TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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Abel's test: if $\sum a_k$ converges and $(b_k)$ is monotone and bounded then $\sum a_k b_k$ converges

Statement

Let $(a_k)$ and $(b_k)$ be sequences of reals. If $\sum a_k$ converges (Series, partial sums, convergence and the sum, divergence, and the tail series) and $(b_k)$ is monotone (Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences) and bounded, then $\sum a_k b_k$ converges, and its sum is

$\sum_{k=0}^{\infty} a_k b_k \;=\; \Bigl(\sum_{k=0}^{\infty} a_k (b_k - b)\Bigr) + b \sum_{k=0}^{\infty} a_k, \qquad b := \lim_k b_k ,$

the limit $b$ existing because a monotone bounded sequence converges (A monotone sequence converges if and only if it is bounded).

Compared with 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 the hypotheses trade places: there $\sum a_k$ need only have bounded partial sums while $(b_k)$ must tend to $0$ ; here $\sum a_k$ must converge while $(b_k)$ need only be monotone with some limit. Neither test implies the other.

Facts & Assumptions

Given: Sequences $(a_k)$ and $(b_k)$ of reals with $\sum a_k$ convergent and $(b_k)$ monotone and bounded, and the partial sums $A_n = \sum_{k<n} a_k$ (Series, partial sums, convergence and the sum, divergence, and the tail series).

[L1]

A monotone sequence of reals converges if and only if it is bounded (A monotone sequence converges if and only if it is bounded).

[L2]

Monotone means nondecreasing or nonincreasing, and these are the only two possibilities (Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences).

[L3]

A convergent sequence of reals is bounded (Every convergent sequence is bounded).

[L4]

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

[L5]

Linearity of series: if $\sum x_k$ and $\sum y_k$ converge then $\sum(x_k + y_k)$ converges to the sum of the sums, and $\sum c\,x_k$ converges to $c$ times the sum (Convergent series add and scale termwise).

[L6]

Algebra of limits: a convergent sequence minus a constant converges to the limit minus that constant, and multiplying a convergent sequence by $-1$ negates the limit (Algebra of limits: sums, scalar multiples, products and quotients, Limits and Cauchy sequences of reals).

Proof

technique · cases

1.1

Assume $(b_k)$ is nonincreasing.

assume-case noninc

1.2

Assume instead $(b_k)$ is nondecreasing.

assume-case nondec

1.3

In either case $(b_k)$ is monotone and bounded, so it converges; write $b$ for its limit and put $c_k := b_k - b$ , a sequence converging to $0$ .

givenL1L6

1.4

The series $\sum a_k$ converges, so its partial sums $A_n$ form a convergent sequence and are therefore bounded.

givenL3

2.1

In the case where $(b_k)$ is nonincreasing, $(c_k)$ is nonincreasing as well, since it differs from $(b_k)$ by the constant $b$ .

step 1.1step 1.3L2

2.2

In the case where $(b_k)$ is nondecreasing, $(-c_k)$ is nonincreasing and converges to $0$ .

step 1.2step 1.3L2L6

3.1

In the nonincreasing case, $(A_n)$ is bounded and $(c_k)$ is nonincreasing with limit $0$ , so $\sum a_k c_k$ converges by Dirichlet's test.

step 1.4step 2.1L4

3.2

In the nondecreasing case, $(A_n)$ is bounded and $(-c_k)$ is nonincreasing with limit $0$ , so $\sum a_k(-c_k)$ converges by Dirichlet's test; multiplying by the constant $-1$ , $\sum a_k c_k$ converges.

step 1.4step 2.2L4L5

4.1

So in both cases $\sum a_k c_k$ converges; and $\sum b\,a_k$ converges, being a constant multiple of the convergent $\sum a_k$ .

step 3.1step 3.2L5

5.1

Since $a_k b_k = a_k c_k + b\,a_k$ for every $k$ , the series $\sum a_k b_k$ converges, with sum $\sum_{k=0}^{\infty} a_k c_k + b \sum_{k=0}^{\infty} a_k$ , which is the displayed formula.

step 1.3step 4.1L5

6.1

A monotone sequence is nonincreasing or nondecreasing and there is no third possibility, so the two cases cover every hypothesis of the theorem.

step 5.1L2cases-exhaustive∎

Remarks

Both monotonicity directions have to be handled, and only one of them is Dirichlet's hypothesis. 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 requires a nonincreasing factor tending to $0$ . For a nondecreasing bounded $(b_k)$ the shifted sequence $b_k - b$ is nondecreasing and nonpositive, so it is $b - b_k$ that Dirichlet's test accepts, and the sign is absorbed afterwards by linearity. Dirichlet's test could equally have been stated with "monotone" in place of "nonincreasing", since the two forms are equivalent for a factor tending to $0$ (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, remarks); the proof below takes the nonincreasing form as given and does the sign bookkeeping explicitly, which is why both directions appear.
Boundedness of $(b_k)$ is used twice. Once through A monotone sequence converges if and only if it is bounded to produce the limit $b$ , and then implicitly in the decomposition $b_k = (b_k - b) + b$ , which would name nothing if the limit did not exist. Monotone and unbounded is one of the two cases the theorem excludes; the other is bounded and not monotone, and it is that one the companion counterexample to Abel's test on the examples page settles, by showing that dropping monotonicity alone already destroys the conclusion.
The hypothesis on $\sum a_k$ cannot be weakened to bounded partial sums. With $a_k = (-1)^k$ and $b_k = 1$ the partial sums of $\sum a_k$ are bounded and $(b_k)$ is monotone and bounded, yet $\sum a_k b_k = \sum (-1)^k$ diverges. What Dirichlet's test adds in that situation is the hypothesis $b_k \to 0$ , which fails here.

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Sources

Abel'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)

Abel's test: if ∑ak\sum a_k∑ak​ converges and (bk)(b_k)(bk​) is monotone and bounded then ∑akbk\sum a_k b_k∑ak​bk​ converges