Dirichlet's Test

The partial sums of sin⁡(kx)\sin(kx) wobble, trapped inside 1/∣sin⁡(x/2)∣1/|\sin(x/2)|, and the 1/n1/n weight sinking to 00 tames them into the convergent series ∑n=1∞sin⁡(nx)n\sum_{n=1}^{\infty} \frac{\sin(nx)}{n} — Dirichlet's test.

∑k=1Nsin⁡(kx)=sin⁡Nx2 sin⁡(N+1)x2sin⁡x2,∣∑k=1Nsin⁡(kx)∣≤1∣sin⁡(x/2)∣\sum_{k=1}^{N}\sin(kx) = \frac{\sin\frac{Nx}{2}\,\sin\frac{(N+1)x}{2}}{\sin\frac{x}{2}},\quad \left|\sum_{k=1}^{N}\sin(kx)\right| \le \frac{1}{|\sin(x/2)|}
∑n=1∞sin⁡(nx)n=π−x2,x∈(0,2π)\sum_{n=1}^{\infty}\frac{\sin(nx)}{n} = \frac{\pi-x}{2},\quad x \in (0, 2\pi)

Part of these learning paths

Examples & Exercises

Show that ∑n=1∞sin⁡(nx)n\sum_{n=1}^{\infty} \frac{\sin(nx)}{n} converges for every xx that is not a multiple of 2π2\pi.

∑k=1Nsin⁡(kx)=sin⁡Nx2 sin⁡(N+1)x2sin⁡x2\sum_{k=1}^{N} \sin(kx) = \frac{\sin\frac{Nx}{2}\,\sin\frac{(N+1)x}{2}}{\sin\frac{x}{2}}
∣∑k=1Nsin⁡(kx)∣≤1∣sin⁡(x/2)∣\left|\sum_{k=1}^{N} \sin(kx)\right| \le \frac{1}{|\sin(x/2)|}
x=1x=1: the partial sums ∣∑k≤Nsin⁡(kx)∣\left\lvert\sum_{k \le N}\sin(kx)\right\rvert stay below 1/sin⁡(0.5)≈2.08581/\sin(0.5) \approx 2.0858.
Show proof
  1. If xx is not a multiple of 2π2\pi then sin⁡(x/2)≠0\sin(x/2) \neq 0. The product-to-sum identity 2sin⁡x2sin⁡(kx)=cos⁡(k−12)x−cos⁡(k+12)x2\sin\frac{x}{2}\sin(kx) = \cos\left(k-\tfrac{1}{2}\right)x - \cos\left(k+\tfrac{1}{2}\right)x, summed over k=1,…,Nk = 1, \dots, N, telescopes to ∑k=1Nsin⁡(kx)=cos⁡x2−cos⁡(N+12)x2sin⁡x2=sin⁡Nx2sin⁡(N+1)x2sin⁡x2\sum_{k=1}^{N} \sin(kx) = \frac{\cos\frac{x}{2} - \cos\left(N+\tfrac{1}{2}\right)x}{2\sin\frac{x}{2}} = \frac{\sin\frac{Nx}{2}\sin\frac{(N+1)x}{2}}{\sin\frac{x}{2}}.
  2. Each factor in the numerator has absolute value at most 11, so the partial sums are bounded uniformly in NN by ∣∑k=1Nsin⁡(kx)∣≤1∣sin⁡(x/2)∣\left\lvert\sum_{k=1}^{N} \sin(kx)\right\rvert \le \frac{1}{|\sin(x/2)|}. At x=1x = 1 this is 1/sin⁡(0.5)≈2.08581/\sin(0.5) \approx 2.0858 — matching the caption's figure.
  3. bn=1/nb_n = 1/n decreases to 00. Since the partial sums of an=sin⁡(nx)a_n = \sin(nx) are bounded, Dirichlet's test (summation by parts) gives convergence of ∑n=1∞anbn=∑n=1∞sin⁡(nx)n\sum_{n=1}^{\infty} a_n b_n = \sum_{n=1}^{\infty} \frac{\sin(nx)}{n}. (In fact this series equals (π−x)/2(\pi - x)/2 for x∈(0,2π)x \in (0, 2\pi).)

Show that the Leibniz alternating series test is a corollary of Dirichlet's test.

∣∑n=1N(−1)n∣≤1\left|\sum_{n=1}^{N} (-1)^n\right| \le 1
The alternating partial sums ∣∑n≤N(−1)n∣\left\lvert\sum_{n \le N}(-1)^n\right\rvert never exceed 11 — the Leibniz test is a corollary of Dirichlet's test.
Show proof
  1. Set an=(−1)na_n = (-1)^n. Its partial sums are 00 when NN is even and −1-1 when NN is odd, so ∣∑n=1Nan∣≤1\left\lvert\sum_{n=1}^{N} a_n\right\rvert \le 1 for every NN — the boundedness condition of Dirichlet's test holds.
  2. The Leibniz hypothesis is bn↓0b_n \downarrow 0. Applying Dirichlet's test to the pair (an,bn)(a_n, b_n) yields convergence of ∑n=1∞(−1)nbn\sum_{n=1}^{\infty} (-1)^n b_n at once — the alternating series test is Dirichlet's test in the case an=(−1)na_n = (-1)^n.

Show that ∑n=1∞(−1)nnn2+1\sum_{n=1}^{\infty} (-1)^n \frac{n}{n^2+1} converges.

bn=nn2+1,bn>bn+1  ⟺  n2+n−1>0b_n = \frac{n}{n^2+1}, \qquad b_n > b_{n+1} \iff n^2+n-1 > 0
b1=0.5>b2=0.4>⋯b_1 = 0.5 > b_2 = 0.4 > \cdots: bn=nn2+1b_n = \frac{n}{n^2+1} decreases monotonically for n≥1n \ge 1 and tends to 00.
Show proof
  1. The partial sums of an=(−1)na_n = (-1)^n have absolute value at most 11 for every NN (as in the previous problem). For bn=nn2+1b_n = \frac{n}{n^2+1} we have 0<bn<1n0 < b_n < \frac{1}{n}, so bn→0b_n \to 0.
  2. Monotonicity of bnb_n: bn>bn+1b_n > b_{n+1} is, after cross-multiplication, n((n+1)2+1)>(n+1)(n2+1)n\left((n+1)^2+1\right) > (n+1)(n^2+1), i.e. n3+2n2+2n>n3+n2+n+1n^3+2n^2+2n > n^3+n^2+n+1, i.e. n2+n−1>0n^2+n-1 > 0. At n=1n = 1 we have 12+1−1=1>01^2+1-1 = 1 > 0, so this holds for all n≥1n \ge 1. Hence bn↓0b_n \downarrow 0.
  3. b1=12=0.5>b2=25=0.4b_1 = \frac{1}{2} = 0.5 > b_2 = \frac{2}{5} = 0.4 — matching the caption's figures. By Dirichlet's test, ∑n=1∞(−1)nnn2+1\sum_{n=1}^{\infty} (-1)^n \frac{n}{n^2+1} converges.

6. Series