Dirichlet's Test

A wobbling series gets tamed by a sinking weight in this scene. Two ingredients meet and produce convergence.

The partial sums of the sine terms wobble around but stay trapped inside a fixed fence. Multiply them by weights that sink to 0. The wobble gets pressed flatter and flatter until the series converges. The test's exact conditions are in the formulas below.

For example, a ⚽ ball bounces around inside a fence and never escapes. Push it more and more gently, and its wandering dies down until it settles.

Examples & Exercises

Show that $\sum_{n=1}^{\infty} \frac{\sin(nx)}{n}$ converges for every $x$ that is not a multiple of $2\pi$ .

\sum_{k=1}^{N} \sin(kx) = \frac{\sin\frac{Nx}{2}\,\sin\frac{(N+1)x}{2}}{\sin\frac{x}{2}}

\left|\sum_{k=1}^{N} \sin(kx)\right| \le \frac{1}{|\sin(x/2)|}

x=1

: the partial sums

\left\lvert\sum_{k \le N}\sin(kx)\right\rvert

stay below

1/\sin(0.5) \approx 2.0858

Show proof

If $x$ is not a multiple of $2\pi$ then $\sin(x/2) \neq 0$ . The product-to-sum identity $2\sin\frac{x}{2}\sin(kx) = \cos\left(k-\tfrac{1}{2}\right)x - \cos\left(k+\tfrac{1}{2}\right)x$ , summed over $k = 1, \dots, N$ , telescopes to $\sum_{k=1}^{N} \sin(kx) = \frac{\cos\frac{x}{2} - \cos\left(N+\tfrac{1}{2}\right)x}{2\sin\frac{x}{2}} = \frac{\sin\frac{Nx}{2}\sin\frac{(N+1)x}{2}}{\sin\frac{x}{2}}$ .
Each factor in the numerator has absolute value at most $1$ , so the partial sums are bounded uniformly in $N$ by $\left\lvert\sum_{k=1}^{N} \sin(kx)\right\rvert \le \frac{1}{|\sin(x/2)|}$ . At $x = 1$ this is $1/\sin(0.5) \approx 2.0858$ — matching the caption's figure.
$b_n = 1/n$ decreases to $0$ .
Since the partial sums of $a_n = \sin(nx)$ are bounded, Dirichlet's test (summation by parts) gives convergence of $\sum_{n=1}^{\infty} a_n b_n = \sum_{n=1}^{\infty} \frac{\sin(nx)}{n}$ .
(In fact this series equals $(\pi - x)/2$ for $x \in (0, 2\pi)$ .)

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

\left|\sum_{n=1}^{N} (-1)^n\right| \le 1

The alternating partial sums

\left\lvert\sum_{n \le N}(-1)^n\right\rvert

never exceed

1

— the Leibniz test is a corollary of Dirichlet's test.

Show proof

Set $a_n = (-1)^n$ . Its partial sums are $0$ when $N$ is even and $-1$ when $N$ is odd, so $\left\lvert\sum_{n=1}^{N} a_n\right\rvert \le 1$ for every $N$ — the boundedness condition of Dirichlet's test holds.
The Leibniz hypothesis is $b_n \downarrow 0$ . Applying Dirichlet's test to the pair $(a_n, b_n)$ yields convergence of $\sum_{n=1}^{\infty} (-1)^n b_n$ at once — the alternating series test is Dirichlet's test in the case $a_n = (-1)^n$ .

Show that $\sum_{n=1}^{\infty} (-1)^n \frac{n}{n^2+1}$ converges.

b_n = \frac{n}{n^2+1}, \qquad b_n > b_{n+1} \iff n^2+n-1 > 0

b_1 = 0.5 > b_2 = 0.4 > \cdots

b_n = \frac{n}{n^2+1}

decreases monotonically for

n \ge 1

and tends to

0

Show proof

The partial sums of $a_n = (-1)^n$ have absolute value at most $1$ for every $N$ (as in the previous problem). For $b_n = \frac{n}{n^2+1}$ we have $0 < b_n < \frac{1}{n}$ , so $b_n \to 0$ .
Monotonicity of $b_n$ : $b_n > b_{n+1}$ is, after cross-multiplication, $n\left((n+1)^2+1\right) > (n+1)(n^2+1)$ , i.e. $n^3+2n^2+2n > n^3+n^2+n+1$ , i.e. $n^2+n-1 > 0$ .
At $n = 1$ we have $1^2+1-1 = 1 > 0$ , so this holds for all $n \ge 1$ .
Hence $b_n \downarrow 0$ .
$b_1 = \frac{1}{2} = 0.5 > b_2 = \frac{2}{5} = 0.4$ — matching the caption's figures. By Dirichlet's test, $\sum_{n=1}^{\infty} (-1)^n \frac{n}{n^2+1}$ converges.

Part of these learning paths

Examples & Exercises