that if $f(a^-)=p$ and $f(a^+)=q$, then its Fourier series converges to $\frac{p+q}2$ at $a$.

However, in the case $f(x)=\tan(x)$ and $a=\frac\pi2$, $p=\infty$ and $q=-\infty$. My questions are

What value does the Fourier series of $\tan(x)$ converge to at $\frac\pi2$? Am I allowed to say that it converges to $\lim_{N\to\infty}\frac{N+(-N)}2=0$?

Moreover,

On the whole complex plane, where does not the Fourier series converges to the original function except (possibly) the poles?

Thanks in advance.

Answer 1

It is well known that if $f(a^-)=p$ and $f(a^+)=q$, then its Fourier series converges to $\frac{p+q}2$ at $a$.

This may be well known, but it is not true. Additional assumptions on $f$ are required, such as: "$f$ is bounded with finitely many intervals of monotonicity" or (more general) "$f$ is a function of bounded variation". In whatever form, this result (a theorem of Dirichlet) does not apply to a highly singular function such as $\tan x$, which is not even integrable.

Neither do other classical theorems on the convergence of Fourier series. And indeed, the series $$ 2\sum^\infty_{n=1}(-1)^{n-1}\sin(2nx) $$ does not converge at any points other than $x=\pi k/2$, $k\in\mathbb{Z}$ (see Find all $x$ such that lim $\sin (nx)$ exists). At these points its sum is indeed $0$, which reflects the symmetries of the tangent function: $$ \tan\left(\frac{\pi k}{2} - x\right) = -\tan\left(\frac{\pi k}{2} + x\right) $$ but is not a consequence of the Dirichlet theorem.

To my surprise, there isn’t much information about the Fourier series of $\tan(x)$ on the internet.

Perhaps the reason is that the aforementioned lack of convergence of the series. Placing the equality sign between it and $\tan x$ is not justified.