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 and be sequences of reals whose series converge (Series, partial sums, convergence and the sum, divergence, and the tail series), and let . Then:
- converges, with ;
- converges, with .
Moreover, for and an arbitrary sequence , whose series is not assumed to converge:
- converges if and only if converges. Equivalently, diverges if and only if diverges.
Claim 3 is the form used whenever a comparison is made against a constant multiple of a known series.
Facts & Assumptions
Given: Sequences , of reals and , with partial sums and (Series, partial sums, convergence and the sum, divergence, and the tail series, Finite sums and finite products, by recursion).
Additivity and scaling of finite sums: and (Laws of finite sums and finite products).
Algebra of limits: if and then and (Algebra of limits: sums, scalar multiples, products and quotients).
Proof
The partial sums of are , and those of are .
Assume and converge, say and .
Then , so converges with sum , which is claim 1.
Likewise , so converges with sum , which is claim 2.
For claim 3, let and let be arbitrary. If converges then converges by claim 2.
Conversely, if converges then applying claim 2 to the sequence and the scalar , which exists since , shows that converges.
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 . 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 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.
Depends on
Used by
- ∏_j ≥ 0 (1 + (-1)ʲ/√j+2) has partial products tending to 0 although ∑_j ≥ 0 (-1)ʲ/√j+2 converges Counterexample
- 1/4 lies in the Cantor set and is the endpoint of no removed interval, so the endpoints do not exhaust it Counterexample
- Two series with aₖ ≤ bₖ for all k, ∑ bₖ convergent and ∑ aₖ divergent, when the terms may be negative Counterexample
- The subspace Γ of directions along which a series converges absolutely, and its orthogonal complement Γ^⊥ Definition
- 0.999… = 1 and 0.4999… = 0.5: the second expansion of a number is exactly an eventually-all-(b-1) digit sequence Example
- A convergent series in ℝ² with Γ a line and Γ^⊥ a line, computed from the definition Example
- The Cantor function takes the value 1/2 on all of [1/3, 2/3], and its values at 1/9, 1/4 and 7/9 Example
- The Cantor set is homeomorphic to {0,1}^ℕ with the product of discrete topologies, the ternary digits being the coordinates Example
- The Hilbert cube [0,1]^ℕ with the product topology is metrizable, by d(x,y) = ∑ₖ |xₖ - yₖ| / 2^ k+1 Example
- Which points of [0,1] lie in the Cantor set, read off their ternary expansions, with 1/4 worked out Example
- FALSE: limsup |aₖ₊₁/aₖ| ≥ 1 implies the series diverges False statement
- For 0<x<1, the Abel transform of a series is (1-x)²∑_n≥0(n+1)σₙxⁿ, where σₙ are the Cesaro means of its partial sums Lemma
- Positive and negative parts: aₖ = aₖ⁺ - aₖ⁻ and |aₖ| = aₖ⁺ + aₖ⁻; a series converges absolutely iff both ∑ aₖ⁺ and ∑ aₖ⁻ converge, and for a conditionally convergent series both diverge to +∞ Lemma
- The Cauchy product of two absolutely convergent complex series converges absolutely to the product of their sums Lemma
- Abel's test: if ∑ aₖ converges and (bₖ) is monotone and bounded then ∑ aₖ bₖ converges Theorem
- Base-b expansions: for an integer b ≥ 2 every x ∈ [0,1) is the sum of ∑_j ≥ 0 dⱼ / b^ j+1 for digits dⱼ < b, and the digit sequence is unique among those that are not eventually constantly b-1 Theorem
- Dirichlet's rearrangement theorem: an absolutely convergent series converges unconditionally, and every rearrangement of it has the same sum Theorem
- Dirichlet's test: if the partial sums of ∑ aₖ are bounded and (bₖ) is nonincreasing with bₖ → 0, then ∑ aₖ bₖ converges Theorem
- Euler's formula: exp(iθ)=cosθ+isinθ for every real θ Theorem
- For aₖ, bₖ > 0 with aₖ/bₖ → L: if L ∈ (0,∞) the two series share their behaviour, while L = 0 and L = ∞ give one implication each Theorem
- Gauss: for positive terms, if aₖ/aₖ₊₁ = 1 + h/k + rₖ with |rₖ| ≤ C k^-1-ε for k ≥ 1, some constant C and some rational ε > 0, the series converges iff h > 1 Theorem
- Kummer: for positive terms aₖ and weights ζₖ > 0, liminf(ζₖ aₖ/aₖ₊₁ - ζₖ₊₁) > 0 gives convergence, and if ∑ 1/ζₖ diverges while that expression is eventually ≤ 0 the series diverges Theorem
- Ratio test: limsup |aₖ₊₁/aₖ| < 1 gives absolute convergence and hence convergence, and liminf |aₖ₊₁/aₖ| > 1 gives divergence Theorem
- The Cantor function is well defined, satisfies c(x) ≤ c(y) whenever x ≤ y, is surjective onto [0,1], and is constant on every interval removed from the Cantor set Theorem
- The Cantor set is compact, perfect, uncountable, nowhere dense and of measure zero, and it contains no interval of positive length, so its only nonempty connected subsets are single points Theorem
- The Cantor set is exactly the set of ∑_k ≥ 1 aₖ 3⁻ᵏ with every aₖ ∈ {0,2}, and this gives a bijection with {0,1}^ℕ Theorem
- Under square summability, the signed product of (1+pₙ) converges iff the series of pₙ converges Theorem
Dependency tree · next 3 levels
Direct dependencies and their dependencies through the next three levels: 62 results over 22 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.
Sources
- Series (mathematics) (Wikipedia) (standard reference, not scraped)
- T. Tao, Analysis I, 3rd ed., §7.2 (standard reference, not scraped)
- John K. Hunter, An Introduction to Real Analysis (standard reference, not scraped)