---
tag: 2138K3VK6D
kind: proof
name: Proof of the Weierstrass M-test
msc: [40A30]
proves: 0NX5NWAFC9
uses: [0MRE6XRE5V, 04Z5G0THN6, 0W3K3QPEWT, 0Y123X48GA, 0K0H5E3Z5N, 0FM6N4R0VB, 03HF5PMY5T, 0N32V0E3XJ, 08FPABDQ4C]
sources:
  - url: https://en.wikipedia.org/wiki/Weierstrass_M-test
    locator: Proof
---

## 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|0MRE6XRE5V]].
2. Fix $x \in S$. Since $0 \le |f_n(x)| \le M_n$ and $\sum M_n$ converges, the [[comparison test|04Z5G0THN6]] shows that $\sum |f_n(x)|$ converges.
3. So $\sum f_n(x)$ [[converges absolutely|0W3K3QPEWT]]. By [[absolutely convergent series converge|0Y123X48GA]], 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|0K0H5E3Z5N]].
5. By the [[monotone convergence theorem|0FM6N4R0VB]], $M$ 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|03HF5PMY5T]] 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|0N32V0E3XJ]].
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|08FPABDQ4C]] 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 $S$.

## Idea

At every point, the tail of the series is at most the tail $\sum_{k > m} M_k$ of a convergent series of numbers. That tail tends to $0$ and does not involve $x$. So one index $m_0$ works for all points at once, which is uniform convergence.