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CorollaryStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11
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Integral test as an equivalence with an improper integral

Statement

Let f:[1,∞)→[0,∞)f:[1,\infty)\to[0,\infty) be nonincreasing and Riemann integrable on every compact interval. Then ∑n=1∞f(n)converges if and only if∫1∞f(x) dxconverges.\sum_{n=1}^{\infty}f(n)\quad\text{converges if and only if}\quad\int_1^\infty f(x)\,dx\quad\text{converges}.

Changing finitely many initial terms or moving the finite lower integration endpoint does not affect this equivalence.

Facts & Assumptions

Given: A nonnegative nonincreasing locally integrable ff.

[L1]

Proper integration preserves order, evaluates the integral of a constant, and is additive over adjacent intervals (If f≤gf \le g on [a,b][a,b] and both are integrable then ∫abf≤∫abg\int_a^b f \le \int_a^b g; and m(b−a)≤∫abf≤M(b−a)m(b-a) \le \int_a^b f \le M(b-a), For a<c<ba<c<b: ff is integrable on [a,b][a,b] if and only if it is integrable on [a,c][a,c] and on [c,b][c,b], and then ∫abf=∫acf+∫cbf\int_a^b f = \int_a^c f + \int_c^b f; with the oriented form for arbitrary a,b,ca,b,c).

[L2]

Nonnegative partial sums and integer truncation integrals are nondecreasing, hence converge exactly when bounded (A monotone sequence converges if and only if it is bounded).

[L3]

Every real truncation is bracketed between two integer truncations (Every complete ordered field is Archimedean).

Proof

technique · direct
1.1

For every integer k≥1k\ge1, monotonicity of ff and [L1, L2] give f(k+1)≤∫kk+1f(x) dx≤f(k)f(k+1)\le\int_k^{k+1}f(x)\,dx\le f(k). Adding these inequalities and using interval additivity yields ∑k=2N+1f(k)≤∫1N+1f(x) dx≤∑k=1Nf(k)\sum_{k=2}^{N+1}f(k)\le\int_1^{N+1}f(x)\,dx\le\sum_{k=1}^{N}f(k). Thus the series partial sums are bounded exactly when the integer truncation integrals are bounded. By [L2], this is exactly convergence of the corresponding two monotone sequences.

L1L2
2.1

Suppose the integer truncations converge to II. Given a sufficiently large integer NN and any real R≥NR\ge N, [L3] supplies an integer M>RM>R. Nonnegativity gives ∫1Nf≤∫1Rf≤∫1Mf\int_1^Nf\le\int_1^Rf\le\int_1^Mf, and both integer bounds tend to II; hence the full real-parameter limit is II. The reverse implication is immediate by restriction to integer truncations. Tail invariance handles finite changes.

L3∎

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