Cauchy, but uniform
Just as a Cauchy sequence of reals lets you prove convergence without naming the limit, there is a uniform analogue. The uniform Cauchy criterion says: (f_n) converges uniformly on E if and only if for every epsilon > 0 there is an N such that for all m, n >= N and all x in E, |f_n(x) - f_m(x)| < epsilon. In sup-norm language this is simply ||f_n - f_m|| < epsilon for m, n >= N — the functions bunch together uniformly.
From criterion to M-test
For an infinite series of functions sum f_k, the partial sums s_n = f_1 + … + f_n are a function sequence, so the uniform Cauchy criterion applies. The difference of two partial sums is a block sum, and bounding it term by term gives the Weierstrass M-test: if |f_k(x)| <= M_k for all x in E, and the constant series sum M_k converges, then sum f_k converges uniformly (and absolutely) on E.
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. QEDUsing the test in practice
The M-test reduces a function-series problem to a comparison test for a single constant series — usually a geometric series or a p-series. You bound each term by its worst-case height over the domain, then check the constants sum.
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.)