{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-absolute-convergence-of-series-01",
  "mko_id": "mko-absolute-convergence-of-series",
  "title": {
    "zh-Hant": "絕對收斂與交錯調和的雙證書",
    "en": "Absolute convergence and the two certificates for alternating harmonic"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-absolute-convergence-of-series-01",
    "source": {
      "mko_id": "mko-absolute-convergence-of-series",
      "commit": "b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f",
      "sha256": "4419bd0abbb51b07728caad18022eaace6aec3c8243880972704e8b1413cc1f8",
      "english_sha256": "faeb3d8b3f3ed278c06df7c7a387e9df82810c62c55a1d90becc7eaa1efaee1c"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "algebra",
      "epsilon",
      "monotone_complete",
      "definition_here"
    ],
    "goal": {
      "zh-Hant": "區分負比幾何級數的絕對收斂與交錯調和的條件收斂。",
      "en": "Distinguish absolute convergence of a negative-ratio geometric series from conditional convergence of alternating harmonic."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "a_n=(-1/2)^n（n≥0）；b_n=(-1)^{n+1}/n（n≥1）。",
          "en": "a_n=(-1/2)^n for n≥0; b_n=(-1)^{n+1}/n for n≥1."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1"
        ],
        "before_tex": "|a_n|=2^{-n}",
        "after_tex": "\\sum_{n=0}^{N}|a_n|=2-2^{-N}",
        "relation": "equivalent",
        "rule": "algebra",
        "conditions": [],
        "explanation": {
          "zh-Hant": "由有限幾何相減公式求全部絕對值部分和。",
          "en": "The finite geometric subtraction identity computes every absolute partial sum."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "G1"
          ],
          "explanation": {
            "zh-Hant": "先逐項取絕對值，不是把有號和最後才取絕對值。",
            "en": "Take absolute values termwise, not only after summing signed terms."
          }
        },
        "leaf_basis": "algebra"
      },
      {
        "id": "S02",
        "uses": [
          "S01"
        ],
        "before_tex": "2^{-N}\\le1/(N+1)\\to0",
        "after_tex": "\\sum_{n=0}^{\\infty}|a_n|=2",
        "relation": "implies",
        "rule": "epsilon",
        "conditions": [],
        "explanation": {
          "zh-Hant": "2^N≥N+1 可歸納證，故餘項全尾趨零，a 級數絕對收斂。",
          "en": "Induction proves 2^N≥N+1, so the whole remainder tail vanishes and the a series converges absolutely."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S01"
          ],
          "explanation": {
            "zh-Hant": "不是僅憑負號交錯判絕對收斂。",
            "en": "Absolute convergence is not inferred merely from alternating signs."
          }
        },
        "leaf_basis": "epsilon"
      },
      {
        "id": "S03",
        "uses": [
          "G1"
        ],
        "before_tex": "B_{2m}=\\sum_{n=1}^{2m}b_n",
        "after_tex": "B_{2(m+1)}-B_{2m}=1/(2m+1)-1/(2m+2)>0",
        "relation": "implies",
        "rule": "algebra",
        "conditions": [],
        "explanation": {
          "zh-Hant": "偶數部分和遞增，因每一對項差為正。",
          "en": "Even partial sums increase because each paired difference is positive."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1"
          ],
          "explanation": {
            "zh-Hant": "m≥1 或初始 B_0=0 皆可，索引一致。",
            "en": "Use m≥1 or initialize B_0=0 consistently."
          }
        },
        "leaf_basis": "algebra"
      },
      {
        "id": "S04",
        "uses": [
          "G1",
          "S03"
        ],
        "before_tex": "B_{2m+1}=B_{2m}+1/(2m+1)",
        "after_tex": "0\\le B_{2m}<B_{2m+1}\\le1",
        "relation": "implies",
        "rule": "algebra",
        "conditions": [],
        "explanation": {
          "zh-Hant": "奇數部分和由 1 起每次減一個正差，所以遞減；兩者互相夾住。",
          "en": "Odd partial sums start at 1 and decrease by a positive difference; the two parity sequences bound each other."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S03"
          ],
          "explanation": {
            "zh-Hant": "交錯性以 1/n 遞減為支撐，不是任意交錯序列。",
            "en": "Alternation is supported by decreasing 1/n magnitudes, not arbitrary signs."
          }
        },
        "leaf_basis": "algebra"
      },
      {
        "id": "S05",
        "uses": [
          "S03",
          "S04"
        ],
        "before_tex": "B_{2m}\\uparrow,\\ B_{2m+1}\\downarrow,\\ B_{2m+1}-B_{2m}=1/(2m+1)\\to0",
        "after_tex": "B_N\\to B,\\quad |B-B_N|\\le1/(N+1)",
        "relation": "implies",
        "rule": "monotone_complete",
        "conditions": [],
        "explanation": {
          "zh-Hant": "兩個有界單調子列各有極限，間距趨零所以極限相同；極限夾在相鄰偶奇和間，給交錯餘項界。",
          "en": "Both bounded monotone parity subsequences have limits; their vanishing gap makes the limits equal. The common limit lies between adjacent parity sums, giving the alternating remainder bound."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "S04"
          ],
          "explanation": {
            "zh-Hant": "餘項界由夾擠產生，不以未知閉式常數取代收斂證明。",
            "en": "The bound follows from bracketing, without substituting an unknown closed form for a proof."
          }
        },
        "leaf_basis": "monotone_complete"
      },
      {
        "id": "S06",
        "uses": [
          "G1"
        ],
        "before_tex": "\\sum_{n=2^{j-1}+1}^{2^j}|b_n|",
        "after_tex": "\\sum_{n=2^{j-1}+1}^{2^j}1/n\\ge2^{j-1}/2^j=1/2",
        "relation": "implies",
        "rule": "algebra",
        "conditions": [
          "j\\ge1"
        ],
        "explanation": {
          "zh-Hant": "每個二進位塊有 2^{j-1} 項，每項至少 1/2^j。",
          "en": "Each dyadic block has 2^{j-1} terms, each at least 1/2^j."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1"
          ],
          "explanation": {
            "zh-Hant": "此下界對所有整數 j≥1，不是有限塊觀察。",
            "en": "The lower bound holds for every integer j≥1, not observed blocks only."
          }
        },
        "leaf_basis": "algebra"
      },
      {
        "id": "S07",
        "uses": [
          "S06"
        ],
        "before_tex": "\\sum_{n=1}^{2^m}|b_n|\\ge1+m/2",
        "after_tex": "\\sum_{n=1}^{\\infty}|b_n|\\text{ diverges}",
        "relation": "implies",
        "rule": "algebra",
        "conditions": [],
        "explanation": {
          "zh-Hant": "塊數 m 可任意大，絕對部分和無界。",
          "en": "The number of blocks m is arbitrary, making absolute partial sums unbounded."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S06"
          ],
          "explanation": {
            "zh-Hant": "不可由有號部分和有界推出絕對部分和有界。",
            "en": "Bounded signed partial sums do not imply bounded absolute partial sums."
          }
        },
        "leaf_basis": "algebra"
      },
      {
        "id": "S08",
        "uses": [
          "S02",
          "S05",
          "S07"
        ],
        "before_tex": "\\sum a_n\\text{ absolutely convergent};\\quad\\sum b_n\\text{ convergent but not absolutely}",
        "after_tex": "\\sum b_n\\text{ conditionally convergent}",
        "relation": "definition",
        "rule": "definition_here",
        "conditions": [],
        "explanation": {
          "zh-Hant": "條件收斂的兩份責任均有獨立論證；絕對收斂則由 Cauchy 尾部三角界推出原級數收斂。",
          "en": "Both obligations for conditional convergence have independent proofs; absolute convergence implies signed convergence through the Cauchy tail triangle bound."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "S05",
            "S07"
          ],
          "explanation": {
            "zh-Hant": "未將有界項、項趨零或單純交錯升格成任何證書。",
            "en": "Bounded terms, vanishing terms and alternation alone have not been promoted to certificates."
          }
        },
        "leaf_basis": "definition_here"
      }
    ],
    "result": {
      "zh-Hant": "(-1/2)^n 級數絕對收斂；交錯調和級數條件收斂。",
      "en": "The (-1/2)^n series is absolutely convergent; alternating harmonic is conditionally convergent."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "交錯就一定絕對收斂。",
          "en": "Alternation guarantees absolute convergence."
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "交錯調和的絕對級數已證發散。",
          "en": "The series of termwise absolute values of the alternating harmonic series was proved divergent."
        }
      },
      {
        "claim": {
          "zh-Hant": "|sum b_n| 等於 sum |b_n|。",
          "en": "|sum b_n| equals sum |b_n|."
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "已在前兩項得到 |1-1/2|=1/2 而 1+1/2=3/2。",
          "en": "Already |1-1/2|=1/2 while 1+1/2=3/2."
        }
      },
      {
        "claim": {
          "zh-Hant": "項趨零足以絕對收斂。",
          "en": "Terms tending to zero suffice for absolute convergence."
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "1/n 趨零而分塊下界使級數發散。",
          "en": "1/n tends to zero while block lower bounds force divergence."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "mko_definition",
        "kind": "cited",
        "statement": {
          "zh-Hant": "幾何級數用|r|<1證絕對收斂；交錯harmonic以alternating remainder證原級數收斂、以harmonic divergence證非絕對。bounded terms、term→0、|partial sum|皆不足。",
          "en": "Geometric series use |r|<1 for absolute convergence. The alternating harmonic example combines an alternating-remainder certificate of convergence with harmonic divergence of the absolute series. Bounded terms, terms tending to zero, or the absolute value of a partial sum do not establish absolute convergence."
        },
        "source": {
          "mko_id": "mko-absolute-convergence-of-series",
          "sha256": "4419bd0abbb51b07728caad18022eaace6aec3c8243880972704e8b1413cc1f8",
          "commit": "b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f"
        }
      },
      {
        "id": "algebra",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數域的等式運算；除法前分母必非零，不等式乘正數保持方向。",
          "en": "Use real-field identities; division requires a nonzero denominator and multiplication by a positive number preserves inequalities."
        }
      },
      {
        "id": "epsilon",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "收斂表示每個正 epsilon 有一個整數 N，使全部 n≥N 的誤差嚴格小於 epsilon；使用實數 Archimedean 性。",
          "en": "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."
        }
      },
      {
        "id": "monotone_complete",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數單調有界數列收斂；非負級數的部分和單調遞增。",
          "en": "Every bounded monotone real sequence converges; partial sums of a nonnegative series are nondecreasing."
        }
      },
      {
        "id": "definition_here",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "絕對收斂是逐項絕對值級數收斂；條件收斂是原級數收斂且逐項絕對值級數不收斂。",
          "en": "Absolute convergence means convergence of the termwise-absolute-value series; conditional convergence means signed convergence without absolute convergence."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "absolute-vs-signed",
        "lhs": "abs(1-1/2)",
        "rhs": "1/2",
        "step": "S08"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "精確幾何尾與調和分塊；條件收斂須同時證原級數收斂及絕對級數發散。",
      "en": "Exact geometric tails and harmonic blocks; conditional convergence needs both signed convergence and absolute divergence."
    },
    "md_symbols": [
      [
        "B_N",
        "交錯調和部分和",
        "alternating-harmonic partial sum"
      ],
      [
        "B",
        "已證存在的極限",
        "proved existing limit"
      ]
    ],
    "md_error_controls": [
      {
        "title": "雙證書 / Two certificates",
        "zh": "條件收斂同時需要有號和收斂及絕對和發散。",
        "en": "Conditional convergence needs signed convergence and absolute divergence."
      },
      {
        "title": "全尾而非符號 / Whole tails, not signs",
        "zh": "以完整單調與分塊公式控制全部項。",
        "en": "Use full monotonicity and block formulas for every term."
      }
    ],
    "md_open_items": []
  },
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    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "reader_review": "142_of_142_source_bound",
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    "scope": "Owner-adopted reader revision. Lead and bilingual review concern the recorded text and premises; no new universal or Lean proof, canonical promotion or historical-label rewrite."
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}