第 21013 題：單調有界數列收斂定理

原題與已知條件

實數列的symbolic全稱單調與相應有界證書。操作=monoseq-decreasing-limit-certificate；資料={"limit":[13,14],"coefficient":13,"shift":2}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。

• G001 — 實數列的symbolic全稱單調與相應有界證書。操作=monoseq-decreasing-limit-certificate；資料={"limit":[13,14],"coefficient":13,"shift":2}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。

定義與適用範圍

沿用原題明示的論域、假設與來源 MKO 定義。本題已完成第二對話數理與中英語義覆核，並由原始參數逐題獨立重算模型、答案與每個算術步驟；原題仍為 candidate。

• mko-monotone-sequence-convergence-theorem

逐步推導

source operation=monoseq-decreasing-limit-certificate;original snapshot and mathematics retained

explicit ordered source output fields=[direction, limit, coefficient, shift, first_three_terms, symbolic_certificate, certificate]

ORIGINAL direction=decreasing,L=13/14,c=13>0,shift=2>=1;WHOLE a_n=L+(1)c/(n+shift) for ALL n>=0

for EVERY n>=0,a_(n+1)-a_n=(-1)*(13)/((n+2)*(n+1+2));positive denominator;all a_n strictly above L=13/14

WHOLE decreasing law with ORIGINAL L=13/14,c=13,shift=2

for ANY epsilon>0 choose natural N with N+2>13/epsilon;ALL n>=N have |a_n-L|<epsilon;L is infimum but NOT attained because c/(n+shift)>0

monoseq illustrative source n=0 denominator := sum(0, 2)

monoseq illustrative source n=0 denominator = 2

monoseq illustrative source n=0 unsigned tail := div(13, 2)

monoseq illustrative source n=0 unsigned tail = 13/2

monoseq illustrative source n=0 signed value := sum(13/14, 13/2)

monoseq illustrative source n=0 signed value = 52/7

monoseq illustrative source n=1 denominator := sum(1, 2)

monoseq illustrative source n=1 denominator = 3

monoseq illustrative source n=1 unsigned tail := div(13, 3)

monoseq illustrative source n=1 unsigned tail = 13/3

monoseq illustrative source n=1 signed value := sum(13/14, 13/3)

monoseq illustrative source n=1 signed value = 221/42

monoseq illustrative source n=2 denominator := sum(2, 2)

monoseq illustrative source n=2 denominator = 4

monoseq illustrative source n=2 unsigned tail := div(13, 4)

monoseq illustrative source n=2 unsigned tail = 13/4

monoseq illustrative source n=2 signed value := sum(13/14, 13/4)

monoseq illustrative source n=2 signed value = 117/28

逐步反向核對

• S012 · premise_and_result_audit — 以精確有理數重算 monoseq illustrative source n=2 signed value，不以小數近似替代。
• S011 · premise_and_result_audit — 以精確有理數重算 monoseq illustrative source n=2 unsigned tail，不以小數近似替代。
• S010 · premise_and_result_audit — 以精確有理數重算 monoseq illustrative source n=2 denominator，不以小數近似替代。
• S009 · premise_and_result_audit — 以精確有理數重算 monoseq illustrative source n=1 signed value，不以小數近似替代。
• S008 · premise_and_result_audit — 以精確有理數重算 monoseq illustrative source n=1 unsigned tail，不以小數近似替代。
• S007 · premise_and_result_audit — 以精確有理數重算 monoseq illustrative source n=1 denominator，不以小數近似替代。
• S006 · premise_and_result_audit — 以精確有理數重算 monoseq illustrative source n=0 signed value，不以小數近似替代。
• S005 · premise_and_result_audit — 以精確有理數重算 monoseq illustrative source n=0 unsigned tail，不以小數近似替代。
• S004 · premise_and_result_audit — 以精確有理數重算 monoseq illustrative source n=0 denominator，不以小數近似替代。
• S003 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S002 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S001 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。

結論

(decreasing, 13/14, 13, 2, [52/7, 221/42, 117/28], 是, nonincreasing_bounded_below_infimum_limit)

依上列完整代值及條件核對得到原題答案；原題編號、格式與狀態保持不變。

本題覆核與來源綁定

覆核日期：2026-10-26

原題、答案、步驟表達式、數值 AST 與既有歷史保持原樣；原來的候選狀態屬於來源歷史，不覆寫為形式化證明。