Use terms a*r^k for k=0,1,... with exact a=5/6, r=-1/2 and |r|<1. Give (a, r, absolute_series_sum, absolutely_convergent, certificate).
G001 — Use terms a*r^k for k=0,1,... with exact a=5/6, r=-1/2 and |r|<1. Give (a, r, absolute_series_sum, absolutely_convergent, certificate).
Definition and scope
Use the original question domain, assumptions and cited MKO definitions. This item has second-context mathematical and bilingual review, with the model, answer and every arithmetic step independently recomputed from the original parameters; the source question remains a candidate.
Preserve the original snapshot's historical interface marker while listing the full ordered fields here,without browser JavaScript or an external lookup.Field labels are presentation metadata,not mathematical evaluation.The fourth term-zero No marks insufficiency,not a mislabeled second convergence flag.
absolute geometric series:original SIGNED a=5/6;SIGNED r=-1/2;absolute ratio=1/2<1;effective terms=(5/6)*(1/2)^k,k=0,1,...
After
candidate sum=5/3;effective denominator1-ratio=1/2 is nonzero
Conditions: none
Retain signed a and r.Change to effective absolute values only when this operation asks for the absolute series.The |r|<1 guard is this certificate's domain;finite samples or an unrelated zero-coefficient exception do not change it.
absolute geometric series:for ANY N>=1,the first N terms start at k0 with leading=5/6,ratio=1/2
After
(1-ratio)*S_N=(5/6)*(1-(1/2)^N);S_N=(5/6)*(1-(1/2)^N)/(1/2);candidate minus S_N=(5/6)*(1/2)^N/(1/2)
Conditions: none
Multiply the finite sum by1−ratio and cancel adjacent powers,leaving endpoints.This identity holds for arbitrary N≥1,not a formula guessed from several partial sums.
geometric power tends to0;remainder tends to0;partial sums converge to the exact candidate
Conditions: none
Powers of a ratio with absolute value strictly below1 tend to0.Fixed finite coefficients and a nonzero denominator preserve this remainder limit,which proves convergence of the partial-sum sequence.
original signed a=5/6;r=-1/2;for EVERY k>=0,abs(a*r^k)=abs(a)*abs(r)^k;absolute-series sum=5/3
After
for every m>n>=1,absolute signed tail<=absolute-series tail=abs(a)*abs(r)^n/(1-abs(r)) tending to0;Cauchy criterion and real completeness give convergence of original series
Conditions: none
Absolute values act termwise,giving |a|/(1−|r|),not |a/(1−r)|.Every signed tail is bounded by an absolute tail tending to0;the Cauchy criterion and real completeness then give original-series convergence,not extrapolation from a finite prefix.
S007 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
S006 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
S005 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
S004 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
S003 · premise_and_result_audit — Recompute absolute geometric series exact infinite candidate sum with exact rational arithmetic, not rounded decimals.
S002 · premise_and_result_audit — Recompute absolute geometric series one minus effective ratio with exact rational arithmetic, not rounded decimals.
S001 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
The full substitutions and premise checks give the original answer; source identity,format and status are unchanged.
This question review and source binding
Review date: 2026-10-26
Original questions, answers, step expressions, numeric ASTs and historical records are retained. Source candidate labels remain historical and are not overwritten as formal proof.