{
  "index": "1969-B-3",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG",
    "NT"
  ],
  "difficulty": "",
  "question": "B-3. The terms of a sequence \\( T_{n} \\) satisfy\n\\[\nT_{n} T_{n+1}=n \\quad(n=1,2,3, \\cdots) \\text { and } \\lim _{n \\rightarrow \\infty} \\frac{T_{n}}{T_{n+1}}=1\n\\]\n\nShow that \\( \\pi T_{1}^{2}=2 \\).",
  "solution": "B-3 The first relation implies that\n\\[\n\\begin{array}{ll}\nT_{n}=\\frac{(n-1)(n-3) \\cdots 3}{(n-2)(n-4) \\cdots 2} \\cdot \\frac{1}{T_{1}} & \\text { if } n \\text { is even } \\\\\nT_{n}=\\frac{(n-1)(n-3) \\cdots 2}{(n-2)(n-4) \\cdots 1} \\cdot T_{1} & \\text { if } n \\text { is odd }\n\\end{array}\n\\]\n\nIf \\( n \\) is odd,\n\\[\n\\frac{T_{n}}{T_{n+1}}=\\left(T_{1}\\right)^{2} \\cdot \\frac{2}{1} \\frac{2}{3} \\frac{4}{3} \\frac{4}{5} \\frac{6}{5} \\cdots \\frac{(n-1)}{(n-2)} \\frac{(n-1)}{n}\n\\]\n\nThe Wallis product is \\( \\pi / 2=\\frac{22}{1} \\frac{24}{3} \\frac{4}{5} \\frac{8}{5} \\cdots \\). After an even number of factors the partial product is less than \\( \\pi / 2 \\) and after an odd number of factors the partial product is greater than \\( \\pi / 2 \\). Thus for the case when \\( n \\) is odd, \\( T_{n} / T_{n+1}<\\frac{1}{2} \\pi T_{1}^{2} \\). A similar calculation shows that, when \\( n \\) is even, \\( T_{n} / T_{n+1}<2 / \\pi T_{1}^{2} \\). Since the limit of \\( T_{n} / T_{n+1}=1,1 \\) is less than or equal to both \\( \\frac{1}{2} \\pi T_{1}^{2} \\) and its reciprocal. This implies that \\( \\pi T_{1}^{2}=2 \\).",
  "vars": [
    "T_n",
    "T_n+1",
    "n"
  ],
  "params": [
    "T_1"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "T_n": "curterm",
        "T_n+1": "nextterm",
        "n": "indexvar",
        "T_1": "firstterm"
      },
      "question": "B-3. The terms of a sequence \\( curterm \\) satisfy\n\\[\ncurterm nextterm=indexvar \\quad(indexvar=1,2,3, \\cdots) \\text { and } \\lim _{indexvar \\rightarrow \\infty} \\frac{curterm}{nextterm}=1\n\\]\n\nShow that \\( \\pi firstterm^{2}=2 \\).",
      "solution": "B-3 The first relation implies that\n\\[\n\\begin{array}{ll}\ncurterm=\\frac{(indexvar-1)(indexvar-3) \\cdots 3}{(indexvar-2)(indexvar-4) \\cdots 2} \\cdot \\frac{1}{firstterm} & \\text { if } indexvar \\text { is even } \\\\\ncurterm=\\frac{(indexvar-1)(indexvar-3) \\cdots 2}{(indexvar-2)(indexvar-4) \\cdots 1} \\cdot firstterm & \\text { if } indexvar \\text { is odd }\n\\end{array}\n\\]\n\nIf \\( indexvar \\) is odd,\n\\[\n\\frac{curterm}{nextterm}=\\left(firstterm\\right)^{2} \\cdot \\frac{2}{1} \\frac{2}{3} \\frac{4}{3} \\frac{4}{5} \\frac{6}{5} \\cdots \\frac{(indexvar-1)}{(indexvar-2)} \\frac{(indexvar-1)}{indexvar}\n\\]\n\nThe Wallis product is \\( \\pi / 2=\\frac{22}{1} \\frac{24}{3} \\frac{4}{5} \\frac{8}{5} \\cdots \\). After an even number of factors the partial product is less than \\( \\pi / 2 \\) and after an odd number of factors the partial product is greater than \\( \\pi / 2 \\). Thus for the case when \\( indexvar \\) is odd, \\( curterm / nextterm<\\frac{1}{2} \\pi firstterm^{2} \\). A similar calculation shows that, when \\( indexvar \\) is even, \\( curterm / nextterm<2 / \\pi firstterm^{2} \\). Since the limit of \\( curterm / nextterm=1,1 \\) is less than or equal to both \\( \\frac{1}{2} \\pi firstterm^{2} \\) and its reciprocal. This implies that \\( \\pi firstterm^{2}=2 \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "T_n": "lilacstone",
        "T_n+1": "crimsonleaf",
        "n": "harborwind",
        "T_1": "orchidtrail"
      },
      "question": "B-3. The terms of a sequence \\( lilacstone \\) satisfy\n\\[\nlilacstone\\, crimsonleaf=harborwind \\quad(harborwind=1,2,3, \\cdots) \\text { and } \\lim _{harborwind \\rightarrow \\infty} \\frac{lilacstone}{crimsonleaf}=1\n\\]\n\nShow that \\( \\pi orchidtrail^{2}=2 \\).",
      "solution": "B-3 The first relation implies that\n\\[\n\\begin{array}{ll}\nlilacstone=\\frac{(harborwind-1)(harborwind-3) \\cdots 3}{(harborwind-2)(harborwind-4) \\cdots 2} \\cdot \\frac{1}{orchidtrail} & \\text { if } harborwind \\text { is even } \\\\\nlilacstone=\\frac{(harborwind-1)(harborwind-3) \\cdots 2}{(harborwind-2)(harborwind-4) \\cdots 1} \\cdot orchidtrail & \\text { if } harborwind \\text { is odd }\n\\end{array}\n\\]\n\nIf \\( harborwind \\) is odd,\n\\[\n\\frac{lilacstone}{crimsonleaf}=\\left(orchidtrail\\right)^{2} \\cdot \\frac{2}{1} \\frac{2}{3} \\frac{4}{3} \\frac{4}{5} \\frac{6}{5} \\cdots \\frac{(harborwind-1)}{(harborwind-2)} \\frac{(harborwind-1)}{harborwind}\n\\]\n\nThe Wallis product is \\( \\pi / 2=\\frac{22}{1} \\frac{24}{3} \\frac{4}{5} \\frac{8}{5} \\cdots \\). After an even number of factors the partial product is less than \\( \\pi / 2 \\) and after an odd number of factors the partial product is greater than \\( \\pi / 2 \\). Thus for the case when \\( harborwind \\) is odd, \\( lilacstone / crimsonleaf<\\frac{1}{2} \\pi orchidtrail^{2} \\). A similar calculation shows that, when \\( harborwind \\) is even, \\( lilacstone / crimsonleaf<2 / \\pi orchidtrail^{2} \\). Since the limit of \\( lilacstone / crimsonleaf=1,1 \\) is less than or equal to both \\( \\frac{1}{2} \\pi orchidtrail^{2} \\) and its reciprocal. This implies that \\( \\pi orchidtrail^{2}=2 \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "T_{n}": "constantvalue",
        "T_{n+1}": "previousvalue",
        "n": "steadystate",
        "T_{1}": "lastterm"
      },
      "question": "B-3. The terms of a sequence \\( constantvalue \\) satisfy\n\\[\nconstantvalue\\,previousvalue = steadystate \\quad(steadystate=1,2,3, \\cdots) \\text { and } \\lim _{steadystate \\rightarrow \\infty} \\frac{constantvalue}{previousvalue}=1\n\\]\n\nShow that \\( \\pi lastterm^{2}=2 \\).",
      "solution": "B-3 The first relation implies that\n\\[\n\\begin{array}{ll}\nconstantvalue=\\frac{(steadystate-1)(steadystate-3) \\cdots 3}{(steadystate-2)(steadystate-4) \\cdots 2} \\cdot \\frac{1}{lastterm} & \\text { if } n \\text { is even } \\\\\nconstantvalue=\\frac{(steadystate-1)(steadystate-3) \\cdots 2}{(steadystate-2)(steadystate-4) \\cdots 1} \\cdot lastterm & \\text { if } n \\text { is odd }\n\\end{array}\n\\]\n\nIf \\( steadystate \\) is odd,\n\\[\n\\frac{constantvalue}{previousvalue}=\\left(lastterm\\right)^{2} \\cdot \\frac{2}{1} \\frac{2}{3} \\frac{4}{3} \\frac{4}{5} \\frac{6}{5} \\cdots \\frac{(steadystate-1)}{(steadystate-2)} \\frac{(steadystate-1)}{steadystate}\n\\]\n\nThe Wallis product is \\( \\pi / 2=\\frac{22}{1} \\frac{24}{3} \\frac{4}{5} \\frac{8}{5} \\cdots \\). After an even number of factors the partial product is less than \\( \\pi / 2 \\) and after an odd number of factors the partial product is greater than \\( \\pi / 2 \\). Thus for the case when \\( steadystate \\) is odd, \\( constantvalue / previousvalue<\\frac{1}{2} \\pi lastterm^{2} \\). A similar calculation shows that, when \\( steadystate \\) is even, \\( constantvalue / previousvalue<2 / \\pi lastterm^{2} \\). Since the limit of \\( constantvalue / previousvalue=1,1 \\) is less than or equal to both \\( \\frac{1}{2} \\pi lastterm^{2} \\) and its reciprocal. This implies that \\( \\pi lastterm^{2}=2 \\)."
    },
    "garbled_string": {
      "map": {
        "T_n": "qzxwvtnp",
        "T_n+1": "hjgrksla",
        "n": "blmkqtrs",
        "T_1": "vprmjzqe"
      },
      "question": "B-3. The terms of a sequence \\( qzxwvtnp \\) satisfy\n\\[\nqzxwvtnp hjgrksla=blmkqtrs \\quad(blmkqtrs=1,2,3, \\cdots) \\text { and } \\lim _{blmkqtrs \\rightarrow \\infty} \\frac{qzxwvtnp}{hjgrksla}=1\n\\]\n\nShow that \\( \\pi vprmjzqe^{2}=2 \\).",
      "solution": "B-3 The first relation implies that\n\\[\n\\begin{array}{ll}\nqzxwvtnp=\\frac{(blmkqtrs-1)(blmkqtrs-3) \\cdots 3}{(blmkqtrs-2)(blmkqtrs-4) \\cdots 2} \\cdot \\frac{1}{vprmjzqe} & \\text { if } blmkqtrs \\text { is even } \\\\\nqzxwvtnp=\\frac{(blmkqtrs-1)(blmkqtrs-3) \\cdots 2}{(blmkqtrs-2)(blmkqtrs-4) \\cdots 1} \\cdot vprmjzqe & \\text { if } blmkqtrs \\text { is odd }\n\\end{array}\n\\]\n\nIf \\( blmkqtrs \\) is odd,\n\\[\n\\frac{qzxwvtnp}{hjgrksla}=\\left(vprmjzqe\\right)^{2} \\cdot \\frac{2}{1} \\frac{2}{3} \\frac{4}{3} \\frac{4}{5} \\frac{6}{5} \\cdots \\frac{(blmkqtrs-1)}{(blmkqtrs-2)} \\frac{(blmkqtrs-1)}{blmkqtrs}\n\\]\n\nThe Wallis product is \\( \\pi / 2=\\frac{22}{1} \\frac{24}{3} \\frac{4}{5} \\frac{8}{5} \\cdots \\). After an even number of factors the partial product is less than \\( \\pi / 2 \\) and after an odd number of factors the partial product is greater than \\( \\pi / 2 \\). Thus for the case when \\( blmkqtrs \\) is odd, \\( qzxwvtnp / hjgrksla<\\frac{1}{2} \\pi vprmjzqe^{2} \\). A similar calculation shows that, when \\( blmkqtrs \\) is even, \\( qzxwvtnp / hjgrksla<2 / \\pi vprmjzqe^{2} \\). Since the limit of \\( qzxwvtnp / hjgrksla=1,1 \\) is less than or equal to both \\( \\frac{1}{2} \\pi vprmjzqe^{2} \\) and its reciprocal. This implies that \\( \\pi vprmjzqe^{2}=2 \\)."
    },
    "kernel_variant": {
      "question": "Fix a real parameter β > 0 and let (aₙ)_{n≥0} be a sequence of positive real numbers satisfying  \n\n (I) aₙ aₙ₊₁ = n + β   (n = 0,1,2,…),  \n\n (II) lim_{n→∞} aₙ / aₙ₊₁ = 1.  \n\nProve the following assertions.\n\n1.  (Square–root law) lim_{n→∞} aₙ / √n = 1.  \n\n2.  (Exact normalisation)   \n  a₀² = 2·[ Γ((β+1)/2) ⁄ Γ(β/2) ]².  \n\n3.  (Full asymptotic expansion) For every m∈ℕ there exist polynomials P₁,…,P_m in β with deg P_k = k such that  \n\n  aₙ = √n · ( 1 + ∑_{k=1}^{m} P_k(β)/n^{k} ) + O(n^{-m-½})  (n→∞).  \n\n The first two polynomials are  \n  P₁(β) = β/2 − 1/4,  P₂(β) = (−4β² + 4β + 1) ⁄ 32,  \n\n and in particular  \n\n  lim_{n→∞} n ( aₙ − √n ) = β/2 − 1/4.\n\n(The general coefficients can be written explicitly in terms of Bernoulli numbers.)\n\n————————————————————————————————————",
      "solution": "Throughout we fix β > 0 and assume aₙ > 0.  We denote n→∞ with n∈ℕ.\n\n0.  Even–odd splitting  \nDefine  \n\n Eₙ := a_{2n},  Oₙ := a_{2n+1}  (n≥0).\n\nFrom (I) we derive  \n\n EₙOₙ = 2n+β,  OₙE_{n+1} = 2n+1+β.  (†)\n\nConsequently  \n\n E_{n+1} = (2n+1+β)/Oₙ,  Oₙ = (2n+β)/Eₙ.  (††)\n\n————————————————————————————————————  \n1.  Closed product formulae  \nIterating (††) yields  \n\n Eₙ = a₀ · ∏_{j=1}^{n} (2j−1+β)/(2j−2+β),  (1a)  \n\n Oₙ = (β/a₀)·∏_{j=1}^{n} (2j+β)/(2j−1+β).  (1b)\n\n(The case n=0 reads a₁=β/a₀.)\n\n————————————————————————————————————  \n2.  Determination of a₀  \nLet  \n\n s := (β+1)/2  (so β=2s−1 and s>½).\n\nUsing the identity  \n  ∏_{j=1}^{n}(j+α)=Γ(n+1+α)/Γ(1+α),  \nformulae (1a)–(1b) become  \n\n Eₙ = a₀·Γ(n+s)/Γ(s)·Γ(s−½)/Γ(n+s−½),  (2a)  \n\n Oₙ = (β/a₀)·Γ(n+s+½)/Γ(s+½)·Γ(s)/Γ(n+s).  (2b)\n\nTaking the quotient we get  \n\n Eₙ/Oₙ = (a₀²/β) · Γ(n+s)² Γ(s−½)Γ(s+½)\n/[ Γ(s)² Γ(n+s−½)Γ(n+s+½) ].  (3)\n\nHypothesis (II) says Eₙ/Oₙ→1, so, letting n→∞ and noting  \n lim_{n→∞} Γ(n+s)² /[Γ(n+s−½)Γ(n+s+½)]=1,  \nwe obtain  \n\n a₀² = β·Γ(s)² /[ Γ(s−½)Γ(s+½) ].  (4)\n\nNow use the Euler shift Γ(z+1)=zΓ(z) with z=s−½:\n\n Γ(s+½)= (s−½)Γ(s−½)= (β/2) Γ(s−½).\n\nInsert this into (4):\n\n a₀² = β·Γ(s)² /( Γ(s−½)·(β/2)Γ(s−½) )\n      = 2·[ Γ(s)/Γ(s−½) ]²,\n\nwhich is exactly Statement 2.\n\n————————————————————————————————————  \n3.  Square–root law  \nWe treat even and odd indices separately.\n\n3.1  Even indices.  \nFrom (2a) and Stirling’s expansion  \n\n Γ(N+α)/Γ(N+β)\n   = N^{α−β} (1+∑_{k≥1} C_k(α,β)/N^{k}),  (5)\n\nwhere C_k is a polynomial of degree k in α,β, we find for large n  \n\n Eₙ = a₀·Γ(s−½)/Γ(s)·n^{½}·(1+C₁/(2n)+O(n^{-2})). (6)\n\nBecause of (2) we have a₀·Γ(s−½)/Γ(s)=√2, hence  \n\n Eₙ = √{2n}·(1+O(n^{-1})).  (7)\n\n3.2  Odd indices.  \nA parallel computation with (2b) gives  \n\n Oₙ = √{2n+1}·(1+O(n^{-1})) = √{2n}·(1+O(n^{-1})). (8)\n\n3.3  Conclusion.  \nSince a_{2n}=Eₙ and a_{2n+1}=Oₙ, (7)–(8) imply  \n\n lim_{n→∞} aₙ/√n = 1,\n\nestablishing Statement 1.\n\n————————————————————————————————————  \n4.  Complete asymptotic expansion  \n\n4.1  Logarithmic form.  \nWrite n=2N or n=2N+1 and keep s=(β+1)/2.  From (2a)–(2b) and the\nclassical asymptotic series for log Γ we obtain  \n\n log(aₙ/√n)=∑_{k≥1} A_k(β)/n^{k},  (9)\n\nwhere A_k is a polynomial in β of degree k.  A straightforward (though\nlengthy) coefficient comparison yields  \n\n A₁=(β−½)/2, A₂=β(1−β)/4, A₃=(4β³−6β²+1)/24. (10)\n\n4.2  Passage to the aₙ–expansion.  \nExponentiating (9) and using the exponential Bell polynomials gives  \n\n aₙ = √n·(1+∑_{k=1}^{m}P_k(β)/n^{k})+O(n^{-m-½}), (11)\n\nwhere  \n\n P₁=A₁, P₂=A₂+A₁²/2, P₃=A₃+A₁A₂+A₁³/6, … .  \n\nFrom (10) we recover  \n\n P₁(β)=β/2−1/4, P₂(β)=(−4β²+4β+1)/32,\n\nmatching Statement 3.  Because the A_k ultimately stem from the\nStirling series, they can be written in closed form via the Bernoulli\nnumbers B_{2j}; the same is therefore true of the P_k, completing the\nproof.\n\n————————————————————————————————————",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.586185",
        "was_fixed": false,
        "difficulty_analysis": "• Variable parameter β.  The original problem deals with the single concrete recurrence aₙ aₙ₊₁ = n+1; here the constant term n+β is left arbitrary, forcing the solver to carry β through every calculation and to master Γ-function identities with symbolic parameters.\n\n• Advanced special-function techniques.  Determining a₀ now requires evaluating an infinite product that no longer collapses to the classical Wallis product; one must manipulate Euler products, use the Gamma duplication formula and Stirling’s asymptotics.\n\n• Higher-order asymptotics.  Beyond the classical √n-law, the problem demands a full asymptotic expansion with explicit coefficients expressed via Bernoulli numbers, something completely absent from the original statement.\n\n• Multiple interacting concepts.  The solution intertwines discrete recurrences, infinite products, special-function theory (Γ, duplication), real analysis (convergence of products and series), and asymptotic analysis (Euler–Maclaurin).\n\n• Substantially longer argument.  Each of the three parts requires a separate, non-trivial chain of reasoning; together they are markedly more intricate than proving π a₀²=2."
      }
    },
    "original_kernel_variant": {
      "question": "Fix a real parameter β > 0 and let (aₙ)_{n≥0} be a sequence of positive real numbers satisfying  \n\n (I) aₙ aₙ₊₁ = n + β   (n = 0,1,2,…),  \n\n (II) lim_{n→∞} aₙ / aₙ₊₁ = 1.  \n\nProve the following assertions.\n\n1.  (Square–root law) lim_{n→∞} aₙ / √n = 1.  \n\n2.  (Exact normalisation)   \n  a₀² = 2·[ Γ((β+1)/2) ⁄ Γ(β/2) ]².  \n\n3.  (Full asymptotic expansion) For every m∈ℕ there exist polynomials P₁,…,P_m in β with deg P_k = k such that  \n\n  aₙ = √n · ( 1 + ∑_{k=1}^{m} P_k(β)/n^{k} ) + O(n^{-m-½})  (n→∞).  \n\n The first two polynomials are  \n  P₁(β) = β/2 − 1/4,  P₂(β) = (−4β² + 4β + 1) ⁄ 32,  \n\n and in particular  \n\n  lim_{n→∞} n ( aₙ − √n ) = β/2 − 1/4.\n\n(The general coefficients can be written explicitly in terms of Bernoulli numbers.)\n\n————————————————————————————————————",
      "solution": "Throughout we fix β > 0 and assume aₙ > 0.  We denote n→∞ with n∈ℕ.\n\n0.  Even–odd splitting  \nDefine  \n\n Eₙ := a_{2n},  Oₙ := a_{2n+1}  (n≥0).\n\nFrom (I) we derive  \n\n EₙOₙ = 2n+β,  OₙE_{n+1} = 2n+1+β.  (†)\n\nConsequently  \n\n E_{n+1} = (2n+1+β)/Oₙ,  Oₙ = (2n+β)/Eₙ.  (††)\n\n————————————————————————————————————  \n1.  Closed product formulae  \nIterating (††) yields  \n\n Eₙ = a₀ · ∏_{j=1}^{n} (2j−1+β)/(2j−2+β),  (1a)  \n\n Oₙ = (β/a₀)·∏_{j=1}^{n} (2j+β)/(2j−1+β).  (1b)\n\n(The case n=0 reads a₁=β/a₀.)\n\n————————————————————————————————————  \n2.  Determination of a₀  \nLet  \n\n s := (β+1)/2  (so β=2s−1 and s>½).\n\nUsing the identity  \n  ∏_{j=1}^{n}(j+α)=Γ(n+1+α)/Γ(1+α),  \nformulae (1a)–(1b) become  \n\n Eₙ = a₀·Γ(n+s)/Γ(s)·Γ(s−½)/Γ(n+s−½),  (2a)  \n\n Oₙ = (β/a₀)·Γ(n+s+½)/Γ(s+½)·Γ(s)/Γ(n+s).  (2b)\n\nTaking the quotient we get  \n\n Eₙ/Oₙ = (a₀²/β) · Γ(n+s)² Γ(s−½)Γ(s+½)\n/[ Γ(s)² Γ(n+s−½)Γ(n+s+½) ].  (3)\n\nHypothesis (II) says Eₙ/Oₙ→1, so, letting n→∞ and noting  \n lim_{n→∞} Γ(n+s)² /[Γ(n+s−½)Γ(n+s+½)]=1,  \nwe obtain  \n\n a₀² = β·Γ(s)² /[ Γ(s−½)Γ(s+½) ].  (4)\n\nNow use the Euler shift Γ(z+1)=zΓ(z) with z=s−½:\n\n Γ(s+½)= (s−½)Γ(s−½)= (β/2) Γ(s−½).\n\nInsert this into (4):\n\n a₀² = β·Γ(s)² /( Γ(s−½)·(β/2)Γ(s−½) )\n      = 2·[ Γ(s)/Γ(s−½) ]²,\n\nwhich is exactly Statement 2.\n\n————————————————————————————————————  \n3.  Square–root law  \nWe treat even and odd indices separately.\n\n3.1  Even indices.  \nFrom (2a) and Stirling’s expansion  \n\n Γ(N+α)/Γ(N+β)\n   = N^{α−β} (1+∑_{k≥1} C_k(α,β)/N^{k}),  (5)\n\nwhere C_k is a polynomial of degree k in α,β, we find for large n  \n\n Eₙ = a₀·Γ(s−½)/Γ(s)·n^{½}·(1+C₁/(2n)+O(n^{-2})). (6)\n\nBecause of (2) we have a₀·Γ(s−½)/Γ(s)=√2, hence  \n\n Eₙ = √{2n}·(1+O(n^{-1})).  (7)\n\n3.2  Odd indices.  \nA parallel computation with (2b) gives  \n\n Oₙ = √{2n+1}·(1+O(n^{-1})) = √{2n}·(1+O(n^{-1})). (8)\n\n3.3  Conclusion.  \nSince a_{2n}=Eₙ and a_{2n+1}=Oₙ, (7)–(8) imply  \n\n lim_{n→∞} aₙ/√n = 1,\n\nestablishing Statement 1.\n\n————————————————————————————————————  \n4.  Complete asymptotic expansion  \n\n4.1  Logarithmic form.  \nWrite n=2N or n=2N+1 and keep s=(β+1)/2.  From (2a)–(2b) and the\nclassical asymptotic series for log Γ we obtain  \n\n log(aₙ/√n)=∑_{k≥1} A_k(β)/n^{k},  (9)\n\nwhere A_k is a polynomial in β of degree k.  A straightforward (though\nlengthy) coefficient comparison yields  \n\n A₁=(β−½)/2, A₂=β(1−β)/4, A₃=(4β³−6β²+1)/24. (10)\n\n4.2  Passage to the aₙ–expansion.  \nExponentiating (9) and using the exponential Bell polynomials gives  \n\n aₙ = √n·(1+∑_{k=1}^{m}P_k(β)/n^{k})+O(n^{-m-½}), (11)\n\nwhere  \n\n P₁=A₁, P₂=A₂+A₁²/2, P₃=A₃+A₁A₂+A₁³/6, … .  \n\nFrom (10) we recover  \n\n P₁(β)=β/2−1/4, P₂(β)=(−4β²+4β+1)/32,\n\nmatching Statement 3.  Because the A_k ultimately stem from the\nStirling series, they can be written in closed form via the Bernoulli\nnumbers B_{2j}; the same is therefore true of the P_k, completing the\nproof.\n\n————————————————————————————————————",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.471600",
        "was_fixed": false,
        "difficulty_analysis": "• Variable parameter β.  The original problem deals with the single concrete recurrence aₙ aₙ₊₁ = n+1; here the constant term n+β is left arbitrary, forcing the solver to carry β through every calculation and to master Γ-function identities with symbolic parameters.\n\n• Advanced special-function techniques.  Determining a₀ now requires evaluating an infinite product that no longer collapses to the classical Wallis product; one must manipulate Euler products, use the Gamma duplication formula and Stirling’s asymptotics.\n\n• Higher-order asymptotics.  Beyond the classical √n-law, the problem demands a full asymptotic expansion with explicit coefficients expressed via Bernoulli numbers, something completely absent from the original statement.\n\n• Multiple interacting concepts.  The solution intertwines discrete recurrences, infinite products, special-function theory (Γ, duplication), real analysis (convergence of products and series), and asymptotic analysis (Euler–Maclaurin).\n\n• Substantially longer argument.  Each of the three parts requires a separate, non-trivial chain of reasoning; together they are markedly more intricate than proving π a₀²=2."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}