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

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If $\sum ∣ a_{k} ∣$ converges then $\sum a_{k}$ converges

Statement

Let $(a_{k})$ be a sequence of reals. If the series $\sum ∣ a_{k} ∣$ converges (Series, partial sums, convergence and the sum, divergence, and the tail series) then the series $\sum a_{k}$ converges.

A series with the property that $\sum ∣ a_{k} ∣$ converges is called absolutely convergent; the lemma says that absolute convergence implies convergence.

The same statement holds for a family from a general starting index $m$ , being this statement applied to the shifted sequence $j \mapsto a_{j + m}$ (Series, partial sums, convergence and the sum, divergence, and the tail series).

The converse is false, and the standard witness is the alternating harmonic series. That witness is not available on this page: its convergence is the alternating series test, which is not proved here. Nothing below asserts a converse, and no item on this page uses one.

Facts & Assumptions

Given: A sequence $(a_{k})$ of reals such that the series $\sum ∣ a_{k} ∣$ converges, with partial sums as in Series, partial sums, convergence and the sum, divergence, and the tail series and finite sums as in Finite sums and finite products, by recursion.

[L1]

The Cauchy criterion for series: $\sum d_{k}$ converges if and only if for every real $ε > 0$ there is $N \in N$ with $∣ \sum_{k = p + 1}^{n} d_{k} ∣ < ε$ for all $n > p \geq N$ (A series converges iff for every $ε > 0$ there is $N$ with $∣ a_{m + 1} + \dots + a_{n} ∣ < ε$ for all $n > m \geq N$ , Series, partial sums, convergence and the sum, divergence, and the tail series).

[L2]

Triangle inequality for finite sums: $∣ \sum_{k < d} u_{k} ∣ \leq \sum_{k < d} ∣ u_{k} ∣$ (Triangle inequality for finite sums); the block $\sum_{k = p + 1}^{n} u_{k}$ is by definition the finite sum $\sum_{t < n - p} u_{p + 1 + t}$ (Finite sums and finite products, by recursion), so applying the inequality to the shifted sequence $t \mapsto u_{p + 1 + t}$ gives $∣ \sum_{k = p + 1}^{n} u_{k} ∣ \leq \sum_{k = p + 1}^{n} ∣ u_{k} ∣$ for all naturals $n > p$ .

[L3]

Monotonicity of finite sums: if $x_{t} \geq 0$ for all $t < d$ then $\sum_{t < d} x_{t} \geq 0$ (Laws of finite sums and finite products).

[L4]

Absolute value: $∣ u ∣ \geq 0$ for every real $u$ , and $∣ u ∣ = u$ whenever $u \geq 0$ (Basic properties of the absolute value).

[L5]

Convergence of a real sequence, and the fact that the real and rational formulations of a tolerance agree (Limits and Cauchy sequences of reals).

Proof

technique · direct

1.1L1L5choose

Let $ε > 0$ be an arbitrary real; since $\sum ∣ a_{k} ∣$ converges, the Cauchy criterion applied to the sequence $(∣ a_{k} ∣)$ supplies $N \in N$ with $∣ \sum_{k = p + 1}^{n} ∣ a_{k} ∣ ∣ < ε$ for all $n > p \geq N$ .

1.2L2L3L4

For all naturals $n > p$ the block $\sum_{k = p + 1}^{n} ∣ a_{k} ∣$ is a finite sum of nonnegative terms, hence nonnegative, hence equal to its own absolute value.

2.1step 1.1step 1.2L2

So for all $n > p \geq N$ one has $∣ \sum_{k = p + 1}^{n} a_{k} ∣ \leq \sum_{k = p + 1}^{n} ∣ a_{k} ∣ = ∣ \sum_{k = p + 1}^{n} ∣ a_{k} ∣ ∣ < ε$ .

3.1step 2.1L1∎

As $ε > 0$ was arbitrary, the sequence $(a_{k})$ satisfies the Cauchy criterion, so $\sum a_{k}$ converges.

Remarks

Nothing here identifies the two sums, and they are in general different. What is proved is that the second series converges, not that it converges to the same value; the bound $∣ \sum_{k = 0}^{\infty} a_{k} ∣ \leq \sum_{k = 0}^{\infty} ∣ a_{k} ∣$ is true and is not needed anywhere on this page, so it is not proved here.
Why the Cauchy criterion is the right instrument. The terms have no sign, so A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum does not apply to $\sum a_{k}$ and boundedness of its partial sums would prove nothing. The Cauchy criterion is the one convergence test on this page that never names a candidate sum and never asks for a sign, and the whole proof is the observation that its hypothesis for $(∣ a_{k} ∣)$ implies its hypothesis for $(a_{k})$ , term by term, through one application of the finite triangle inequality.
What this unlocks on this page. The root test (Root test: $lim sup ∣ a_{k} ∣^{1 / k} < 1$ gives absolute convergence and hence convergence, $> 1$ gives divergence, and $= 1$ decides nothing) and the ratio test (Ratio test: $lim sup ∣ a_{k + 1} / a_{k} ∣ < 1$ gives absolute convergence and hence convergence, and $lim inf ∣ a_{k + 1} / a_{k} ∣ > 1$ gives divergence) each produce convergence of $\sum ∣ a_{k} ∣$ directly, by comparison with a geometric series; with this lemma both reach their standard conclusion, the convergence of $\sum a_{k}$ itself. Without it their convergence halves would be strictly weaker than the classical statements.
The systematic theory is elsewhere. Rearrangement, the Riemann series theorem, conditional convergence and products of series all belong with absolute convergence and are developed on a later page of this track. This lemma is only the one implication those two tests need.

If $\sum ∣ a_{k} ∣$ converges then $\sum a_{k}$ converges

Statement

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Proof

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Depends on

Dependency tree · 2 levels

Sources

If ∑∣ak∣∑∣ak​∣ converges then ∑ak∑ak​ converges

Statement

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Proof

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Dependency tree · 2 levels

Sources

If $\sum ∣ a_{k} ∣$ converges then $\sum a_{k}$ converges