derivation-mko-ratio-test-series-01 · ocme-new142-reader-adoption-20261021
Use complete rational ratios for L<1 convergence and L>1 divergence, then audit indecision at L=1.
Qualify eventual nonzero terms and absolute ratio limits, not limits inferred from finite ratios.
algebraepsilonreal_completeG1 — a_n=(1/2)^n/(n+1), b_n=2^n/(n+1) for every n≥0; also c_n=1/(n+1), d_n=1/(n+1)^2.L — proved absolute-ratio limitq — domination ratio between L and oneSome rule statements have reader clarifications in this revision: they are paraphrases, not verbatim quotations. Referenced original source bytes and hashes are unchanged.
mko_definition · cited — Declare the existing limit L=lim|a_(n+1)/a_n| on an eventually nonzero tail. When L<1 choose L<q<1 and establish eventual geometric domination to obtain absolute convergence. When L>1 the term magnitudes eventually fail to tend to 0, so the series diverges. L=1 gives no decision: reciprocal-square and harmonic tails have opposite statuses with that same limit. Use absolute ratios and an actual limit argument, not a finite favorable prefix. Zero denominators and missing limits are separate boundary cases. · mko-ratio-test-seriesalgebra · prerequisite — Use real-field identities; division requires a nonzero denominator and multiplication by a positive number preserves inequalities.epsilon · prerequisite — Convergence requires for every positive epsilon an integer N controlling every n≥N by a strict error bound; use the Archimedean property of the reals.real_complete · prerequisite — Every nonempty bounded-above real set has a supremum, so bounded monotone real sequences converge, including bounded partial sums of nonnegative series.Before
a_n>0\quad(n\ge0)
After
\left|a_{n+1}/a_n\right|=\tfrac12\frac{n+1}{n+2}=\tfrac12-\frac1{2(n+2)}Conditions: none
Rule: algebra · Uses: G1
Leaf basis: algebra
All terms are nonzero, and the full absolute-ratio identity directly supplies its limit.
Before
\frac1{2(n+2)}\to0After
L_a=1/2<q=3/4<1
Conditions: none
Rule: epsilon · Uses: S01
Leaf basis: epsilon
Taking n>1/(2epsilon) controls the error, proving the actual limit and an explicit intermediate q.
Before
a_{n+1}/a_n<1/2<3/4\quad(n\ge0)After
0<a_n\le(3/4)^n a_0=(3/4)^n
Conditions: none
Rule: algebra · Uses: S01, S02
Leaf basis: algebra
Induct from a_0=1 to obtain geometric domination of the entire tail.
Before
\sum_{n=0}^{N}a_n\le\sum_{n=0}^{N}(3/4)^n\le4After
\sum a_n\text{ converges absolutely}Conditions: \text{bounded monotone real sequences converge}
Rule: real_complete · Uses: S03, real_complete
Leaf basis: real_complete
The nonnegative partial sums increase and are bounded; real completeness gives convergence, which is absolute here.
Before
\left|b_{n+1}/b_n\right|=2\frac{n+1}{n+2}=2-\frac2{n+2}After
L_b=2>1,\quad b_{n+1}/b_n\ge3/2\ (n\ge2)Conditions: none
Rule: algebra · Uses: G1
Leaf basis: algebra
The full difference formula proves limit two, and for n≥2 the ratio is at least 3/2.
Before
b_n\ge b_2(3/2)^{n-2}\quad(n\ge2)After
b_n\not\to0;\quad\sum b_n\text{ diverges}Conditions: none
Rule: algebra · Uses: S05
Leaf basis: algebra
Induction gives geometric growth and unbounded terms, violating the necessary term condition for series convergence.
Before
c_{n+1}/c_n=\frac{n+1}{n+2},\quad d_{n+1}/d_n=\left(\frac{n+1}{n+2}\right)^2After
L_c=L_d=1
Conditions: none
Rule: algebra · Uses: G1
Leaf basis: algebra
The errors of the two rational ratios relative to 1 tend to zero, giving the same limit 1; their denominators do not tend to zero, and neither ratio is constant.
Before
\sum c_n\text{ diverges by dyadic blocks};\quad\frac1{(n+1)^2}\le\frac1{n(n+1)}\ (n\ge1)After
\sum d_n\text{ converges by telescoping upper comparison};\quad L=1\text{ is inconclusive}Conditions: none
Rule: epsilon · Uses: G1, S07
Leaf basis: epsilon
Each harmonic dyadic block is at least 1/2; reciprocal-square tails are bounded above by a telescoping series of sum one, giving opposite statuses at the same L.
No substeps in this guide.
S08 · premise_audit · G1, S07 — An inconclusive test does not mean the series has no convergence status.S07 · premise_audit · G1 — A limit of one is not confused with every ratio equaling one.S06 · premise_audit · S05 — A single ratio above one is not treated as a divergence proof.S05 · premise_audit · G1 — The greater-than-one bound holds on the entire tail.S04 · premise_audit · S03, real_complete — Bounded-monotone real convergence is assumed, with the upper bound independently supplied by a geometric sum.S03 · premise_audit · S01, S02 — Domination covers every n, not merely one favorable ratio.S02 · premise_audit · S01 — All comparisons in L<q<1 are strict.S01 · inverse_equivalence · G1 — Absolute values are retained and no zero denominator is used.Qualify a nonzero tail before proving its absolute-ratio limit.
At L=1 use independent arguments instead of forcing a decision.
reject — c is divergent harmonic.reject — The later tail may differ; a limit or whole-tail domination is required.reject — A zero denominator makes the ratio undefined and requires separate treatment.a converges absolutely, b diverges, and harmonic versus reciprocal-square series give opposite outcomes at L=1.
1 declared checks for this guide; only the listed expressions are covered.
b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9fc7fc6484daf7954a4c99bdd9232847aea95535b66a3afb08b3aaccb4de9bca45145cf07dee2db5902e0aee1cd90128f627537e1e28080372018beb930eeeb3c5SCOPED_LEAD_AND_BILINGUAL_CONTEXT_REVIEWSame-source full JSON (formulas, ASTs, controls and revision log)