Cauchy，但是一致的
正如實數的 Cauchy 序列能讓你不指明極限就證明收斂，這裡也有一致版本。一致 Cauchy 判據說：(f_n) 在 E 上一致收斂當且僅當對每個 epsilon > 0 存在 N，使得對一切 m, n >= N 與一切 x ∈ E 有 |f_n(x) - f_m(x)| < epsilon。用上確界範數說，就是當 m, n >= N 時 ||f_n - f_m|| < epsilon——這些函數一致地聚攏到一起。
從判據到 M-判別法
對於函數的無窮級數 sum f_k，其部分和 s_n = f_1 + … + f_n 構成函數序列，故可用一致 Cauchy 判據。兩部分和之差是一段塊和，逐項放縮它便得到 Weierstrass M-判別法：若對一切 x ∈ E 有 |f_k(x)| <= M_k，且常數級數 sum M_k 收斂，則 sum f_k 在 E 上一致（且絕對）收斂。
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在實踐中使用判別法
M-判別法把函數級數問題化為對單個常數級數的比較判別法——通常是幾何級數或 p-級數。先用每項在定義域上的最壞高度作界，再檢驗這些常數之和。
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.)