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

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Convergent series add and scale termwise

Statement

Let $(a_{k})$ and $(b_{k})$ be sequences of reals whose series converge (Series, partial sums, convergence and the sum, divergence, and the tail series), and let $c \in R$ . Then:

$\sum (a_{k} + b_{k})$ converges, with $\sum_{k = 0}^{\infty} (a_{k} + b_{k}) = \sum_{k = 0}^{\infty} a_{k} + \sum_{k = 0}^{\infty} b_{k}$ ;
$\sum c a_{k}$ converges, with $\sum_{k = 0}^{\infty} c a_{k} = c \sum_{k = 0}^{\infty} a_{k}$ .

Moreover, for $c \neq 0$ and an arbitrary sequence $(a_{k})$ , whose series is not assumed to converge:

$\sum c a_{k}$ converges if and only if $\sum a_{k}$ converges. Equivalently, $\sum a_{k}$ diverges if and only if $\sum c a_{k}$ diverges.

Claim 3 is the form used whenever a comparison is made against a constant multiple of a known series.

Facts & Assumptions

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

[L1]

Additivity and scaling of finite sums: $\sum_{k < n} (a_{k} + b_{k}) = \sum_{k < n} a_{k} + \sum_{k < n} b_{k}$ and $\sum_{k < n} c a_{k} = c \sum_{k < n} a_{k}$ (Laws of finite sums and finite products).

[L2]

Algebra of limits: if $x_{n} \to x$ and $y_{n} \to y$ then $x_{n} + y_{n} \to x + y$ and $c x_{n} \to c x$ (Algebra of limits: sums, scalar multiples, products and quotients).

Proof

technique · direct

1.1givenL1

The partial sums of $\sum (a_{k} + b_{k})$ are $\sum_{k < n} (a_{k} + b_{k}) = s_{n} + t_{n}$ , and those of $\sum c a_{k}$ are $\sum_{k < n} c a_{k} = c s_{n}$ .

1.2given

Assume $\sum a_{k}$ and $\sum b_{k}$ converge, say $s_{n} \to S$ and $t_{n} \to T$ .

2.1step 1.2step 1.1L2

Then $s_{n} + t_{n} \to S + T$ , so $\sum (a_{k} + b_{k})$ converges with sum $S + T$ , which is claim 1.

2.2step 1.2step 1.1L2

Likewise $c s_{n} \to c S$ , so $\sum c a_{k}$ converges with sum $c S$ , which is claim 2.

3.1step 2.2

For claim 3, let $c \neq 0$ and let $(a_{k})$ be arbitrary. If $\sum a_{k}$ converges then $\sum c a_{k}$ converges by claim 2.

3.2step 2.2algebra

Conversely, if $\sum c a_{k}$ converges then applying claim 2 to the sequence $(c a_{k})$ and the scalar $1 / c$ , which exists since $c \neq 0$ , shows that $\sum (1 / c) (c a_{k}) = \sum a_{k}$ converges.

4.1step 3.1step 3.2∎

The two implications are claim 3, and its contrapositive form is the statement about divergence.

Remarks

There is no product rule here, and there is no rule for $\sum a_{k} b_{k}$ . The proof works because a finite sum is additive and homogeneous, and neither property has an analogue for products. Multiplying series is a genuinely harder question, requiring absolute convergence, and it is not treated on this page.
Claim 3 needs $c \neq 0$ and nothing else. In particular it does not need either series to converge, which is what makes it usable in the divergence direction: scaling a divergent series by a nonzero constant leaves it divergent.

Convergent series add and scale termwise

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources