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

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A series converges iff for every $ε > 0$ there is $N$ with $∣ a_{m + 1} + \dots + a_{n} ∣ < ε$ for all $n > m \geq N$

Statement

Let $(a_{k})$ be a sequence of reals, with partial sums $s_{n} = \sum_{k < n} a_{k}$ (Series, partial sums, convergence and the sum, divergence, and the tail series). Then $\sum a_{k}$ converges if and only if

$for every real ε > 0 there is N \in N such that ∣ \sum_{k = m + 1}^{n} a_{k} ∣ < ε for all n > m \geq N .$

The block $\sum_{k = m + 1}^{n} a_{k}$ is the finite sum $a_{m + 1} + \dots + a_{n}$ of Finite sums and finite products, by recursion, and it equals $s_{n + 1} - s_{m + 1}$ .

This is the Cauchy criterion transported from sequences to series. Its value is that it decides convergence without producing, or even naming, the sum.

Facts & Assumptions

Given: A sequence $(a_{k})$ of reals with partial sums $s_{n} = \sum_{k < n} a_{k}$ (Series, partial sums, convergence and the sum, divergence, and the tail series, Finite sums and finite products, by recursion).

[L1]

Splitting of finite sums: if $m^{'} \leq n^{'}$ then $\sum_{k < n^{'}} a_{k} = \sum_{k < m^{'}} a_{k} + \sum_{k = m^{'}}^{n^{'} - 1} a_{k}$ (Laws of finite sums and finite products, Finite sums and finite products, by recursion).

[L2]

Every convergent sequence of reals is Cauchy (Every convergent sequence is Cauchy).

[L3]

Every Cauchy sequence of reals converges (The Cauchy criterion from the least-upper-bound property: in a complete ordered field every Cauchy sequence converges).

[L4]

Cauchyness reads: for every rational $ε > 0$ there is $K$ with $∣ s_{p} - s_{q} ∣ < ε$ for all $p, q \geq K$ (Limits and Cauchy sequences of reals). Rational and real tolerances give the same condition, because every real $ε > 0$ exceeds some rational $1 / n$ with $n \geq 1$ natural (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ), and conversely every positive rational is a positive real.

Proof

technique · direct

1.1

For all naturals $m < n$ , splitting $s_{n + 1}$ at the index $m + 1 \leq n + 1$ gives $s_{n + 1} = s_{m + 1} + \sum_{k = m + 1}^{n} a_{k}$ , hence $\sum_{k = m + 1}^{n} a_{k} = s_{n + 1} - s_{m + 1}$ .

L1algebra

1.2

Conversely, every pair $p > q \geq 1$ of naturals is of the form $p = n + 1$ , $q = m + 1$ with $n = p - 1 > m = q - 1 \geq 0$ .

givenalgebra

1.3

Suppose $\sum a_{k}$ converges, that is $(s_{n})$ converges; then $(s_{n})$ is Cauchy.

givenL2

1.4

Suppose conversely that the stated condition holds, and let a rational $ε > 0$ be given; take $N$ for $ε$ as in the condition and put $K : = N + 1$ .

givenchoose

2.1

Let a real $ε > 0$ be given, choose a rational $ε^{'}$ with $0 < ε^{'} < ε$ , and take $K$ for $ε^{'}$ as in the Cauchy condition; put $N : = K$ .

step 1.3L4choose

2.2

Let $p, q \geq K$ . If $p = q$ then $∣ s_{p} - s_{q} ∣ = 0 < ε$ ; otherwise one of them exceeds the other, and by symmetry we may take $p > q \geq K \geq 1$ .

step 1.4algebra

3.1

For all $n > m \geq N$ one has $n + 1 > m + 1 \geq K$ , so $∣ \sum_{k = m + 1}^{n} a_{k} ∣ = ∣ s_{n + 1} - s_{m + 1} ∣ < ε^{'} < ε$ , which is the stated condition.

step 2.1step 1.1L4

3.2

Writing $p = n + 1$ and $q = m + 1$ gives $n > m$ and $m = q - 1 \geq K - 1 = N$ , so the condition applies and $∣ s_{p} - s_{q} ∣ = ∣ \sum_{k = m + 1}^{n} a_{k} ∣ < ε$ .

step 2.2step 1.2step 1.1step 1.4

4.1

So $(s_{n})$ is Cauchy, hence converges, hence $\sum a_{k}$ converges.

step 2.2step 3.2L4L3

5.1

The two implications together are the stated equivalence.

step 3.1step 4.1∎

Remarks

The criterion is stated over blocks, not over partial sums, on purpose. In applications one estimates a run of consecutive terms $a_{m + 1}, \dots, a_{n}$ directly; the translation into $∣ s_{n + 1} - s_{m + 1} ∣$ is step 1.1 and is done once here so that no later proof has to repeat it.
Taking $n = m + 1$ recovers the term test. The single-term block gives $∣ a_{m + 1} ∣ < ε$ for all $m \geq N$ , which is $a_{k} \to 0$ ; so If a series converges then its terms tend to $0$ is the weakest consequence of this criterion. The criterion is strictly stronger, since it constrains arbitrarily long blocks and not only single terms.

Depends on

Used by

Dependency tree · two levels

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Sources

Cauchy's convergence test (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 3 (standard reference, not scraped)
John K. Hunter, An Introduction to Real Analysis (standard reference, not scraped)

A series converges iff for every ε>0ε>0 there is NN with ∣am+1+⋯+an∣<ε∣am+1​+⋯+an​∣<ε for all n>m≥Nn>m≥N