The Fourier series converges absolutely $\implies$ it converges uniformly. Let $S_N(f)$ be the $N$th partial sum of the Fourier series for $f$.  I.e. 
$$S_N(f) = \sum_{n = -N}^{N} \hat{f}(n) e^{2\pi i n x / L}$$
Suppose that the Fourier series converges absolutely, i.e.
$$\sum_{n = -\infty}^{\infty} \left ( |\hat{f}(n)e^{2\pi i n x/L}| = |\hat{f}(n)|\right ) \lt \infty.$$
Then sequence of functions that are the partial sums converges uniformly.
I think this can be generalized to:
Let $f_n$ be a sequence of functions $f_n:S \rightarrow T$, let $S_N = \sum_{n = -N}^{N} f_n$,  Then if the series $\sum_{N = -\infty}^{\infty} f_n$ converges absolutely, then the sequence of functions $S_N$ converges uniformly.
Attempted Proof .  Let $L = \sum_{N=-\infty}^{\infty} |f_n|$.  Then for all $x\in S$, $\epsilon \gt 0$, there exists $M$ such that $N\gt M \implies d_T(\sum_{n = -N}^{N} |f_n(x)|, L(x)) \lt \epsilon$.  We want to show that there exists a $f:S\rightarrow T$, such that for all $\epsilon \gt 0$, there exists $M$ such that for all $x\in S$, $N \gt M$, $d_T(S_N(x), f(x)) \lt \epsilon$.  What next? 
 A: It is not true. Let $S=[0,1)$ and $f_n(x)=x^n$ if $n>0$, $f_n(x)=0$ if $n\le0$. The series converges absolutely but not uniformly.
A: Let $f_n : S \rightarrow T$ be a sequence of continuous functions $f_n : S \rightarrow T$.  Suppose $\lim_{N\rightarrow \infty} \sum_{k = -N}^{N} |f_n| = L(x)$ for each $x \in S$.  Then for all $\epsilon \gt 0$ there's $M$ such that $N\gt M \implies |\sum|f_n(x)| - L(x)| \lt \epsilon$.  But absolute convergence of the sequence of numbers $|f_n(x)|$ implies convergence when $T$ is Cauchy-complete.  So assume $T$ is and $\sum f_n \rightarrow f$ for some map $f$.  We want to show that $\sum f_n \rightarrow f$ uniformly.
According to Julián Aguirre, this is not true generally.  So there's something about the fourier series ($g_n(x) = \hat{g}(n)e^{-i 2 \pi n x / L}$) that makes this so.
For all $\epsilon \gt 0$ does there exist $N$ such that for all $M \gt N$, $x\in S$, $ \ |\sum_{n = -M}^{M} g_n(x) - g(x)| \lt \epsilon$?
I want to say that $g(x) - \sum^M g_n(x) = \sum_{n = -\infty}^{-(M+1)} g_n(x) + \sum_{n=M+1}^{\infty} g_n(x)$.  Since the series is absolutely convergent then any sub-sum is too.
$$
|\sum_{n = -M}^{M} g_n(x) - g(x)| \leq |\sum_{n = -\infty}^{-(M+1)} g_n(x)| + |\sum_{n=M+1}^{\infty} g_n(x)| \leq \epsilon
$$
since the two sums on the right go to zero.  But this proves it in general, so something's wrong.
