Proof of the Weierstrass M-test

Proof

We bound the distance from each partial sum to the sum by a tail of $\sum M_n$, which does not depend on $x$.

1. Let $S$, $f_n$ and $M_n$ be as in the statement. Let $s_N = \sum_{n=0}^{N} f_n$ be the partial sums of the series of functions.
2. Fix $x \in S$. Since $0 \le |f_n(x)| \le M_n$ and $\sum M_n$ converges, the comparison test shows that $\sum |f_n(x)|$ converges.
3. So $\sum f_n(x)$ converges absolutely. By absolutely convergent series converge, it converges. Let $s(x) = \sum_{n=0}^{\infty} f_n(x)$.
4. Let $M = \sum_{n=0}^{\infty} M_n$ and $T_m = \sum_{n=0}^{m} M_n$. Since $M_n \ge 0$, the sequence $(T_m)$ is increasing, and $T_m \to M$ by the definition of the sum of a series.
5. By the monotone convergence theorem, $M$ is the supremum of the $T_m$. So $T_N \le M$ for all $N$.
6. Let $\tau_m = M - T_m$. By step 4, $\tau_m \to 0$. By step 5, $\sum_{k=m+1}^{N} M_k = T_N - T_m \le \tau_m$ for all $N > m$.
7. Let $x \in S$ and $N > m$. By the triangle inequality and step 6, $$|s_N(x) - s_m(x)| = \Big| \sum_{k=m+1}^{N} f_k(x) \Big| \le \sum_{k=m+1}^{N} |f_k(x)| \le \sum_{k=m+1}^{N} M_k \le \tau_m.$$
8. So $-\tau_m \le s_N(x) - s_m(x) \le \tau_m$ for all $N > m$. Letting $N \to \infty$ and using step 3, we get $|s(x) - s_m(x)| \le \tau_m$, since limits preserve non-strict inequalities.
9. Let $\varepsilon > 0$. By step 6, there is $m_0$ with $\tau_m < \varepsilon$ for all $m \ge m_0$. By step 8, $|s_m(x) - s(x)| < \varepsilon$ for all $m \ge m_0$ and all $x \in S$.
10. So $s_m \to s$ uniformly on $S$, and the series converges uniformly on $S$.
11. The functions $|f_n|$ satisfy the same hypotheses, with the same $M_n$. So steps 1 to 10 also show that $\sum |f_n|$ converges uniformly on $S$.
12. By steps 3, 10 and 11, the series $\sum_{n=0}^{\infty} f_n$ converges absolutely and uniformly on $S$.