{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-infinite-series-01",
  "mko_id": "mko-infinite-series",
  "title": {
    "zh-Hant": "幾何與望遠鏡級數的完整尾項",
    "en": "Complete tails for geometric and telescoping series"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-infinite-series-01",
    "source": {
      "mko_id": "mko-infinite-series",
      "commit": "b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f",
      "sha256": "f2c97043e047957807c3f303d9bf37bbe3d188f1721ae0dcfbab1537a45b8d0b",
      "english_sha256": "d540bafe796e5c209c969a5f34c55ef21c4244a2805d9558a5cbc8938ef09592"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "algebra",
      "epsilon",
      "definition_here"
    ],
    "goal": {
      "zh-Hant": "用精確部分和與餘項證兩個級數收斂。",
      "en": "Prove convergence of two series with exact partial sums and remainders."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "S_N=sum_{n=0}^N 2^{-n}，T_N=sum_{n=1}^N 1/[n(n+1)]，N≥1。",
          "en": "S_N=sum from n=0 to N of 2^{-n}; T_N=sum from n=1 to N of 1/[n(n+1)], N≥1."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1"
        ],
        "before_tex": "S_N=\\sum_{n=0}^{N}2^{-n}",
        "after_tex": "S_N-\\tfrac12S_N=1-2^{-(N+1)}",
        "relation": "equivalent",
        "rule": "algebra",
        "conditions": [],
        "explanation": {
          "zh-Hant": "縮半後相減，除首尾外對應項完全抵銷，索引從 0 保留首項 1。",
          "en": "Subtract the halved sum; matched terms cancel, preserving first term 1 from the zero start."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "G1"
          ],
          "explanation": {
            "zh-Hant": "右尾指數是 N+1，不是 N。",
            "en": "The remaining last exponent is N+1, not N."
          }
        },
        "leaf_basis": "algebra"
      },
      {
        "id": "S02",
        "uses": [
          "S01"
        ],
        "before_tex": "\\tfrac12 S_N=1-2^{-(N+1)}",
        "after_tex": "S_N=2-2^{-N}",
        "relation": "equivalent",
        "rule": "algebra",
        "conditions": [],
        "explanation": {
          "zh-Hant": "乘 2 得精確部分和。",
          "en": "Multiply by two to obtain the exact partial sum."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "S01"
          ],
          "explanation": {
            "zh-Hant": "乘數非零，故可反向還原。",
            "en": "The nonzero multiplier makes this reversible."
          }
        },
        "leaf_basis": "algebra"
      },
      {
        "id": "S03",
        "uses": [
          "S02"
        ],
        "before_tex": "R_N=2-S_N",
        "after_tex": "R_N=2^{-N}>0",
        "relation": "equivalent",
        "rule": "algebra",
        "conditions": [],
        "explanation": {
          "zh-Hant": "餘項仍保留而非直接把有限和當無窮和。",
          "en": "Retain the remainder rather than equating the finite sum with the infinite sum."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "S02"
          ],
          "explanation": {
            "zh-Hant": "N 有限時餘項非零。",
            "en": "The remainder is nonzero for finite N."
          }
        },
        "leaf_basis": "algebra"
      },
      {
        "id": "S04",
        "uses": [
          "S03"
        ],
        "before_tex": "2^N\\ge N+1\\quad(N\\ge0)",
        "after_tex": "0<R_N\\le1/(N+1)\\to0",
        "relation": "implies",
        "rule": "epsilon",
        "conditions": [],
        "explanation": {
          "zh-Hant": "初始 1≥1；歸納時 2(N+1)≥N+2。任意 epsilon 取 N>1/epsilon，故部分和趨 2。",
          "en": "The base is 1≥1; induction uses 2(N+1)≥N+2. For any epsilon choose N>1/epsilon, proving partial sums tend to 2."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03"
          ],
          "explanation": {
            "zh-Hant": "全 N 歸納及全尾 epsilon 界，而非大 N 抽樣。",
            "en": "This is an all-N induction and all-tail epsilon bound, not a large-N sample."
          }
        },
        "leaf_basis": "epsilon"
      },
      {
        "id": "S05",
        "uses": [
          "G1"
        ],
        "before_tex": "\\frac1{n(n+1)}",
        "after_tex": "\\frac1n-\\frac1{n+1}",
        "relation": "equivalent",
        "rule": "algebra",
        "conditions": [
          "n\\ge1"
        ],
        "explanation": {
          "zh-Hant": "通分分子是 (n+1)-n=1，n≥1 使兩分母非零。",
          "en": "A common denominator gives numerator (n+1)-n=1, and n≥1 keeps both denominators nonzero."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "G1"
          ],
          "explanation": {
            "zh-Hant": "索引 n=0 不在此級數中。",
            "en": "Index zero is excluded from this series."
          }
        },
        "leaf_basis": "algebra"
      },
      {
        "id": "S06",
        "uses": [
          "G1",
          "S05"
        ],
        "before_tex": "T_N=\\sum_{n=1}^{N}\\left(\\frac1n-\\frac1{n+1}\\right)",
        "after_tex": "T_N=1-\\frac1{N+1}",
        "relation": "equivalent",
        "rule": "algebra",
        "conditions": [],
        "explanation": {
          "zh-Hant": "每個負項與下一個正項相消，只留首項和最後餘項；可由 N 到 N+1 歸納核。",
          "en": "Each negative term cancels the following positive term, leaving the first and final terms; induction from N to N+1 also verifies the formula."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "G1",
            "S05"
          ],
          "explanation": {
            "zh-Hant": "留最後負項 -1/(N+1)，不能刪掉。",
            "en": "The final negative term -1/(N+1) must remain."
          }
        },
        "leaf_basis": "algebra"
      },
      {
        "id": "S07",
        "uses": [
          "S06"
        ],
        "before_tex": "1-T_N=1/(N+1)",
        "after_tex": "\\forall\\varepsilon>0\\ \\exists M\\ \\forall N\\ge M:\\ |T_N-1|<\\varepsilon",
        "relation": "implies",
        "rule": "epsilon",
        "conditions": [],
        "explanation": {
          "zh-Hant": "取整數 M>1/epsilon，得全部尾端嚴格界。",
          "en": "Choose an integer M>1/epsilon to obtain the strict bound on the whole tail."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S06"
          ],
          "explanation": {
            "zh-Hant": "只因單項趨零並不足夠，此處已證部分和極限。",
            "en": "Terms tending to zero would not suffice; here the partial-sum limit is established."
          }
        },
        "leaf_basis": "epsilon"
      },
      {
        "id": "S08",
        "uses": [
          "S04",
          "S07"
        ],
        "before_tex": "\\lim S_N=2,\\quad\\lim T_N=1",
        "after_tex": "\\sum_{n=0}^{\\infty}2^{-n}=2,\\quad\\sum_{n=1}^{\\infty}\\frac1{n(n+1)}=1",
        "relation": "definition",
        "rule": "definition_here",
        "conditions": [],
        "explanation": {
          "zh-Hant": "無窮級數按部分和極限定義，兩個索引起點不同所以各自保留。",
          "en": "Define infinite series by partial-sum limits, preserving the different starting indices."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04",
            "S07"
          ],
          "explanation": {
            "zh-Hant": "沒有對 |r|≥1 使用幾何和收斂公式。",
            "en": "No convergent geometric formula is used for |r|≥1."
          }
        },
        "leaf_basis": "definition_here"
      }
    ],
    "result": {
      "zh-Hant": "兩級數分別為 2 與 1；精確尾誤差分別為 2^{-N} 與 1/(N+1)。",
      "en": "The sums are 2 and 1, with exact tail errors 2^{-N} and 1/(N+1)."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "有限 S_N 已恰等於 2。",
          "en": "Finite S_N already equals 2."
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "餘項 2^{-N}>0。",
          "en": "The remainder 2^{-N} is positive."
        }
      },
      {
        "claim": {
          "zh-Hant": "單項趨零必使級數收斂。",
          "en": "Terms tending to zero force convergence."
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "調和級數分塊每塊至少 1/2，仍發散。",
          "en": "Harmonic blocks each contribute at least 1/2 and diverge."
        }
      },
      {
        "claim": {
          "zh-Hant": "r=1 的幾何級數和仍為 1/(1-r)。",
          "en": "At r=1 the geometric series still sums to 1/(1-r)."
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "分母零且部分和 N+1 發散。",
          "en": "The denominator vanishes and partial sums N+1 diverge."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "mko_definition",
        "kind": "cited",
        "statement": {
          "zh-Hant": "本輪處理精確幾何級數與望遠鏡級數。保留起始index、partial sum與remainder；幾何級數只在|r|<1使用sum公式。長有限prefix與a_n→0皆不足以單獨證收斂。",
          "en": "This slice handles exact geometric and telescoping series. Retain the starting-index convention, partial sum and remainder. The geometric sum formula is used only for |r|<1. A long finite prefix or terms tending to zero cannot alone establish convergence."
        },
        "source": {
          "mko_id": "mko-infinite-series",
          "sha256": "f2c97043e047957807c3f303d9bf37bbe3d188f1721ae0dcfbab1537a45b8d0b",
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        }
      },
      {
        "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": "definition_here",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "無窮級數的和定義為依指定起點取的有限部分和之極限。",
          "en": "An infinite-series sum is the limit of finite partial sums with the specified starting index."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "geometric-N3",
        "lhs": "1+1/2+1/4+1/8",
        "rhs": "2-1/8",
        "step": "S02"
      },
      {
        "id": "telescoping-N3",
        "lhs": "1/2+1/6+1/12",
        "rhs": "1-1/4",
        "step": "S06"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "只處理明示全公式：從 0 起的 2^{-n} 及從 1 起的 1/[n(n+1)]。",
      "en": "Only full declared laws: 2^{-n} starting at zero and 1/[n(n+1)] starting at one."
    },
    "md_symbols": [
      [
        "S_N,T_N",
        "兩種部分和",
        "two partial sums"
      ],
      [
        "R_N",
        "幾何餘項",
        "geometric remainder"
      ]
    ],
    "md_error_controls": [
      {
        "title": "索引起點 / Starting indices",
        "zh": "幾何從 0、望遠鏡從 1，首項不能混。",
        "en": "The geometric law starts at zero and the telescoping law at one."
      },
      {
        "title": "部分和與極限 / Partial sums versus limits",
        "zh": "先給精確餘項再證趨零。",
        "en": "Give the exact remainder before proving it tends to zero."
      }
    ],
    "md_open_items": []
  },
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    "wording_review": "original_Lead_decisions_resolved",
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}