Cauchy，但是一致的
正如实数的 Cauchy 序列能让你不指明极限就证明收敛，这里也有一致版本。一致 Cauchy 判据说：(f_n) 在 E 上一致收敛当且仅当对每个 epsilon > 0 存在 N，使得对一切 m, n >= N 与一切 x ∈ E 有 |f_n(x) - f_m(x)| < epsilon。用上确界范数说，就是当 m, n >= N 时 ||f_n - f_m|| < epsilon——这些函数一致地聚拢到一起。
从判据到 M-判别法
对于函数的无穷级数 sum f_k，其部分和 s_n = f_1 + … + f_n 构成函数序列，故可用一致 Cauchy 判据。两部分和之差是一段块和，逐项放缩它便得到 Weierstrass M-判别法：若对一切 x ∈ E 有 |f_k(x)| <= M_k，且常数级数 sum M_k 收敛，则 sum f_k 在 E 上一致（且绝对）收敛。
Proof of the M-test via the uniform Cauchy criterion.
Hypothesis: |f_k(x)| <= M_k for all x in E, and sum M_k < infinity.
For m > n, the block of partial sums obeys, for EVERY x:
|s_m(x) - s_n(x)| = | sum_{k=n+1}^{m} f_k(x) |
<= sum_{k=n+1}^{m} |f_k(x)| (triangle inequality)
<= sum_{k=n+1}^{m} M_k.
The right side has no x in it. Since sum M_k converges, its tails
are a real Cauchy sequence: given epsilon > 0 there is N with
sum_{k=n+1}^{m} M_k < epsilon for all m > n >= N.
Therefore ||s_m - s_n|| = sup_x |s_m(x) - s_n(x)| <= sum_{k=n+1}^m M_k
< epsilon for all m > n >= N.
The partial sums are uniformly Cauchy => sum f_k converges
UNIFORMLY on E. QED在实践中使用判别法
M-判别法把函数级数问题化为对单个常数级数的比较判别法——通常是几何级数或 p-级数。先用每项在定义域上的最坏高度作界，再检验这些常数之和。
Example 1: sum_{k>=1} (cos(kx)) / k^2 on all of R.
|cos(kx)/k^2| <= 1/k^2 = M_k, and sum 1/k^2 = pi^2/6 < infinity.
M-test => converges UNIFORMLY on R. (Limit is continuous, by Guide 4.)
Example 2: sum_{k>=0} x^k on [-r, r] with 0 < r < 1.
|x^k| <= r^k = M_k, and sum r^k = 1/(1-r) < infinity.
M-test => uniform on [-r, r]. (But NOT on (-1,1): there
sup|x^k| = 1, so no summable M_k exists — uniformity is local.)