TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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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 ${A_{n} : n \in N}$ is bounded (Lower bound, bounded below, bounded set), that is there is a real $M \geq 0$ with $∣ A_{n} ∣ \leq M$ 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 $b_{k} \geq 0$ (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 $n \geq 1$ , $\sum_{k < n} a_{k} b_{k} = A_{n} b_{n - 1} - \sum_{k < n - 1} A_{k + 1} (b_{k + 1} - b_{k})$ (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 $n \geq 1$ ).

[L2]

Nonincreasing means $b_{j} \geq b_{k}$ whenever $j \leq k$ (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 $0 \leq x_{k} \leq y_{k}$ from some index on and $\sum y_{k}$ converges, then $\sum x_{k}$ converges (If $0 \leq a_{k} \leq b_{k}$ 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: $∣ x y ∣ = ∣ x ∣ ∣ y ∣$ , $∣ x ∣ \geq 0$ , 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

1.1givenL12choose

Fix a real $M \geq 0$ with $∣ A_{n} ∣ \leq M$ for every $n \in N$ .

1.2givenL2L3

For each fixed $k$ the inequality $b_{m} \leq b_{k}$ holds for all $m \geq k$ , and $(b_{m})_{m}$ converges to $0$ while the constant sequence with value $b_{k}$ converges to $b_{k}$ ; hence $0 \leq b_{k}$ .

1.3given

Put $d_{k} : = b_{k} - b_{k + 1}$ and $c_{k} : = A_{k + 1} (b_{k + 1} - b_{k})$ for $k \in N$ , and let $s_{n} : = \sum_{k < n} a_{k} b_{k}$ , $t_{n} : = \sum_{k < n} c_{k}$ and $u_{n} : = A_{n + 1} b_{n}$ .

1.4givenL2L4

Each $d_{k} \geq 0$ , since $(b_{k})$ is nonincreasing; and $\sum d_{k}$ converges, with sum $b_{0} - 0 = b_{0}$ , because $(b_{k})$ converges to $0$ .

2.1step 1.1step 1.3step 1.4L11

For every $k$ , $∣ c_{k} ∣ = ∣ A_{k + 1} ∣ ∣ b_{k + 1} - b_{k} ∣ = ∣ A_{k + 1} ∣ d_{k} \leq M d_{k}$ , using $b_{k + 1} - b_{k} = - d_{k}$ and $d_{k} \geq 0$ .

2.2step 1.1step 1.3givenL8

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 $0$ .

2.3step 1.4L7

The series $\sum M d_{k}$ converges, by step 1.4 and linearity.

2.4step 1.3L1

For every $n \in N$ , applying [L1] at the index $n + 1 \geq 1$ gives $s_{n + 1} = A_{n + 1} b_{n} - \sum_{k < n} A_{k + 1} (b_{k + 1} - b_{k}) = u_{n} - t_{n}$ .

3.1step 2.1step 2.3L5L6

Since $0 \leq ∣ c_{k} ∣ \leq M d_{k}$ for every $k$ , the series $\sum ∣ c_{k} ∣$ converges by comparison, and therefore $\sum c_{k}$ converges; write $T$ for its sum, so that $t_{n} \to T$ .

4.1step 2.2step 3.1step 2.4L9

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

5.1step 4.1L10∎

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

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 $b_{k} \geq 0$ 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.

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

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

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

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

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