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Dirichlet's test: if the partial sums of $\sum a_k$ are bounded and $(b_k)$ is nonincreasing with $b_k \to 0$ , then $\sum a_k b_k$ converges

Statement

Let $(a_k)$ and $(b_k)$ be sequences of reals, and let $A_n = \sum_{k<n} a_k$ be the partial sums of $\sum a_k$ (Series, partial sums, convergence and the sum, divergence, and the tail series). Suppose that

the range $\{\, A_n : n \in \mathbb{N} \,\}$ is bounded (Lower bound, bounded below, bounded set), that is there is a real $M \ge 0$ with $|A_n| \le M$ for every $n$ ; and
$(b_k)$ is nonincreasing (Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences) and converges to $0$ (Limits and Cauchy sequences of reals).

Then $\sum a_k b_k$ converges.

Under hypothesis 2 the terms $b_k$ are automatically nonnegative, and the proof says so before using it: a nonincreasing sequence is bounded below by each of its own later terms, and passing to the limit gives $b_k \ge 0$ (Limits preserve non-strict inequalities).

Nothing is assumed about $\sum a_k$ itself. Its partial sums need only stay bounded; they need not converge. That is what makes this test the source of the alternating series test (The alternating series test: if $(b_k)$ is nonincreasing with $b_k \to 0$ then $\sum_{k} (-1)^{k} b_k$ converges, the sum lies between any two consecutive partial sums, and the error after $n$ terms is at most $b_n$ ) and of examples whose sign pattern is not alternating at all.

Facts & Assumptions

Given: Sequences $(a_k)$ and $(b_k)$ of reals with $A_n = \sum_{k<n} a_k$ bounded in absolute value, and $(b_k)$ nonincreasing with $b_k \to 0$ .

[L1]

Abel summation by parts: for every $n \ge 1$ , $\sum_{k<n} a_k b_k = A_n b_{n-1} - \sum_{k<n-1} A_{k+1}(b_{k+1} - b_k)$ (Abel summation by parts: with $A_n = \sum_{k<n} a_k$ one has $\sum_{k<n} a_k b_k = A_n b_{n-1} - \sum_{k < n-1} A_{k+1}\,(b_{k+1} - b_k)$ for every $n \ge 1$ ).

[L2]

Nonincreasing means $b_j \ge b_k$ whenever $j \le k$ (Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences).

[L3]

Limits preserve non-strict inequalities holding eventually (Limits preserve non-strict inequalities, Limits and Cauchy sequences of reals).

[L4]

Telescoping: with $d_k := b_k - b_{k+1}$ , the partial sums of $\sum d_k$ are $b_0 - b_n$ , and $\sum d_k$ converges if and only if $(b_k)$ converges, with sum $b_0 - \lim_k b_k$ ( $\sum (b_k - b_{k+1})$ converges iff $(b_k)$ converges, with sum $b_0 - \lim b_k$ ).

[L5]

Direct comparison: if $0 \le x_k \le y_k$ from some index on and $\sum y_k$ converges, then $\sum x_k$ converges (If $0 \le a_k \le b_k$ eventually, convergence of $\sum b_k$ gives convergence of $\sum a_k$ , and divergence of $\sum a_k$ gives divergence of $\sum b_k$ ).

[L6]

If $\sum |x_k|$ converges then $\sum x_k$ converges (If $\sum |a_k|$ converges then $\sum a_k$ converges).

[L7]

Linearity: if $\sum x_k$ converges then so does $\sum c\,x_k$ for every real $c$ (Convergent series add and scale termwise).

[L8]

A null sequence times a bounded sequence is null (A null sequence times a bounded sequence is null).

[L9]

Algebra of limits for differences of convergent sequences (Algebra of limits: sums, scalar multiples, products and quotients).

[L10]

A sequence converges to $x$ if and only if some tail of it converges to $x$ (Convergence depends only on the tail).

[L11]

Absolute value: $|xy| = |x||y|$ , $|x| \ge 0$ , and $|-x| = |x|$ (Basic properties of the absolute value).

[L12]

A bounded set of reals admits a bound in absolute value (Lower bound, bounded below, bounded set).

Proof

technique · direct

1.1

Fix a real $M \ge 0$ with $|A_n| \le M$ for every $n \in \mathbb{N}$ .

givenL12choose

1.2

For each fixed $k$ the inequality $b_m \le b_k$ holds for all $m \ge k$ , and $(b_m)_m$ converges to $0$ while the constant sequence with value $b_k$ converges to $b_k$ ; hence $0 \le b_k$ .

givenL2L3

1.3

Put $d_k := b_k - b_{k+1}$ and $c_k := A_{k+1}(b_{k+1} - b_k)$ for $k \in \mathbb{N}$ , and let $s_n := \sum_{k<n} a_k b_k$ , $t_n := \sum_{k<n} c_k$ and $u_n := A_{n+1} b_n$ .

given

1.4

Each $d_k \ge 0$ , since $(b_k)$ is nonincreasing; and $\sum d_k$ converges, with sum $b_0 - 0 = b_0$ , because $(b_k)$ converges to $0$ .

givenL2L4

2.1

For every $k$ , $|c_k| = |A_{k+1}|\,|b_{k+1} - b_k| = |A_{k+1}|\, d_k \le M d_k$ , using $b_{k+1} - b_k = -d_k$ and $d_k \ge 0$ .

step 1.1step 1.3step 1.4L11

2.2

The sequence $(A_{n+1})_{n}$ is bounded by $M$ and $(b_n)$ converges to $0$ , so $u_n = A_{n+1} b_n$ converges to $0$ .

step 1.1step 1.3givenL8

2.3

The series $\sum M d_k$ converges, by step 1.4 and linearity.

step 1.4L7

2.4

For every $n \in \mathbb{N}$ , applying [L1] at the index $n+1 \ge 1$ gives $s_{n+1} = A_{n+1} b_n - \sum_{k<n} A_{k+1}(b_{k+1}-b_k) = u_n - t_n$ .

step 1.3L1

3.1

Since $0 \le |c_k| \le M d_k$ for every $k$ , the series $\sum |c_k|$ converges by comparison, and therefore $\sum c_k$ converges; write $T$ for its sum, so that $t_n \to T$ .

step 2.1step 2.3L5L6

4.1

By step 2.2, step 3.1 and the algebra of limits, $s_{n+1} \to 0 - T = -T$ as $n \to \infty$ .

step 2.2step 3.1step 2.4L9

5.1

The sequence $(s_{n+1})_{n \in \mathbb{N}}$ is the first tail of $(s_n)$ , so $(s_n)$ itself converges to $-T$ ; that is, $\sum a_k b_k$ converges, with sum $-T$ .

step 4.1L10∎

Remarks

Where each hypothesis is used, and none is decorative. Boundedness of $(A_n)$ is used twice: once to bound $|c_k|$ in step 2.1, and once to kill the boundary term in step 2.2. Monotonicity of $(b_k)$ is what makes $|b_{k+1} - b_k|$ equal to $b_k - b_{k+1}$ , so that the bound in step 2.1 telescopes; without it the differences need not sum to anything. And $b_k \to 0$ is used both in the telescoping sum of step 1.4 and in the boundary term of step 2.2.
Why nonincreasing and not monotone, although either would do. Hypothesis 2 could equally be stated with "monotone", and the theorem would still be true: a nondecreasing $(b_k)$ converging to $0$ is nonpositive, so $(-b_k)$ is nonincreasing and converges to $0$ , and applying the theorem to it gives convergence of $\sum a_k(-b_k)$ and hence of $\sum a_k b_k$ (Convergent series add and scale termwise). What "monotone" may not be weakened to is "monotone and bounded": a monotone $(b_k)$ with a nonzero limit is not covered, and for such a factor the conclusion fails in general. The nonincreasing form is chosen here because it is the form the proof uses, and because it makes $b_k \ge 0$ immediate. Abel's test: if $\sum a_k$ converges and $(b_k)$ is monotone and bounded then $\sum a_k b_k$ converges is the result that handles monotone bounded factors, and it has a different hypothesis on $\sum a_k$ .
The sum is not computed. The proof produces the limit as $-T$ , where $T$ is the sum of a series that the argument only proves convergent. This is a convergence test and nothing more.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 81 results over 25 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Dirichlet's test (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 3 (standard reference, not scraped)
Thomson, Bruckner, and Bruckner, Elementary Real Analysis (standard reference, not scraped)

Dirichlet's test: if the partial sums of ∑ak\sum a_k∑ak​ are bounded and (bk)(b_k)(bk​) is nonincreasing with bk→0b_k \to 0bk​→0, then ∑akbk\sum a_k b_k∑ak​bk​ converges