This question is originated from S/S Fourier Analysis Chapter 2 Exercise 8.
Problem says show sawtooth function$$ f(x)= \begin{cases} -\frac{\pi}{2}-\frac{x}{2}, -\pi<x<0\\ \frac{\pi}{2}-\frac{x}{2}, 0<x<\pi\\ \end{cases} $$ has fourier series $$\frac{1}{2i}\sum_{n\neq 0} \frac{e^{inx}}{n}$$
It's just a simple calculation and we can show the series converges by Dirichlet's test.
But what about the converse, $$\frac{1}{2i}\sum_{n\neq 0} \frac{e^{inx}}{n}$$ (pointwise) converges to $$f(x)= \begin{cases} -\frac{\pi}{2}-\frac{x}{2}, -\pi<x<0\\ \frac{\pi}{2}-\frac{x}{2}, 0<x<\pi\\ \end{cases} $$
For me, with only tools learned until Chapter 2, I think the only way to showing is using uniqueness of the Fourier series with uniform convergence condition.
I think the series $$\frac{1}{2i}\sum_{n\neq 0} \frac{e^{inx}}{n}$$ will uniformly convergent on $[\pi+\delta,-\delta]$ and $[\delta,\pi-\delta]$
then we can show the series has same fourier series by interchanging sum and integral and it's done. But Weierstrass M-test fails on this series.
Is the conjecture true? Otherwise, What's the another way to showing convergence to sawtooth function?