The Weierstrass M-Test

If a majorant MnM_n dominates every term at all xx and ∑Mn\sum M_n converges, the series converges uniformly — here ∣sin⁡(nx)n2∣≤1n2=:Mn\left|\frac{\sin(nx)}{n^2}\right| \le \frac{1}{n^2} =: M_n and ∑Mn=π26\sum M_n = \frac{\pi^2}{6}, so the partial-sum error band tightens across the whole xx-interval at once.

∣fn(x)∣≤Mn ∀x,∑Mn<∞ ⇒ ∑fn converges 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)−SN(x)∣≤∑k>N1k2,N=8: ≈0.1175\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 ∑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)n2∣≤1n2=:Mn,∑n=1∞Mn=π26\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 ≤∑k>81/k2≈0.1175\le \sum_{k>8} 1/k^2 \approx 0.1175 for every xx — the band tightens across the whole interval at once.
Show proof
  1. For each nn and every xx, ∣sin⁡(nx)n2∣≤1n2=:Mn\left|\frac{\sin(nx)}{n^2}\right| \le \frac{1}{n^2} =: M_n (since ∣sin⁡(nx)∣≤1|\sin(nx)| \le 1).
  2. ∑n=1∞Mn=∑n=1∞1n2=π26≈1.6449<∞\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, ∑sin⁡(nx)n2\sum \frac{\sin(nx)}{n^2} converges uniformly on all of R\mathbb{R}. At N=8N = 8 the tail bound ∑k>81k2=π26−∑k=181k2≈1.6449−1.5274=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 — matching the caption's figure.

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

x≥0⇒0<e−nx≤1,∣e−nxn2∣≤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 ≤∑k>81/k2≈0.1175\le \sum_{k>8} 1/k^2 \approx 0.1175 across all of [0,∞)[0, \infty).
Show proof
  1. For x≥0x \ge 0, −nx≤0-nx \le 0 gives 0<e−nx≤10 < e^{-nx} \le 1, so ∣e−nxn2∣=e−nxn2≤1n2=:Mn\left|\frac{e^{-nx}}{n^2}\right| = \frac{e^{-nx}}{n^2} \le \frac{1}{n^2} =: M_n.
  2. Since ∑n=1∞Mn=∑n=1∞1n2=π26≈1.6449<∞\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 π26−∑k=181k2≈0.1175\frac{\pi^2}{6} - \sum_{k=1}^{8} \frac{1}{k^2} \approx 0.1175, applying simultaneously to every x≥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-ε\varepsilon argument).

∣f(x)−f(x0)∣≤∣f(x)−fN(x)∣+∣fN(x)−fN(x0)∣+∣fN(x0)−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 ±0.1175\pm 0.1175 band.
Show proof
  1. Let ε>0\varepsilon > 0. Uniform convergence gives an NN with ∣f(x)−fN(x)∣<ε3|f(x) - f_N(x)| < \frac{\varepsilon}{3} for every xx.
  2. Since fNf_N is continuous at x0x_0, there is δ>0\delta > 0 such that ∣x−x0∣<δ|x - x_0| < \delta implies ∣fN(x)−fN(x0)∣<ε3|f_N(x) - f_N(x_0)| < \frac{\varepsilon}{3}.
  3. If ∣x−x0∣<δ|x - x_0| < \delta, the triangle inequality gives ∣f(x)−f(x0)∣≤∣f(x)−fN(x)∣+∣fN(x)−fN(x0)∣+∣fN(x0)−f(x0)∣<ε3+ε3+ε3=ε|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.

8. Series of Functions & Fourier