{
  "index": "1982-A-2",
  "type": "ANA",
  "tag": [
    "ANA",
    "NT"
  ],
  "difficulty": "",
  "question": "Problem A-2\nFor positive real \\( x \\), let\n\\[\nB_{n}(x)=1^{x}+2^{x}+3^{x}+\\cdots+n^{x}\n\\]\n\nProve or disprove the convergence of\n\\[\n\\sum_{n=2}^{\\infty} \\frac{B_{n}\\left(\\log _{n} 2\\right)}{\\left(n \\log _{2} n\\right)^{2}}\n\\]",
  "solution": "A-2.\nSince \\( x=\\log _{n} 2>0, B_{n}(x)=1^{x}+2^{x}+\\cdots+n^{x} \\leqslant n \\cdot n^{x} \\) and\n\\[\n0 \\leqslant \\frac{B_{n}\\left(\\log _{n} 2\\right)}{\\left(n \\log _{2} n\\right)^{2}} \\leqslant \\frac{n \\cdot n^{\\log _{n} 2}}{\\left(n \\log _{2} n\\right)^{2}}=\\frac{2}{n\\left(\\log _{2} n\\right)^{2}} .\n\\]\n\nAs \\( \\sum_{n=2}^{x}\\left[2 / n\\left(\\log _{2} n\\right)^{i}\\right] \\) converges by the integral test, the given series converges by the comparison test.",
  "vars": [
    "x",
    "n",
    "B_n"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "exponent",
        "n": "indexvar",
        "B_n": "partialsum"
      },
      "question": "Problem A-2\nFor positive real \\( exponent \\), let\n\\[\npartialsum_{indexvar}(exponent)=1^{exponent}+2^{exponent}+3^{exponent}+\\cdots+indexvar^{exponent}\n\\]\n\nProve or disprove the convergence of\n\\[\n\\sum_{indexvar=2}^{\\infty} \\frac{partialsum_{indexvar}\\left(\\log _{indexvar} 2\\right)}{\\left(indexvar \\log _{2} indexvar\\right)^{2}}\n\\]",
      "solution": "A-2.\nSince \\( exponent=\\log _{indexvar} 2>0, partialsum_{indexvar}(exponent)=1^{exponent}+2^{exponent}+\\cdots+indexvar^{exponent} \\leqslant indexvar \\cdot indexvar^{exponent} \\) and\n\\[\n0 \\leqslant \\frac{partialsum_{indexvar}\\left(\\log _{indexvar} 2\\right)}{\\left(indexvar \\log _{2} indexvar\\right)^{2}} \\leqslant \\frac{indexvar \\cdot indexvar^{\\log _{indexvar} 2}}{\\left(indexvar \\log _{2} indexvar\\right)^{2}}=\\frac{2}{indexvar\\left(\\log _{2} indexvar\\right)^{2}} .\n\\]\n\nAs \\( \\sum_{indexvar=2}^{exponent}\\left[2 / indexvar\\left(\\log _{2} indexvar\\right)^{i}\\right] \\) converges by the integral test, the given series converges by the comparison test."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "lanterns",
        "n": "compassrose",
        "B_n": "tapestry"
      },
      "question": "Problem A-2\nFor positive real \\( lanterns \\), let\n\\[\n\\tapestry_{compassrose}(lanterns)=1^{lanterns}+2^{lanterns}+3^{lanterns}+\\cdots+compassrose^{lanterns}\n\\]\n\nProve or disprove the convergence of\n\\[\n\\sum_{compassrose=2}^{\\infty} \\frac{\\tapestry_{compassrose}\\left(\\log _{compassrose} 2\\right)}{\\left(compassrose \\log _{2} compassrose\\right)^{2}}\n\\]",
      "solution": "A-2.\nSince \\( lanterns=\\log _{compassrose} 2>0, \\tapestry_{compassrose}(lanterns)=1^{lanterns}+2^{lanterns}+\\cdots+compassrose^{lanterns} \\leqslant compassrose \\cdot compassrose^{lanterns} \\) and\n\\[\n0 \\leqslant \\frac{\\tapestry_{compassrose}\\left(\\log _{compassrose} 2\\right)}{\\left(compassrose \\log _{2} compassrose\\right)^{2}} \\leqslant \\frac{compassrose \\cdot compassrose^{\\log _{compassrose} 2}}{\\left(compassrose \\log _{2} compassrose\\right)^{2}}=\\frac{2}{compassrose\\left(\\log _{2} compassrose\\right)^{2}} .\n\\]\n\nAs \\( \\sum_{compassrose=2}^{lanterns}\\left[2 / compassrose\\left(\\log _{2} compassrose\\right)^{i}\\right] \\) converges by the integral test, the given series converges by the comparison test."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "negativequantity",
        "n": "continuousvalue",
        "B_n": "sparseelement"
      },
      "question": "Problem A-2\nFor positive real \\( negativequantity \\), let\n\\[\nsparseelement_{continuousvalue}(negativequantity)=1^{negativequantity}+2^{negativequantity}+3^{negativequantity}+\\cdots+continuousvalue^{negativequantity}\n\\]\n\nProve or disprove the convergence of\n\\[\n\\sum_{continuousvalue=2}^{\\infty} \\frac{sparseelement_{continuousvalue}\\left(\\log _{continuousvalue} 2\\right)}{\\left(continuousvalue \\log _{2} continuousvalue\\right)^{2}}\n\\]",
      "solution": "A-2.\nSince \\( negativequantity=\\log _{continuousvalue} 2>0, sparseelement_{continuousvalue}(negativequantity)=1^{negativequantity}+2^{negativequantity}+\\cdots+continuousvalue^{negativequantity} \\leqslant continuousvalue \\cdot continuousvalue^{negativequantity} \\) and\n\\[\n0 \\leqslant \\frac{sparseelement_{continuousvalue}\\left(\\log _{continuousvalue} 2\\right)}{\\left(continuousvalue \\log _{2} continuousvalue\\right)^{2}} \\leqslant \\frac{continuousvalue \\cdot continuousvalue^{\\log _{continuousvalue} 2}}{\\left(continuousvalue \\log _{2} continuousvalue\\right)^{2}}=\\frac{2}{continuousvalue\\left(\\log _{2} continuousvalue\\right)^{2}} .\n\\]\n\nAs \\( \\sum_{continuousvalue=2}^{negativequantity}\\left[2 / continuousvalue\\left(\\log _{2} continuousvalue\\right)^{i}\\right] \\) converges by the integral test, the given series converges by the comparison test."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "n": "hjgrksla",
        "B_n": "mufjcqro"
      },
      "question": "Problem A-2\nFor positive real \\( qzxwvtnp \\), let\n\\[\nmufjcqro(qzxwvtnp)=1^{qzxwvtnp}+2^{qzxwvtnp}+3^{qzxwvtnp}+\\cdots+hjgrksla^{qzxwvtnp}\n\\]\n\nProve or disprove the convergence of\n\\[\n\\sum_{hjgrksla=2}^{\\infty} \\frac{mufjcqro\\left(\\log _{hjgrksla} 2\\right)}{\\left(hjgrksla \\log _{2} hjgrksla\\right)^{2}}\n\\]",
      "solution": "A-2.\nSince \\( qzxwvtnp=\\log _{hjgrksla} 2>0, mufjcqro(qzxwvtnp)=1^{qzxwvtnp}+2^{qzxwvtnp}+\\cdots+hjgrksla^{qzxwvtnp} \\leqslant hjgrksla \\cdot hjgrksla^{qzxwvtnp} \\) and\n\\[\n0 \\leqslant \\frac{mufjcqro\\left(\\log _{hjgrksla} 2\\right)}{\\left(hjgrksla \\log _{2} hjgrksla\\right)^{2}} \\leqslant \\frac{hjgrksla \\cdot hjgrksla^{\\log _{hjgrksla} 2}}{\\left(hjgrksla \\log _{2} hjgrksla\\right)^{2}}=\\frac{2}{hjgrksla\\left(\\log _{2} hjgrksla\\right)^{2}} .\n\\]\n\nAs \\( \\sum_{hjgrksla=2}^{qzxwvtnp}\\left[2 / hjgrksla\\left(\\log _{2} hjgrksla\\right)^{i}\\right] \\) converges by the integral test, the given series converges by the comparison test."
    },
    "kernel_variant": {
      "question": "Let $a>1$ be a fixed real constant and, for $x>0$, define  \n\\[\nB_{n}(x)=\\sum_{k=1}^{n}k^{x}\\qquad(n\\ge 1).\n\\]\nFor real parameters $(p,q)\\in\\mathbb R^{2}$ consider the positive–term series  \n\\[\nS(p,q):=\\sum_{n=3}^{\\infty}\n        \\frac{B_{n}\\bigl(\\log_{n}a\\bigr)}\n             {\\,n^{p}\\bigl(\\log_{a}n\\bigr)^{q}} .\n\\]\nDetermine precisely for which pairs $(p,q)\\in\\mathbb R^{2}$ the series $S(p,q)$ converges and for which it diverges.  (“$\\log$’’ always denotes the natural logarithm.)",
      "solution": "Throughout write $c:=\\ln a>0$.\n\n1.  Asymptotics of $B_{n}\\!\\bigl(\\log_{n}a\\bigr)$  \n\nPut  \n\\[\n\\alpha_{n}:=\\log_{n}a=\\frac{c}{\\ln n}\\qquad(n\\ge 3).\n\\]\nEuler’s integral bounds  \n\\[\n\\int_{1}^{n}t^{\\alpha}\\,dt\\;\\le\\;\n\\sum_{k=1}^{n}k^{\\alpha}\n\\;\\le\\;\n1+\\int_{1}^{n+1}t^{\\alpha}\\,dt\n\\qquad(\\alpha>-1)\n\\]\ngive, with $\\alpha=\\alpha_{n}>0$,  \n\\[\n\\frac{n^{\\alpha_{n}+1}-1}{\\alpha_{n}+1}\\;\\le\\;\nB_{n}(\\alpha_{n})\n\\;\\le\\;\n1+\\frac{(n+1)^{\\alpha_{n}+1}-1}{\\alpha_{n}+1}.\n\\]\nSince\n\\[\nn^{\\alpha_{n}}=\\exp\\!\\bigl(\\alpha_{n}\\ln n\\bigr)=\\exp(c)=a,\n\\]\nwe have $n^{\\alpha_{n}+1}=a\\,n$ and\n\\[\n\\frac1{\\alpha_{n}+1}\n      =\\frac1{1+\\dfrac{c}{\\ln n}}\n      =1-\\frac{c}{\\ln n}+O\\!\\Bigl(\\frac1{(\\ln n)^{2}}\\Bigr).\n\\]\nHence\n\\[\nB_{n}\\bigl(\\log_{n}a\\bigr)\n      =a\\,n\\Bigl[1+O\\!\\bigl(\\tfrac1{\\ln n}\\bigr)\\Bigr].\n\\tag{1}\n\\]\n\n2.  The general term of $S(p,q)$  \n\nUsing $\\log_{a}n=\\dfrac{\\ln n}{c}$, estimate (1) yields  \n\\[\nT_{n}(p,q)\n       :=\\frac{B_{n}(\\log_{n}a)}\n                 {n^{p}(\\log_{a}n)^{q}}\n        =a\\,c^{\\,q}\\;\n          n^{\\,1-p}(\\ln n)^{-q}\n          \\Bigl[1+O\\!\\bigl(\\tfrac1{\\ln n}\\bigr)\\Bigr].\n\\tag{2}\n\\]\n\n3.  Reduction to a classical $(p,\\log)$–series  \n\nDiscarding the harmless factor $a\\,c^{q}$, the convergence of $S(p,q)$ is equivalent to that of  \n\\[\n\\sum_{n=3}^{\\infty}u_{n},\\qquad\nu_{n}:=n^{\\,1-p}(\\ln n)^{-q}.\n\\]\n\n4.  Complete classification  \n\n4.1  The case $p>2$.  \n\nFix $p>2$ and choose  \n\\[\n\\varepsilon:=\\frac{p-2}{2}>0.\n\\]\nBy the elementary inequality $(\\ln n)^{\\lvert q\\rvert}\\le n^{\\varepsilon/2}$, valid for all $n\\ge N(q,\\varepsilon)$, we obtain  \n\\[\nu_{n}=n^{\\,1-p}(\\ln n)^{-q}\n     \\le n^{\\,1-p+\\varepsilon/2}\n     =n^{\\tfrac{2-3p}{4}}\n     \\qquad(n\\ge N(q,\\varepsilon)).\n\\]\nBecause $p>2$ implies $\\tfrac{2-3p}{4}<-1$, the tail  \n$\\sum_{n\\ge N(q,\\varepsilon)}u_{n}$ is dominated by a convergent $p$-series, and a finite number of initial terms does not affect convergence.  Hence $S(p,q)$ converges for every $q\\in\\mathbb R$ when $p>2$.\n\n4.2  The case $p<2$.  \n\nNow $2-p>0$.  For any fixed $N\\ge 3$ and all $n\\ge N$,\n\\[\nu_{n}\\ge n^{\\,1-p}(\\ln n)^{-\\lvert q\\rvert}\\;>0.\n\\]\nComparing with the integral\n\\[\n\\int_{N}^{\\infty}x^{\\,1-p}(\\ln x)^{-\\lvert q\\rvert}\\,dx,\n\\qquad x=e^{y}\\;(y=\\ln x),\n\\]\ngives  \n\\[\n\\int_{\\ln N}^{\\infty}e^{(2-p)y}\\,y^{-\\lvert q\\rvert}\\,dy.\n\\]\nBecause $2-p>0$, the exponential factor forces this improper integral—and hence the series—to diverge, regardless of the value of $q$.  Therefore $S(p,q)$ diverges for every $q$ when $p<2$.\n\n4.3  The case $p=2$.  \n\nHere $u_{n}=n^{-1}(\\ln n)^{-q}$.  By the Cauchy condensation test (equivalently, the integral test),\n\\[\n\\sum_{n\\ge 3}n^{-1}(\\ln n)^{-q}\n\\text{ converges }\\Longleftrightarrow q>1.\n\\]\nConsequently $S(2,q)$ converges if $q>1$ and diverges if $q\\le 1$.\n\n5.  Conclusion  \n\nCombining 4.1–4.3 we obtain  \n\\[\n\\boxed{\\;\nS(p,q)\\text{ converges}\n      \\Longleftrightarrow\n      \\bigl(p>2\\bigr)\\;\\text{or}\\;\n      \\bigl(p=2,\\;q>1\\bigr).\n\\;}\n\\]\nBecause all terms are non–negative, this describes absolute convergence.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.661519",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimensionality: the problem now involves TWO continuous parameters (p,q) instead of deciding a single fixed series.\n\n2. Deeper analysis: solving it forces the contestant to\n   • extract a precise asymptotic expansion for Bₙ(logₙ a) where the exponent itself tends to 0;  \n   • handle a non–trivial Riemann-sum/integral squeeze to get (8);  \n   • translate that asymptotic into a complete classification in the (p,q)-plane.\n\n3. Multiple concepts: asymptotics, improper-integral comparison, Euler sums, logarithmic refinements of p-series, and uniform error control all interact.\n\n4. No single ad-hoc comparison works; a full limiting analysis (Steps 2–4) is indispensable, followed by bifurcation into separate parameter regimes (Step 5).\n\nHence the variant is substantially harder than both the original problem (a single comparison test) and the current kernel variant, as it demands a global, parameter-dependent convergence theory rather than a yes/no answer for one series."
      }
    },
    "original_kernel_variant": {
      "question": "Let $a>1$ be a fixed real constant and, for $x>0$, define  \n\\[\nB_{n}(x)=\\sum_{k=1}^{n}k^{x}\\qquad(n\\ge 1).\n\\]\nFor real parameters $(p,q)\\in\\mathbb R^{2}$ consider the positive–term series  \n\\[\nS(p,q):=\\sum_{n=3}^{\\infty}\n        \\frac{B_{n}\\bigl(\\log_{n}a\\bigr)}\n             {\\,n^{p}\\bigl(\\log_{a}n\\bigr)^{q}} .\n\\]\nDetermine precisely for which pairs $(p,q)\\in\\mathbb R^{2}$ the series $S(p,q)$ converges and for which it diverges.  (“$\\log$’’ always denotes the natural logarithm.)",
      "solution": "Throughout write $c:=\\ln a>0$.\n\n1.  Asymptotics of $B_{n}\\!\\bigl(\\log_{n}a\\bigr)$  \n\nPut  \n\\[\n\\alpha_{n}:=\\log_{n}a=\\frac{c}{\\ln n}\\qquad(n\\ge 3).\n\\]\nEuler’s integral bounds  \n\\[\n\\int_{1}^{n}t^{\\alpha}\\,dt\\;\\le\\;\n\\sum_{k=1}^{n}k^{\\alpha}\n\\;\\le\\;\n1+\\int_{1}^{n+1}t^{\\alpha}\\,dt\n\\qquad(\\alpha>-1)\n\\]\ngive, with $\\alpha=\\alpha_{n}>0$,  \n\\[\n\\frac{n^{\\alpha_{n}+1}-1}{\\alpha_{n}+1}\\;\\le\\;\nB_{n}(\\alpha_{n})\n\\;\\le\\;\n1+\\frac{(n+1)^{\\alpha_{n}+1}-1}{\\alpha_{n}+1}.\n\\]\nSince\n\\[\nn^{\\alpha_{n}}=\\exp\\!\\bigl(\\alpha_{n}\\ln n\\bigr)=\\exp(c)=a,\n\\]\nwe have $n^{\\alpha_{n}+1}=a\\,n$ and\n\\[\n\\frac1{\\alpha_{n}+1}\n      =\\frac1{1+\\dfrac{c}{\\ln n}}\n      =1-\\frac{c}{\\ln n}+O\\!\\Bigl(\\frac1{(\\ln n)^{2}}\\Bigr).\n\\]\nHence\n\\[\nB_{n}\\bigl(\\log_{n}a\\bigr)\n      =a\\,n\\Bigl[1+O\\!\\bigl(\\tfrac1{\\ln n}\\bigr)\\Bigr].\n\\tag{1}\n\\]\n\n2.  The general term of $S(p,q)$  \n\nUsing $\\log_{a}n=\\dfrac{\\ln n}{c}$, estimate (1) yields  \n\\[\nT_{n}(p,q)\n       :=\\frac{B_{n}(\\log_{n}a)}\n                 {n^{p}(\\log_{a}n)^{q}}\n        =a\\,c^{\\,q}\\;\n          n^{\\,1-p}(\\ln n)^{-q}\n          \\Bigl[1+O\\!\\bigl(\\tfrac1{\\ln n}\\bigr)\\Bigr].\n\\tag{2}\n\\]\n\n3.  Reduction to a classical $(p,\\log)$–series  \n\nDiscarding the harmless factor $a\\,c^{q}$, the convergence of $S(p,q)$ is equivalent to that of  \n\\[\n\\sum_{n=3}^{\\infty}u_{n},\\qquad\nu_{n}:=n^{\\,1-p}(\\ln n)^{-q}.\n\\]\n\n4.  Complete classification  \n\n4.1  The case $p>2$.  \n\nFix $p>2$ and choose  \n\\[\n\\varepsilon:=\\frac{p-2}{2}>0.\n\\]\nBy the elementary inequality $(\\ln n)^{\\lvert q\\rvert}\\le n^{\\varepsilon/2}$, valid for all $n\\ge N(q,\\varepsilon)$, we obtain  \n\\[\nu_{n}=n^{\\,1-p}(\\ln n)^{-q}\n     \\le n^{\\,1-p+\\varepsilon/2}\n     =n^{\\tfrac{2-3p}{4}}\n     \\qquad(n\\ge N(q,\\varepsilon)).\n\\]\nBecause $p>2$ implies $\\tfrac{2-3p}{4}<-1$, the tail  \n$\\sum_{n\\ge N(q,\\varepsilon)}u_{n}$ is dominated by a convergent $p$-series, and a finite number of initial terms does not affect convergence.  Hence $S(p,q)$ converges for every $q\\in\\mathbb R$ when $p>2$.\n\n4.2  The case $p<2$.  \n\nNow $2-p>0$.  For any fixed $N\\ge 3$ and all $n\\ge N$,\n\\[\nu_{n}\\ge n^{\\,1-p}(\\ln n)^{-\\lvert q\\rvert}\\;>0.\n\\]\nComparing with the integral\n\\[\n\\int_{N}^{\\infty}x^{\\,1-p}(\\ln x)^{-\\lvert q\\rvert}\\,dx,\n\\qquad x=e^{y}\\;(y=\\ln x),\n\\]\ngives  \n\\[\n\\int_{\\ln N}^{\\infty}e^{(2-p)y}\\,y^{-\\lvert q\\rvert}\\,dy.\n\\]\nBecause $2-p>0$, the exponential factor forces this improper integral—and hence the series—to diverge, regardless of the value of $q$.  Therefore $S(p,q)$ diverges for every $q$ when $p<2$.\n\n4.3  The case $p=2$.  \n\nHere $u_{n}=n^{-1}(\\ln n)^{-q}$.  By the Cauchy condensation test (equivalently, the integral test),\n\\[\n\\sum_{n\\ge 3}n^{-1}(\\ln n)^{-q}\n\\text{ converges }\\Longleftrightarrow q>1.\n\\]\nConsequently $S(2,q)$ converges if $q>1$ and diverges if $q\\le 1$.\n\n5.  Conclusion  \n\nCombining 4.1–4.3 we obtain  \n\\[\n\\boxed{\\;\nS(p,q)\\text{ converges}\n      \\Longleftrightarrow\n      \\bigl(p>2\\bigr)\\;\\text{or}\\;\n      \\bigl(p=2,\\;q>1\\bigr).\n\\;}\n\\]\nBecause all terms are non–negative, this describes absolute convergence.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.520463",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimensionality: the problem now involves TWO continuous parameters (p,q) instead of deciding a single fixed series.\n\n2. Deeper analysis: solving it forces the contestant to\n   • extract a precise asymptotic expansion for Bₙ(logₙ a) where the exponent itself tends to 0;  \n   • handle a non–trivial Riemann-sum/integral squeeze to get (8);  \n   • translate that asymptotic into a complete classification in the (p,q)-plane.\n\n3. Multiple concepts: asymptotics, improper-integral comparison, Euler sums, logarithmic refinements of p-series, and uniform error control all interact.\n\n4. No single ad-hoc comparison works; a full limiting analysis (Steps 2–4) is indispensable, followed by bifurcation into separate parameter regimes (Step 5).\n\nHence the variant is substantially harder than both the original problem (a single comparison test) and the current kernel variant, as it demands a global, parameter-dependent convergence theory rather than a yes/no answer for one series."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}