# Question 185: A Nonconstant Absolute Ratio Is Controlled below One by a Declared Rational Limit

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This teaching derivation has second-context AI mathematical and bilingual semantic review, limited to the model and assumptions stated in this question. It has no human review or new Lean proof and does not promote the original question or MKO to Canonical truth.

P250-0185 · q-20261015-advance-20261117-r02-ratio-test-series-0010



## Original question and givens

Use exact scales A=10,B=10, rational ratio r=5/11. Set A=10>0,r=5/11∈(0,1), shift=10, alternating=true. The sequence is a_n=A*(r)^n/(n+shift), multiplied by (−1)^n exactly when alternating is true. Give (scale, ratio, shift, alternating, term_description, absolute_ratio_description, limit_description, dominating_ratio, ratio_bracket, domination_statement, conclusion).


- G001 — Use exact scales A=10,B=10, rational ratio r=5/11. Set A=10>0,r=5/11∈(0,1), shift=10, alternating=true. The sequence is a_n=A*(r)^n/(n+shift), multiplied by (−1)^n exactly when alternating is true. Give (scale, ratio, shift, alternating, term_description, absolute_ratio_description, limit_description, dominating_ratio, ratio_bracket, domination_statement, conclusion).





## Definition and scope

Require A>0, 0<r<1, s≥1 and n≥0. The explicit error bound, induction and geometric bound apply to the entire tail.


- [mko-ratio-test-series](https://ocme.evemisslab.com/en/mko/mko-ratio-test-series/)





## Step-by-step derivation

S001 · impliesBefore

a_n=10*(-1)^n*(5/11)^n/(n+10)

After

|a_n|=10·(5/11)^n/(n+10)>0 for every integer n≥0

Conditions: n≥0; A=10>0; 0<r=5/11<1; s=10≥1

The whole rational base is raised to n; an alternating sign, if present, has absolute value 1.

Step identifiers and dependenciesRule: absolute_value · Uses: G001





S002 · impliesBefore

|a_n|=10·(5/11)^n/(n+10)

After

|a_(n+1)|=10·(5/11)^(n+1)/(n+11)

Conditions: none

Replace every occurrence of n by n+1, including the shifted denominator.

Step identifiers and dependenciesRule: index_shift · Uses: G001, S001





S003 · impliesBefore

|a_(n+1)| / |a_n|

After

abs_ratio=5/11*(n+10)/(n+11)

Conditions: none

Multiply by the reciprocal and cancel A and r^n; their nonzero status was established above.

Step identifiers and dependenciesRule: positive_factor_cancellation · Uses: G001, S002





S004 · impliesBefore

abs_ratio=5/11*(n+10)/(n+11)

After

R_n=5/11−(5/11)/(n+11); |R_n−r|=(5/11)/(n+11)

Conditions: none

Write the numerator as the denominator minus 1 to obtain the explicit positive error.

Step identifiers and dependenciesRule: rational_error_identity · Uses: G001, S003





S005 · impliesBefore

epsilon>0; N=floor(r/epsilon)+1

After

n≥N ⇒ n+11>r/epsilon ⇒ |R_n−r|<epsilon

Conditions: epsilon>0

The Archimedean property supplies this integer N. Taking reciprocals of positive numbers reverses the inequality and controls the entire tail, not a finite sample.

Step identifiers and dependenciesRule: archimedean_tail · Uses: G001, S004





S006 · calculationBefore

q−r := sub(8/11, 5/11)

After

q−r = 3/11

Conditions: none

Set q=(r+1)/2 and check that q is strictly greater than r.

Step identifiers and dependenciesRule: exact_rational_arithmetic · Uses: G001, S005





S007 · calculationBefore

1−q := sub(1, 8/11)

After

1−q = 3/11

Conditions: none

Then check that q is strictly below 1.

Step identifiers and dependenciesRule: exact_rational_arithmetic · Uses: G001, S006





S008 · calculationBefore

|a_0| := div(10, 10)

After

|a_0| = 1

Conditions: none

At n=0 the positive base to the zeroth power is 1; compute the initial domination constant.

Step identifiers and dependenciesRule: exact_rational_arithmetic · Uses: G001, S007





S009 · impliesBefore

0<R_n<r=5/11<q=8/11<1

After

|a_n|≤(1)·(8/11)^n for every n≥0

Conditions: none

Equality holds at n=0. If the bound holds at n, multiplying by R_n<q proves it at n+1. Induction gives the bound for every n.

Step identifiers and dependenciesRule: inductive_geometric_domination · Uses: G001, S008





S010 · impliesBefore

sum from k=0 to M of |a_k| ≤ (1)·(1−(8/11)^(M+1))/(1−(8/11))

After

all absolute partial sums ≤ 11/3

Conditions: none

Since 0<q<1 the finite geometric sum has this bound. Absolute partial sums are increasing and bounded, so completeness of the reals gives convergence.

Step identifiers and dependenciesRule: finite_geometric_sum · Uses: G001, S009





S011 · impliesBefore

M>N≥0

After

sum from k=N+1 to M of |a_k| ≤ (1)·(8/11)^(N+1)/(1−(8/11)) → 0

Conditions: none

The same geometric bound controls differences of tail partial sums, explaining absolute convergence of the entire series.

Step identifiers and dependenciesRule: tail_bound · Uses: G001, S010









## Reverse checks


- S011 · premise_and_result_audit — The conclusion holds with or without alternating signs; no unrequested exact series sum is claimed.
- S010 · premise_and_result_audit — Monotone bounded convergence and the geometric-sum identity are explicit prerequisites, not a claim of a Lean proof.
- S009 · premise_and_result_audit — This is an explicit induction argument, not an inference from finitely many checked ratios.
- S008 · premise_and_result_audit — Recompute |a_0| with exact rational arithmetic, not rounded decimals.
- S007 · premise_and_result_audit — Recompute 1−q with exact rational arithmetic, not rounded decimals.
- S006 · premise_and_result_audit — Recompute q−r with exact rational arithmetic, not rounded decimals.
- S005 · premise_and_result_audit — N depends only on epsilon and the fixed parameters, not on a later n.
- S004 · premise_and_result_audit — The error tends to zero while the denominator grows.
- S003 · premise_and_result_audit — Keep the absolute ratio, not a potentially negative signed ratio.
- S002 · premise_and_result_audit — Do not shift the exponent while leaving the denominator unchanged.
- S001 · premise_and_result_audit — Positive A, r and n+s ensure every following ratio denominator is nonzero.





## Result

(10, 5/11, 10, Yes, a_n=10*(-1)^n*(5/11)^n/(n+10), abs_ratio=5/11*(n+10)/(n+11), limit=5/11, 8/11, Yes, eventual_geometric_domination, absolutely_convergent)

The series converges absolutely, with an explicit geometric tail bound.





## This question review and source binding

Review date: 2026-10-24

Original questions, answers, step expressions, numeric ASTs and historical records are retained. Source candidate labels remain historical and are not overwritten as formal proof.

Source hashes and review receipts
- input commit: 0f7be61e2c22317a4075561dc822c87d881a50db
- original question SHA-256: 6b3f90c11a59f5691ed8b4fa77cac3641e6243ba8452a7ffe3f39a1d44a357fa
- candidate record SHA-256: 781a52ad3b410a64920c8ef4baf2867aa6d55446f8e7b8074fadb36eabaf75b5
- semantic review receipt SHA-256: 4727d1bd957a9b7f6cadb7972ddb5ab80a31f5259dc5394c1dff2b71944ed36b
- question-first witness SHA-256: 51e18649133238ac992229f6e16542a5f3af53ca73b2262ab0524508ea937dee
- Lean: NOT_RUN · human review: NOT_PERFORMED

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