Given that a series of functions (continuous) $f_k$ converge to a function $f$, and if we can bound $f_k < M_k$ for some sequence $M_k$, and show that $\sum M_k$ converges, then by the M-test, the series $\sum f_k$ converges uniformly.
However, this is obviously not true if $f$ is a discontinuous function since for a series of continuous functions, given that it converges uniformly, then $f$ must be continuous. However, in the statement of the theorem, I can't see any mention of this case.
Using: http://mathworld.wolfram.com/WeierstrassM-Test.html
The reason I ask this is because I found the fourier series of a discontinuous function, and the fourier series is composed of continuous functions, which I can bound above by some sequence $M_k$ and showed it to converge by the p-test, and concluded that then it must converge to $f$, however I realised that $f$ is discontinuous.
The function $f$ is defined as:
$$f(x) = \begin{cases} 1, \quad 0<x<\pi \\ -1, \quad -\pi<x<0. \end{cases}$$
and the function $f$ is $2\pi$-periodic.
The fourier series I found (from memory) was
$$Sf(x) = \frac{4}{\pi}\sum_{k=1}^{\infty} \frac{\sin ((2k-1)x)}{(2k-1)^2}$$
where the summand can be bounded above by
$$M_k := \frac{1}{k^2}$$ which will converge by the $p$ test.
