The Weierstrass M-Test

One test that settles uniform convergence of a series of functions all at once.

The key is a cap that dominates each term at every point. If the caps themselves sum finitely, the tail shrinks across the whole interval together. So the error band tightens everywhere at once.

For example, think of πŸ“¦ box lids covering every item. If the lid sizes add up finitely, nothing can poke out anywhere.

∣fn(x)βˆ£β‰€MnΒ βˆ€x,βˆ‘Mn<βˆžΒ β‡’Β βˆ‘fnΒ convergesΒ uniformly|f_n(x)| \le M_n\ \forall x,\quad \sum M_n<\infty\ \Rightarrow\ \sum f_n\ \text{converges uniformly}
sup⁑x∣S(x)βˆ’SN(x)βˆ£β‰€βˆ‘k>N1k2,N=8:Β β‰ˆ0.1175\sup_x|S(x)-S_N(x)| \le \sum_{k>N}\frac{1}{k^2},\quad N=8:\ \approx 0.1175

Part of these learning paths

Examples & Exercises

Show that βˆ‘n=1∞sin⁑(nx)n2\sum_{n=1}^{\infty} \frac{\sin(nx)}{n^2} converges uniformly for every real xx, by the M-test.

∣sin⁑(nx)n2βˆ£β‰€1n2=:Mn,βˆ‘n=1∞Mn=Ο€26\left|\frac{\sin(nx)}{n^2}\right| \le \frac{1}{n^2} =: M_n, \qquad \sum_{n=1}^{\infty} M_n = \frac{\pi^2}{6}
N=8N=8: error β‰€βˆ‘k>81/k2β‰ˆ0.1175\le \sum_{k>8} 1/k^2 \approx 0.1175 for every xx β€” the band tightens across the whole interval at once.
Show proof

  1. For each nn and every xx, ∣sin⁑(nx)n2βˆ£β‰€1n2=:Mn\left|\frac{\sin(nx)}{n^2}\right| \le \frac{1}{n^2} =: M_n (since ∣sin⁑(nx)βˆ£β‰€1|\sin(nx)| \le 1).

  2. βˆ‘n=1∞Mn=βˆ‘n=1∞1n2=Ο€26β‰ˆ1.6449<∞\sum_{n=1}^{\infty} M_n = \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} \approx 1.6449 < \infty (the Basel problem).

  3. By the M-test, βˆ‘sin⁑(nx)n2\sum \frac{\sin(nx)}{n^2} converges uniformly on all of R\mathbb{R}. At N=8N = 8 the tail bound βˆ‘k>81k2=Ο€26βˆ’βˆ‘k=181k2β‰ˆ1.6449βˆ’1.5274=0.1175\sum_{k>8} \frac{1}{k^2} = \frac{\pi^2}{6} - \sum_{k=1}^{8} \frac{1}{k^2} \approx 1.6449 - 1.5274 = 0.1175 works for every xx at once β€” matching the caption's figure.

Show βˆ‘n=1∞eβˆ’nxn2\sum_{n=1}^{\infty} \frac{e^{-nx}}{n^2} converges uniformly on [0,∞)[0, \infty), by the M-test.

xβ‰₯0β‡’0<eβˆ’nx≀1,∣eβˆ’nxn2βˆ£β‰€1n2=:Mnx \ge 0 \Rightarrow 0 < e^{-nx} \le 1, \qquad \left|\frac{e^{-nx}}{n^2}\right| \le \frac{1}{n^2} =: M_n
N=8N=8: error β‰€βˆ‘k>81/k2β‰ˆ0.1175\le \sum_{k>8} 1/k^2 \approx 0.1175 across all of [0,∞)[0, \infty).
Show proof

  1. For xβ‰₯0x \ge 0, βˆ’nx≀0-nx \le 0 gives 0<eβˆ’nx≀10 < e^{-nx} \le 1, so ∣eβˆ’nxn2∣=eβˆ’nxn2≀1n2=:Mn\left|\frac{e^{-nx}}{n^2}\right| = \frac{e^{-nx}}{n^2} \le \frac{1}{n^2} =: M_n.

  2. Since βˆ‘n=1∞Mn=βˆ‘n=1∞1n2=Ο€26β‰ˆ1.6449<∞\sum_{n=1}^{\infty} M_n = \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} \approx 1.6449 < \infty, the M-test gives uniform convergence on [0,∞)[0, \infty).

  3. Likewise, at N=8N = 8 the tail bound is Ο€26βˆ’βˆ‘k=181k2β‰ˆ0.1175\frac{\pi^2}{6} - \sum_{k=1}^{8} \frac{1}{k^2} \approx 0.1175, applying simultaneously to every xβ‰₯0x \ge 0.

Prove that if each fnf_n is continuous at x0x_0 and fn→ff_n \to f uniformly, then ff is continuous at x0x_0 (the 3-Ρ\varepsilon argument).

∣f(x)βˆ’f(x0)βˆ£β‰€βˆ£f(x)βˆ’fN(x)∣+∣fN(x)βˆ’fN(x0)∣+∣fN(x0)βˆ’f(x0)∣|f(x)-f(x_0)| \le |f(x)-f_N(x)| + |f_N(x)-f_N(x_0)| + |f_N(x_0)-f(x_0)|
N=8N=8: the partial sum hugs the limit function uniformly inside the Β±0.1175\pm 0.1175 band.
Show proof

  1. Let Ξ΅>0\varepsilon > 0. Uniform convergence gives an NN with ∣f(x)βˆ’fN(x)∣<Ξ΅3|f(x) - f_N(x)| < \frac{\varepsilon}{3} for every xx.

  2. Since fNf_N is continuous at x0x_0, there is Ξ΄>0\delta > 0 such that ∣xβˆ’x0∣<Ξ΄|x - x_0| < \delta implies ∣fN(x)βˆ’fN(x0)∣<Ξ΅3|f_N(x) - f_N(x_0)| < \frac{\varepsilon}{3}.

  3. If ∣xβˆ’x0∣<Ξ΄|x - x_0| < \delta, the triangle inequality gives ∣f(x)βˆ’f(x0)βˆ£β‰€βˆ£f(x)βˆ’fN(x)∣+∣fN(x)βˆ’fN(x0)∣+∣fN(x0)βˆ’f(x0)∣<Ξ΅3+Ξ΅3+Ξ΅3=Ξ΅|f(x) - f(x_0)| \le |f(x)-f_N(x)| + |f_N(x)-f_N(x_0)| + |f_N(x_0)-f(x_0)| < \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon. Hence ff is continuous at x0x_0.

9. Series of Functions & Fourier