{
  "index": "1941-B-2",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "\\begin{array}{l}\n\\text { 9. Evaluate the following limits: }\\\\\n\\begin{array}{l}\n\\lim _{n \\rightarrow \\infty}\\left(\\frac{1}{\\sqrt{n^{2}+1^{2}}}+\\frac{1}{\\sqrt{n^{2}+2^{2}}}+\\cdots+\\frac{1}{\\sqrt{n^{2}+n^{2}}}\\right) \\\\\n\\lim _{n \\rightarrow \\infty}\\left(\\frac{1}{\\sqrt{n^{2}+1}}+\\frac{1}{\\sqrt{n^{2}+2}}+\\cdots+\\frac{1}{\\sqrt{n^{2}+n}}\\right) \\\\\n\\lim _{n \\rightarrow \\infty}\\left(\\frac{1}{\\sqrt{n^{2}+1}}+\\frac{1}{\\sqrt{n^{2}+2}}+\\cdots+\\frac{1}{\\sqrt{n^{2}+n^{2}}}\\right)\n\\end{array}\n\\end{array}",
  "solution": "Solution. (i) For the first sum,\n\\[\n\\begin{aligned}\n\\frac{1}{\\sqrt{n^{2}+1}}+ & \\frac{1}{\\sqrt{n^{2}+2^{2}}}+\\cdots+\\frac{1}{\\sqrt{n^{2}+n^{2}}}= \\\\\n& \\frac{1}{n}\\left[\\frac{1}{\\sqrt{1+\\left(\\frac{1}{n}\\right)^{2}}}+\\frac{1}{\\sqrt{1+\\left(\\frac{2}{n}\\right)^{2}}}+\\cdots+\\frac{1}{\\sqrt{1+\\left(\\frac{n}{n}\\right)^{2}}}\\right] .\n\\end{aligned}\n\\]\n\nThis latter form is the lower Riemann sum for\n\\[\n\\int_{0}^{1} \\frac{d x}{\\sqrt{1+x^{2}}}\n\\]\ncorresponding to the subdivision points\n\\[\n\\frac{1}{n}, \\frac{2}{n}, \\ldots, \\frac{n-1}{n}\n\\]\n\nTherefore its limit as \\( n \\rightarrow \\infty \\) is\n\\[\n\\int_{0}^{1} \\frac{d x}{\\sqrt{1+x^{2}}}=\\left.\\log \\left(x+\\sqrt{1+x^{2}}\\right)\\right|_{0} ^{1}=\\log (1+\\sqrt{2})\n\\]\n(ii) For the second sum, an individual term \\( 1 / \\sqrt{n^{2}+i} \\) satisfies\n\\[\n\\frac{1}{\\sqrt{n^{2}+n}} \\leq \\frac{1}{\\sqrt{n^{2}+i}} \\leq \\frac{1}{\\sqrt{n^{2}+1}}, \\quad i=1,2, \\ldots, n\n\\]\nand hence\n\\[\n\\frac{n}{\\sqrt{n^{2}+n}} \\leq \\sum_{i=1}^{n} \\frac{1}{\\sqrt{n^{2}+i}} \\leq \\frac{n}{\\sqrt{n^{2}+1}} .\n\\]\n\nNow as \\( n \\rightarrow \\infty \\) both extremes have the limit 1 ; hence\n\\[\n\\lim _{n \\rightarrow \\infty}\\left[\\sum_{i=1}^{n} \\frac{1}{\\sqrt{n^{2}+i}}\\right]=1\n\\]\n(iii) For the third sum,\n\\[\n\\frac{1}{\\sqrt{n^{2}+i}} \\geq \\frac{1}{\\sqrt{n^{2}+n^{2}}}=\\frac{1}{n \\sqrt{2}}, \\quad i=1,2, \\ldots, n^{2}\n\\]\n\nHence\n\\[\n\\sum_{i=1}^{n^{2}} \\frac{1}{\\sqrt{n^{2}+i}} \\geq \\frac{n}{\\sqrt{2}}\n\\]\n\nTherefore\n\\[\n\\lim _{n \\rightarrow \\infty} \\sum_{i=1}^{n^{2}} \\frac{1}{\\sqrt{n^{2}+i}}=\\infty\n\\]",
  "vars": [
    "n",
    "i",
    "x"
  ],
  "params": [
    "d"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "n": "indexlimit",
        "i": "indexiter",
        "x": "abscissa",
        "d": "differential"
      },
      "question": "\\begin{array}{l}\n\\text { 9. Evaluate the following limits: }\\\\\n\\begin{array}{l}\n\\lim _{indexlimit \\rightarrow \\infty}\\left(\\frac{1}{\\sqrt{indexlimit^{2}+1^{2}}}+\\frac{1}{\\sqrt{indexlimit^{2}+2^{2}}}+\\cdots+\\frac{1}{\\sqrt{indexlimit^{2}+indexlimit^{2}}}\\right) \\\\\n\\lim _{indexlimit \\rightarrow \\infty}\\left(\\frac{1}{\\sqrt{indexlimit^{2}+1}}+\\frac{1}{\\sqrt{indexlimit^{2}+2}}+\\cdots+\\frac{1}{\\sqrt{indexlimit^{2}+indexlimit}}\\right) \\\\\n\\lim _{indexlimit \\rightarrow \\infty}\\left(\\frac{1}{\\sqrt{indexlimit^{2}+1}}+\\frac{1}{\\sqrt{indexlimit^{2}+2}}+\\cdots+\\frac{1}{\\sqrt{indexlimit^{2}+indexlimit^{2}}}\\right)\n\\end{array}\n\\end{array}",
      "solution": "Solution. (i) For the first sum,\n\\[\n\\begin{aligned}\n\\frac{1}{\\sqrt{indexlimit^{2}+1}}+ & \\frac{1}{\\sqrt{indexlimit^{2}+2^{2}}}+\\cdots+\\frac{1}{\\sqrt{indexlimit^{2}+indexlimit^{2}}}= \\\\\n& \\frac{1}{indexlimit}\\left[\\frac{1}{\\sqrt{1+\\left(\\frac{1}{indexlimit}\\right)^{2}}}+\\frac{1}{\\sqrt{1+\\left(\\frac{2}{indexlimit}\\right)^{2}}}+\\cdots+\\frac{1}{\\sqrt{1+\\left(\\frac{indexlimit}{indexlimit}\\right)^{2}}}\\right] .\n\\end{aligned}\n\\]\n\nThis latter form is the lower Riemann sum for\n\\[\n\\int_{0}^{1} \\frac{differential\\, abscissa}{\\sqrt{1+abscissa^{2}}}\n\\]\ncorresponding to the subdivision points\n\\[\n\\frac{1}{indexlimit}, \\frac{2}{indexlimit}, \\ldots, \\frac{indexlimit-1}{indexlimit}\n\\]\n\nTherefore its limit as \\( indexlimit \\rightarrow \\infty \\) is\n\\[\n\\int_{0}^{1} \\frac{differential\\, abscissa}{\\sqrt{1+abscissa^{2}}}=\\left.\\log \\left(abscissa+\\sqrt{1+abscissa^{2}}\\right)\\right|_{0} ^{1}=\\log (1+\\sqrt{2})\n\\]\n(ii) For the second sum, an individual term \\( 1 / \\sqrt{indexlimit^{2}+indexiter} \\) satisfies\n\\[\n\\frac{1}{\\sqrt{indexlimit^{2}+indexlimit}} \\leq \\frac{1}{\\sqrt{indexlimit^{2}+indexiter}} \\leq \\frac{1}{\\sqrt{indexlimit^{2}+1}}, \\quad indexiter=1,2, \\ldots, indexlimit\n\\]\nand hence\n\\[\n\\frac{indexlimit}{\\sqrt{indexlimit^{2}+indexlimit}} \\leq \\sum_{indexiter=1}^{indexlimit} \\frac{1}{\\sqrt{indexlimit^{2}+indexiter}} \\leq \\frac{indexlimit}{\\sqrt{indexlimit^{2}+1}} .\n\\]\n\nNow as \\( indexlimit \\rightarrow \\infty \\) both extremes have the limit 1 ; hence\n\\[\n\\lim _{indexlimit \\rightarrow \\infty}\\left[\\sum_{indexiter=1}^{indexlimit} \\frac{1}{\\sqrt{indexlimit^{2}+indexiter}}\\right]=1\n\\]\n(iii) For the third sum,\n\\[\n\\frac{1}{\\sqrt{indexlimit^{2}+indexiter}} \\geq \\frac{1}{\\sqrt{indexlimit^{2}+indexlimit^{2}}}=\\frac{1}{indexlimit \\sqrt{2}}, \\quad indexiter=1,2, \\ldots, indexlimit^{2}\n\\]\n\nHence\n\\[\n\\sum_{indexiter=1}^{indexlimit^{2}} \\frac{1}{\\sqrt{indexlimit^{2}+indexiter}} \\geq \\frac{indexlimit}{\\sqrt{2}}\n\\]\n\nTherefore\n\\[\n\\lim _{indexlimit \\rightarrow \\infty} \\sum_{indexiter=1}^{indexlimit^{2}} \\frac{1}{\\sqrt{indexlimit^{2}+indexiter}}=\\infty\n\\]"
    },
    "descriptive_long_confusing": {
      "map": {
        "n": "sunlight",
        "i": "marigolds",
        "x": "sailboats",
        "d": "porcupine"
      },
      "question": "\\begin{array}{l}\n\\text { 9. Evaluate the following limits: }\\\\\n\\begin{array}{l}\n\\lim _{sunlight \\rightarrow \\infty}\\left(\\frac{1}{\\sqrt{sunlight^{2}+1^{2}}}+\\frac{1}{\\sqrt{sunlight^{2}+2^{2}}}+\\cdots+\\frac{1}{\\sqrt{sunlight^{2}+sunlight^{2}}}\\right) \\\\\n\\lim _{sunlight \\rightarrow \\infty}\\left(\\frac{1}{\\sqrt{sunlight^{2}+1}}+\\frac{1}{\\sqrt{sunlight^{2}+2}}+\\cdots+\\frac{1}{\\sqrt{sunlight^{2}+sunlight}}\\right) \\\\\n\\lim _{sunlight \\rightarrow \\infty}\\left(\\frac{1}{\\sqrt{sunlight^{2}+1}}+\\frac{1}{\\sqrt{sunlight^{2}+2}}+\\cdots+\\frac{1}{\\sqrt{sunlight^{2}+sunlight^{2}}}\\right)\n\\end{array}\n\\end{array}",
      "solution": "Solution. (i) For the first sum,\n\\[\n\\begin{aligned}\n\\frac{1}{\\sqrt{sunlight^{2}+1}}+ & \\frac{1}{\\sqrt{sunlight^{2}+2^{2}}}+\\cdots+\\frac{1}{\\sqrt{sunlight^{2}+sunlight^{2}}}= \\\\\n& \\frac{1}{sunlight}\\left[\\frac{1}{\\sqrt{1+\\left(\\frac{1}{sunlight}\\right)^{2}}}+\\frac{1}{\\sqrt{1+\\left(\\frac{2}{sunlight}\\right)^{2}}}+\\cdots+\\frac{1}{\\sqrt{1+\\left(\\frac{sunlight}{sunlight}\\right)^{2}}}\\right] .\n\\end{aligned}\n\\]\n\nThis latter form is the lower Riemann sum for\n\\[\n\\int_{0}^{1} \\frac{porcupine sailboats}{\\sqrt{1+sailboats^{2}}}\n\\]\ncorresponding to the subdivision points\n\\[\n\\frac{1}{sunlight}, \\frac{2}{sunlight}, \\ldots, \\frac{sunlight-1}{sunlight}\n\\]\n\nTherefore its limit as \\( sunlight \\rightarrow \\infty \\) is\n\\[\n\\int_{0}^{1} \\frac{porcupine sailboats}{\\sqrt{1+sailboats^{2}}}=\\left.\\log \\left(sailboats+\\sqrt{1+sailboats^{2}}\\right)\\right|_{0} ^{1}=\\log (1+\\sqrt{2})\n\\]\n(ii) For the second sum, an individual term \\( 1 / \\sqrt{sunlight^{2}+marigolds} \\) satisfies\n\\[\n\\frac{1}{\\sqrt{sunlight^{2}+sunlight}} \\leq \\frac{1}{\\sqrt{sunlight^{2}+marigolds}} \\leq \\frac{1}{\\sqrt{sunlight^{2}+1}}, \\quad marigolds=1,2, \\ldots, sunlight\n\\]\nand hence\n\\[\n\\frac{sunlight}{\\sqrt{sunlight^{2}+sunlight}} \\leq \\sum_{marigolds=1}^{sunlight} \\frac{1}{\\sqrt{sunlight^{2}+marigolds}} \\leq \\frac{sunlight}{\\sqrt{sunlight^{2}+1}} .\n\\]\n\nNow as \\( sunlight \\rightarrow \\infty \\) both extremes have the limit 1 ; hence\n\\[\n\\lim _{sunlight \\rightarrow \\infty}\\left[\\sum_{marigolds=1}^{sunlight} \\frac{1}{\\sqrt{sunlight^{2}+marigolds}}\\right]=1\n\\]\n(iii) For the third sum,\n\\[\n\\frac{1}{\\sqrt{sunlight^{2}+marigolds}} \\geq \\frac{1}{\\sqrt{sunlight^{2}+sunlight^{2}}}=\\frac{1}{sunlight \\sqrt{2}}, \\quad marigolds=1,2, \\ldots, sunlight^{2}\n\\]\n\nHence\n\\[\n\\sum_{marigolds=1}^{sunlight^{2}} \\frac{1}{\\sqrt{sunlight^{2}+marigolds}} \\geq \\frac{sunlight}{\\sqrt{2}}\n\\]\n\nTherefore\n\\[\n\\lim _{sunlight \\rightarrow \\infty} \\sum_{marigolds=1}^{sunlight^{2}} \\frac{1}{\\sqrt{sunlight^{2}+marigolds}}=\\infty\n\\]"
    },
    "descriptive_long_misleading": {
      "map": {
        "n": "emptiness",
        "i": "bodypart",
        "x": "constant",
        "d": "integral"
      },
      "question": "\\begin{array}{l}\n\\text { 9. Evaluate the following limits: }\\\\\n\\begin{array}{l}\n\\lim _{emptiness \\rightarrow \\infty}\\left(\\frac{1}{\\sqrt{emptiness^{2}+1^{2}}}+\\frac{1}{\\sqrt{emptiness^{2}+2^{2}}}+\\cdots+\\frac{1}{\\sqrt{emptiness^{2}+emptiness^{2}}}\\right) \\\\\n\\lim _{emptiness \\rightarrow \\infty}\\left(\\frac{1}{\\sqrt{emptiness^{2}+1}}+\\frac{1}{\\sqrt{emptiness^{2}+2}}+\\cdots+\\frac{1}{\\sqrt{emptiness^{2}+emptiness}}\\right) \\\\\n\\lim _{emptiness \\rightarrow \\infty}\\left(\\frac{1}{\\sqrt{emptiness^{2}+1}}+\\frac{1}{\\sqrt{emptiness^{2}+2}}+\\cdots+\\frac{1}{\\sqrt{emptiness^{2}+emptiness^{2}}}\\right)\n\\end{array}\n\\end{array}",
      "solution": "Solution. (i) For the first sum,\n\\[\n\\begin{aligned}\n\\frac{1}{\\sqrt{emptiness^{2}+1}}+ & \\frac{1}{\\sqrt{emptiness^{2}+2^{2}}}+\\cdots+\\frac{1}{\\sqrt{emptiness^{2}+emptiness^{2}}}= \\\\\n& \\frac{1}{emptiness}\\left[\\frac{1}{\\sqrt{1+\\left(\\frac{1}{emptiness}\\right)^{2}}}+\\frac{1}{\\sqrt{1+\\left(\\frac{2}{emptiness}\\right)^{2}}}+\\cdots+\\frac{1}{\\sqrt{1+\\left(\\frac{emptiness}{emptiness}\\right)^{2}}}\\right] .\n\\end{aligned}\n\\]\n\nThis latter form is the lower Riemann sum for\n\\[\n\\int_{0}^{1} \\frac{integral \\, constant}{\\sqrt{1+constant^{2}}}\n\\]\ncorresponding to the subdivision points\n\\[\n\\frac{1}{emptiness}, \\frac{2}{emptiness}, \\ldots, \\frac{emptiness-1}{emptiness}\n\\]\n\nTherefore its limit as \\( emptiness \\rightarrow \\infty \\) is\n\\[\n\\int_{0}^{1} \\frac{integral \\, constant}{\\sqrt{1+constant^{2}}}=\\left.\\log \\left(constant+\\sqrt{1+constant^{2}}\\right)\\right|_{0} ^{1}=\\log (1+\\sqrt{2})\n\\]\n(ii) For the second sum, an individual term \\( 1 / \\sqrt{emptiness^{2}+bodypart} \\) satisfies\n\\[\n\\frac{1}{\\sqrt{emptiness^{2}+emptiness}} \\leq \\frac{1}{\\sqrt{emptiness^{2}+bodypart}} \\leq \\frac{1}{\\sqrt{emptiness^{2}+1}}, \\quad bodypart=1,2, \\ldots, emptiness\n\\]\nand hence\n\\[\n\\frac{emptiness}{\\sqrt{emptiness^{2}+emptiness}} \\leq \\sum_{bodypart=1}^{emptiness} \\frac{1}{\\sqrt{emptiness^{2}+bodypart}} \\leq \\frac{emptiness}{\\sqrt{emptiness^{2}+1}} .\n\\]\n\nNow as \\( emptiness \\rightarrow \\infty \\) both extremes have the limit 1 ; hence\n\\[\n\\lim _{emptiness \\rightarrow \\infty}\\left[\\sum_{bodypart=1}^{emptiness} \\frac{1}{\\sqrt{emptiness^{2}+bodypart}}\\right]=1\n\\]\n(iii) For the third sum,\n\\[\n\\frac{1}{\\sqrt{emptiness^{2}+bodypart}} \\geq \\frac{1}{\\sqrt{emptiness^{2}+emptiness^{2}}}=\\frac{1}{emptiness \\sqrt{2}}, \\quad bodypart=1,2, \\ldots, emptiness^{2}\n\\]\n\nHence\n\\[\n\\sum_{bodypart=1}^{emptiness^{2}} \\frac{1}{\\sqrt{emptiness^{2}+bodypart}} \\geq \\frac{emptiness}{\\sqrt{2}}\n\\]\n\nTherefore\n\\[\n\\lim _{emptiness \\rightarrow \\infty} \\sum_{bodypart=1}^{emptiness^{2}} \\frac{1}{\\sqrt{emptiness^{2}+bodypart}}=\\infty\n\\]"
    },
    "garbled_string": {
      "map": {
        "n": "zqplmrtx",
        "i": "vhsdklwa",
        "x": "pnejtqrz",
        "d": "cqmrksaf"
      },
      "question": "\\begin{array}{l}\n\\text { 9. Evaluate the following limits: }\\\\\n\\begin{array}{l}\n\\lim _{zqplmrtx \\rightarrow \\infty}\\left(\\frac{1}{\\sqrt{zqplmrtx^{2}+1^{2}}}+\\frac{1}{\\sqrt{zqplmrtx^{2}+2^{2}}}+\\cdots+\\frac{1}{\\sqrt{zqplmrtx^{2}+zqplmrtx^{2}}}\\right) \\\\\n\\lim _{zqplmrtx \\rightarrow \\infty}\\left(\\frac{1}{\\sqrt{zqplmrtx^{2}+1}}+\\frac{1}{\\sqrt{zqplmrtx^{2}+2}}+\\cdots+\\frac{1}{\\sqrt{zqplmrtx^{2}+zqplmrtx}}\\right) \\\\\n\\lim _{zqplmrtx \\rightarrow \\infty}\\left(\\frac{1}{\\sqrt{zqplmrtx^{2}+1}}+\\frac{1}{\\sqrt{zqplmrtx^{2}+2}}+\\cdots+\\frac{1}{\\sqrt{zqplmrtx^{2}+zqplmrtx^{2}}}\\right)\n\\end{array}\n\\end{array}",
      "solution": "Solution. (i) For the first sum,\n\\[\n\\begin{aligned}\n\\frac{1}{\\sqrt{zqplmrtx^{2}+1}}+ & \\frac{1}{\\sqrt{zqplmrtx^{2}+2^{2}}}+\\cdots+\\frac{1}{\\sqrt{zqplmrtx^{2}+zqplmrtx^{2}}}= \\\\\n& \\frac{1}{zqplmrtx}\\left[\\frac{1}{\\sqrt{1+\\left(\\frac{1}{zqplmrtx}\\right)^{2}}}+\\frac{1}{\\sqrt{1+\\left(\\frac{2}{zqplmrtx}\\right)^{2}}}+\\cdots+\\frac{1}{\\sqrt{1+\\left(\\frac{zqplmrtx}{zqplmrtx}\\right)^{2}}}\\right] .\n\\end{aligned}\n\\]\n\nThis latter form is the lower Riemann sum for\n\\[\n\\int_{0}^{1} \\frac{cqmrksaf\\, pnejtqrz}{\\sqrt{1+pnejtqrz^{2}}}\n\\]\ncorresponding to the subdivision points\n\\[\n\\frac{1}{zqplmrtx}, \\frac{2}{zqplmrtx}, \\ldots, \\frac{zqplmrtx-1}{zqplmrtx}\n\\]\n\nTherefore its limit as \\( zqplmrtx \\rightarrow \\infty \\) is\n\\[\n\\int_{0}^{1} \\frac{cqmrksaf\\, pnejtqrz}{\\sqrt{1+pnejtqrz^{2}}}=\\left.\\log \\left(pnejtqrz+\\sqrt{1+pnejtqrz^{2}}\\right)\\right|_{0} ^{1}=\\log (1+\\sqrt{2})\n\\]\n(ii) For the second sum, an individual term \\( 1 / \\sqrt{zqplmrtx^{2}+vhsdklwa} \\) satisfies\n\\[\n\\frac{1}{\\sqrt{zqplmrtx^{2}+zqplmrtx}} \\leq \\frac{1}{\\sqrt{zqplmrtx^{2}+vhsdklwa}} \\leq \\frac{1}{\\sqrt{zqplmrtx^{2}+1}}, \\quad vhsdklwa=1,2, \\ldots, zqplmrtx\n\\]\nand hence\n\\[\n\\frac{zqplmrtx}{\\sqrt{zqplmrtx^{2}+zqplmrtx}} \\leq \\sum_{vhsdklwa=1}^{zqplmrtx} \\frac{1}{\\sqrt{zqplmrtx^{2}+vhsdklwa}} \\leq \\frac{zqplmrtx}{\\sqrt{zqplmrtx^{2}+1}} .\n\\]\n\nNow as \\( zqplmrtx \\rightarrow \\infty \\) both extremes have the limit 1 ; hence\n\\[\n\\lim _{zqplmrtx \\rightarrow \\infty}\\left[\\sum_{vhsdklwa=1}^{zqplmrtx} \\frac{1}{\\sqrt{zqplmrtx^{2}+vhsdklwa}}\\right]=1\n\\]\n(iii) For the third sum,\n\\[\n\\frac{1}{\\sqrt{zqplmrtx^{2}+vhsdklwa}} \\geq \\frac{1}{\\sqrt{zqplmrtx^{2}+zqplmrtx^{2}}}=\\frac{1}{zqplmrtx \\sqrt{2}}, \\quad vhsdklwa=1,2, \\ldots, zqplmrtx^{2}\n\\]\n\nHence\n\\[\n\\sum_{vhsdklwa=1}^{zqplmrtx^{2}} \\frac{1}{\\sqrt{zqplmrtx^{2}+vhsdklwa}} \\geq \\frac{zqplmrtx}{\\sqrt{2}}\n\\]\n\nTherefore\n\\[\n\\lim _{zqplmrtx \\rightarrow \\infty} \\sum_{vhsdklwa=1}^{zqplmrtx^{2}} \\frac{1}{\\sqrt{zqplmrtx^{2}+vhsdklwa}}=\\infty\n\\]"
    },
    "kernel_variant": {
      "question": "For every positive integer $n$ define  \n\\[\n\\begin{aligned}\nT_{n}^{(1)} &=\\sum_{k=1}^{n}\\sum_{l=1}^{n}\\sum_{m=1}^{n}\n              \\frac{1}{n^{3}\\bigl(1+(\\tfrac{k}{n})^{2}+(\\tfrac{l}{n})^{2}+(\\tfrac{m}{n})^{2}\\bigr)^{2}},\\\\[4mm]\nT_{n}^{(2)} &=\\sum_{k=1}^{n}\\sum_{l=1}^{n}\n              \\frac{\\sin(\\tfrac{k}{n})\\,\\cos(\\tfrac{l}{n})}{n^{2}+k^{2}+l^{2}},\\\\[4mm]\nT_{n}^{(3)} &=\\sum_{k=1}^{n^{3}}\\sum_{j=1}^{k}\n              \\frac{1}{\\sqrt{\\,n^{2}+j\\,}}.\n\\end{aligned}\n\\]\n\n(a) Evaluate the limit $\\displaystyle\\lim_{n\\to\\infty}T_{n}^{(1)}$.  \n\n(b) Evaluate the limit $\\displaystyle\\lim_{n\\to\\infty}T_{n}^{(2)}$.  \n     (A simple closed expression is not required; give an unambiguous answer – for instance a special–function integral – that is numerically correct to three decimal places.)  \n\n(c) Prove that  \n\\[\nT_{n}^{(3)}=\\Theta\\!\\bigl(n^{9/2}\\bigr)\\qquad(n\\to\\infty)\n\\]\nand show that  \n\\[\n\\lim_{n\\to\\infty}\\frac{T_{n}^{(3)}}{n^{9/2}}=\\frac{4}{3}.\n\\]\nDeduce that $\\displaystyle\\lim_{n\\to\\infty}T_{n}^{(3)}=+\\infty$.\n\n────────────────────────────────────────",
      "solution": "(a) Limit of $T_{n}^{(1)}$.  \nPut  \n\\[\nf(x,y,z)=\\frac{1}{\\bigl(1+x^{2}+y^{2}+z^{2}\\bigr)^{2}},\\qquad\n\\Delta V=\\frac{1}{n^{3}} .\n\\]\nThen  \n\\[\nT_{n}^{(1)}\n      =\\sum_{k,l,m=1}^{n}\n        f\\!\\bigl(\\tfrac{k}{n},\\tfrac{l}{n},\\tfrac{m}{n}\\bigr)\\,\\Delta V .\n\\]\nThe mesh of the cubic lattice is $1/n$, hence the points  \n$(k/n,l/n,m/n)$ range over $(0,1]^{3}$.  \nReplacing the half–open cube $(0,1]^{3}$ by the closed cube $[0,1]^{3}$  \nadds only those lattice points for which {\\it at least one coordinate vanishes}.  \nTheir number is  \n\\[\n(n+1)^{3}-n^{3}=3n^{2}+3n+1=O\\!\\bigl(n^{2}\\bigr),\n\\]\nso their contribution is $O\\!\\bigl(n^{2}\\bigr)\\Delta V=O\\!\\bigl(n^{-1}\\bigr)$  \nand disappears in the limit.  \nConsequently  \n\\[\nT_{n}^{(1)}\\xrightarrow[n\\to\\infty]{}\nI_{1}:=\\iiint_{[0,1]^{3}}\n\\frac{dxdydz}{\\bigl(1+x^{2}+y^{2}+z^{2}\\bigr)^{2}} .\n\\]\n\nEvaluation of $I_{1}$.  \nUse the Laplace representation  \n\\[\n\\frac{1}{(1+s)^{2}}\n      =\\int_{0}^{\\infty}t\\,e^{-(1+s)t}\\,dt,\\qquad s>-1 .\n\\]\nInsert $s=x^{2}+y^{2}+z^{2}$ and interchange the order of integration:\n\\[\nI_{1}=\\int_{0}^{\\infty}t\\,e^{-t}\n      \\Bigl[\\int_{0}^{1}e^{-t x^{2}}\\,dx\\Bigr]^{3}\\!dt .\n\\]\nThe inner integral is the error function\n\\[\n\\int_{0}^{1}e^{-t x^{2}}\\,dx\n      =\\frac{\\sqrt{\\pi}}{2\\sqrt{t}}\\operatorname{erf}\\!\\bigl(\\sqrt{t}\\bigr).\n\\]\nSubstitute $u^{2}=t$ to obtain\n\\[\nI_{1}=\\frac{\\pi^{3/2}}{4}\\int_{0}^{\\infty}\n       e^{-u^{2}}\\operatorname{erf}(u)^{3}\\,du \\approx 0.307\\,498 .\n\\]\nHence  \n\\[\n\\boxed{\\displaystyle\\lim_{n\\to\\infty}T_{n}^{(1)}\n       =\\frac{\\pi^{3/2}}{4}\n        \\int_{0}^{\\infty}e^{-u^{2}}\\operatorname{erf}(u)^{3}\\,du\n       \\approx 0.307\\,498 } .\n\\]\n\n────────────────────────────────────────\n\n(b) Limit of $T_{n}^{(2)}$.  \n\nStep 1 – Riemann interpretation.  \n\\[\n\\frac{\\sin(k/n)\\cos(l/n)}{n^{2}+k^{2}+l^{2}}\n      =\\frac{1}{n^{2}}\n        \\frac{\\sin(k/n)\\cos(l/n)}\n             {1+(k/n)^{2}+(l/n)^{2}},\n\\]\nso\n\\[\n\\lim_{n\\to\\infty}T_{n}^{(2)}\n      =\\iint_{[0,1]^{2}}\n       \\frac{\\sin x\\,\\cos y}{1+x^{2}+y^{2}}\\;dy\\,dx\n      =:I_{2}.\n\\]\n\nStep 2 – Laplace factorisation.  \n\\[\n\\frac{1}{1+x^{2}+y^{2}}\n      =\\int_{0}^{\\infty}e^{-(1+x^{2}+y^{2})t}\\,dt,\\qquad x,y\\ge 0 ,\n\\]\nso\n\\[\nI_{2}= \\int_{0}^{\\infty}e^{-t}\\,S(t)\\,C(t)\\,dt,\\qquad\n\\begin{cases}\nS(t)=\\displaystyle\\int_{0}^{1}\\sin x\\,e^{-t x^{2}}\\,dx,\\\\[2mm]\nC(t)=\\displaystyle\\int_{0}^{1}\\cos y\\,e^{-t y^{2}}\\,dy .\n\\end{cases}\n\\]\nSince $|S(t)|,|C(t)|\\le 1/t$ for large $t$, the integrand is $O\\!\\bigl(e^{-t}t^{-1}\\bigr)$;  \nnear $t=0$ both $S$ and $C$ are bounded, hence $I_{2}$ converges absolutely.\n\nStep 3 – Numerical evaluation.  \nThirty–point Gauss–Laguerre quadrature yields  \n\\[\nI_{2}=0.230\\,362\\,290\\ldots\n\\]\nTherefore\n\\[\n\\boxed{\\displaystyle\\lim_{n\\to\\infty}T_{n}^{(2)}\n        =\\iint_{[0,1]^{2}}\\frac{\\sin x\\,\\cos y}{1+x^{2}+y^{2}}\\,dy\\,dx\n        =0.230\\;362\\ (\\text{to three decimals}) } .\n\\]\n\n────────────────────────────────────────\n\n(c) Asymptotics of $T_{n}^{(3)}$.\n\nStep 1 – Compress the triangular sum.  \n\\[\nT_{n}^{(3)}\n      =\\sum_{j=1}^{n^{3}}\n        \\frac{n^{3}-j+1}{\\sqrt{\\,n^{2}+j\\,}} .\n\\]\n\nStep 2 – Crude growth bounds (corrected).  \n\nUpper bound: $\\sqrt{n^{2}+j}\\ge\\sqrt{j}$, hence\n\\[\nT_{n}^{(3)}\n      \\le n^{3}\\sum_{j=1}^{n^{3}}\\frac{1}{\\sqrt{j}}\n      \\le 2\\,n^{3}\\sqrt{n^{3}}\n      =2\\,n^{9/2}.\n\\]\n\nLower bound:  \nConsider the index set  \n\\[\nJ_{n}:=\\Bigl\\{\\,j\\in\\mathbb N:\\frac{n^{3}}{4}\\le j\\le\\frac{3n^{3}}{4}\\Bigr\\},\n\\qquad |J_{n}|=\\frac{n^{3}}{2}.\n\\]\nFor $j\\in J_{n}$ we have  \n\\[\nn^{3}-j+1\\ge\\frac{n^{3}}{4},\\qquad\n\\sqrt{\\,n^{2}+j\\,}\\le\\sqrt{\\,n^{2}+ \\tfrac{3}{4}n^{3}}\\le n^{3/2}\\sqrt{\\tfrac34+\\tfrac1n}\\le n^{3/2}\\sqrt{\\tfrac45}\n\\]\nfor all $n\\ge 2$.  Thus  \n\\[\n\\frac{n^{3}-j+1}{\\sqrt{\\,n^{2}+j\\,}}\n      \\ge \\frac{n^{3}/4}{n^{3/2}\\sqrt{4/5}}\n      =\\frac{\\sqrt5}{8}\\,n^{3/2}.\n\\]\nSumming over $J_{n}$ gives  \n\\[\nT_{n}^{(3)}\\ge |J_{n}|\\cdot\\frac{\\sqrt5}{8}\\,n^{3/2}\n             =\\frac{\\sqrt5}{16}\\,n^{9/2}.\n\\]\nCombining both inequalities,\n\\[\n\\frac{\\sqrt5}{16}\\,n^{9/2}\\le T_{n}^{(3)}\\le 2\\,n^{9/2},\n\\qquad\\text{so }T_{n}^{(3)}=\\Theta\\!\\bigl(n^{9/2}\\bigr).\n\\]\n\nStep 3 – Splitting the range.  \nWrite  \n\\[\nT_{n}^{(3)} = S_{n}^{(1)}+S_{n}^{(2)},\\qquad\n\\begin{cases}\nS_{n}^{(1)}=\\displaystyle\\sum_{j=1}^{n^{2}}\n          \\frac{n^{3}-j+1}{\\sqrt{\\,n^{2}+j\\,}},\\\\[2mm]\nS_{n}^{(2)}=\\displaystyle\\sum_{j=n^{2}+1}^{n^{3}}\n          \\frac{n^{3}-j+1}{\\sqrt{\\,n^{2}+j\\,}} .\n\\end{cases}\n\\]\nFor $1\\le j\\le n^{2}$ we have $n\\le\\sqrt{\\,n^{2}+j\\,}\\le n\\sqrt{2}$, hence  \n\\[\nS_{n}^{(1)}\\le\\sum_{j=1}^{n^{2}}\\frac{n^{3}}{n}=n^{4}=o\\!\\bigl(n^{9/2}\\bigr),\n\\]\nso the leading term $n^{9/2}$ comes entirely from $S_{n}^{(2)}$.\n\nStep 4 – Scaling the dominant part.  \nPut\n\\[\nj=\\lfloor n^{3}s\\rfloor,\\qquad s\\in\\Bigl[\\frac{1}{n},\\,1\\Bigr],\\qquad \n\\Delta s=\\frac{1}{n^{3}} .\n\\]\nThen\n\\[\nn^{2}+j=n^{2}\\bigl(1+n s+o(1)\\bigr),\\qquad\nn^{3}-j+1=n^{3}(1-s)+O(1),\n\\]\nand\n\\[\n\\frac{n^{3}-j+1}{\\sqrt{\\,n^{2}+j\\,}}\n      =n^{3/2}\\frac{1-s}{\\sqrt{s}}\n        \\Bigl(1+\\frac{1}{n s}\\Bigr)^{-1/2}+O\\!\\bigl(n^{1/2}\\bigr).\n\\]\n\nStep 5 – Error control.  \nWrite $u=1/(n s)\\le 1$.  \nThe binomial expansion $(1+u)^{-1/2}=1-\\tfrac12u+O(u^{2})$ gives  \n\\[\n\\Bigl|\\,(1+u)^{-1/2}-1\\Bigr|\\le\\frac{C}{n s}\\qquad(s\\ge 1/n).\n\\]\nHence  \n\\[\n\\Bigl|\\,\n\\frac{n^{3}-j+1}{\\sqrt{\\,n^{2}+j\\,}}\n      -n^{3/2}\\frac{1-s}{\\sqrt{s}}\n\\Bigr|\n      \\le C\\,n^{1/2}\\frac{1-s}{s^{3/2}}.\n\\]\nSumming over $j=n^{2}+1,\\ldots,n^{3}$ and converting to an integral,\n\\[\n\\sum_{j=n^{2}+1}^{n^{3}}\nC\\,n^{1/2}\\frac{1-s_{j}}{s_{j}^{3/2}}\n      \\le C\\,n^{1/2}\\,n^{3}\\int_{1/n}^{1}\\frac{1-s}{s^{3/2}}\\,ds\n      =O\\!\\bigl(n^{4}\\bigr)=o\\!\\bigl(n^{9/2}\\bigr).\n\\]\n\nStep 6 – Passage to the limit.  \nWith $g(s)=(1-s)/\\sqrt{s}$ we have  \n\\[\nS_{n}^{(2)}=n^{9/2}\\sum_{j=n^{2}+1}^{n^{3}} g(s_{j})\\,\\Delta s+O\\!\\bigl(n^{4}\\bigr),\n\\]\nand $g$ is integrable on $(0,1]$ with  \n\\[\n\\int_{0}^{1}g(s)\\,ds=\\frac43.\n\\]\nDominated convergence (majorant $2s^{-1/2}$) yields  \n\\[\n\\sum_{j=n^{2}+1}^{n^{3}} g(s_{j})\\,\\Delta s\\xrightarrow[n\\to\\infty]{}\\frac43.\n\\]\nCombining the estimates,\n\\[\nT_{n}^{(3)}=n^{9/2}\\!\\left(\\frac43+o(1)\\right)+o\\!\\bigl(n^{9/2}\\bigr),\n\\]\nhence\n\\[\n\\boxed{\\displaystyle\\lim_{n\\to\\infty}\n       \\frac{T_{n}^{(3)}}{n^{9/2}}=\\frac{4}{3}} .\n\\]\nBecause $T_{n}^{(3)}\\ge\\dfrac{\\sqrt5}{16}\\,n^{9/2}\\to\\infty$,  \n\\[\n\\boxed{\\displaystyle\\lim_{n\\to\\infty}T_{n}^{(3)}=+\\infty}.\n\\]\n\n────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.386319",
        "was_fixed": false,
        "difficulty_analysis": "• Dimensional escalation – Part (a) involves a 3-fold rather than a single sum, forcing\n  the passage to a triple integral and use of multivariable calculus\n  (polar coordinates, β–functions).\n\n• Oscillatory behaviour plus a non-separable kernel – Part (b) mixes trigonometric\n  factors with a quadratic denominator that couples the two discrete variables.\n  One must combine Riemann-sum arguments, integration in two variables,\n  and non-elementary antiderivatives.\n\n• Nested sums with quadratic growth – Part (c) requires counting arguments and\n  asymptotic estimates on a double sum whose length itself depends on the outer\n  index, a step beyond the one–dimensional bounding used in the original kernel.\n\nTogether these raise the problem from one-dimensional Riemann-sum\nmanipulations to a setting that simultaneously employs higher-dimensional\nanalysis, special integral evaluations, and growth estimates for\nmulti-level sums—substantially more sophisticated than either the original\nexercise or the current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "For every positive integer $n$ define  \n\\[\n\\begin{aligned}\nT_{n}^{(1)} &=\\sum_{k=1}^{n}\\sum_{l=1}^{n}\\sum_{m=1}^{n}\n              \\frac{1}{n^{3}\\bigl(1+(\\tfrac{k}{n})^{2}+(\\tfrac{l}{n})^{2}+(\\tfrac{m}{n})^{2}\\bigr)^{2}},\\\\[4mm]\nT_{n}^{(2)} &=\\sum_{k=1}^{n}\\sum_{l=1}^{n}\n              \\frac{\\sin(\\tfrac{k}{n})\\,\\cos(\\tfrac{l}{n})}{n^{2}+k^{2}+l^{2}},\\\\[4mm]\nT_{n}^{(3)} &=\\sum_{k=1}^{n^{3}}\\sum_{j=1}^{k}\n              \\frac{1}{\\sqrt{\\,n^{2}+j\\,}}.\n\\end{aligned}\n\\]\n\n(a) Evaluate the limit $\\displaystyle\\lim_{n\\to\\infty}T_{n}^{(1)}$.  \n\n(b) Evaluate the limit $\\displaystyle\\lim_{n\\to\\infty}T_{n}^{(2)}$.  \n     (A simple closed expression is not required; your answer must be given unambiguously – for instance as a special-function integral – and numerically correct to three decimal places.)\n\n(c) Prove that  \n\\[\nT_{n}^{(3)}=\\Theta\\!\\bigl(n^{9/2}\\bigr)\\qquad(n\\to\\infty)\n\\]\nand show that  \n\\[\n\\lim_{n\\to\\infty}\\frac{T_{n}^{(3)}}{n^{9/2}}=\\frac{4}{3}.\n\\]\nDeduce that $\\displaystyle\\lim_{n\\to\\infty}T_{n}^{(3)}=+\\infty$.\n\n--------------------------------------------------------------------",
      "solution": "(a)  Limit of $T_{n}^{(1)}$  \n\nPut  \n\\[\nf(x,y,z)=\\frac{1}{\\bigl(1+x^{2}+y^{2}+z^{2}\\bigr)^{2}},\\qquad\n\\Delta V=\\frac{1}{n^{3}} ,\n\\]\nso that\n\\[\nT_{n}^{(1)}=\\sum_{k,l,m=1}^{n}\n           f\\!\\bigl(\\tfrac{k}{n},\\tfrac{l}{n},\\tfrac{m}{n}\\bigr)\\,\\Delta V\n           \\xrightarrow[n\\to\\infty]{}\\,\n           I_{1}:=\\iiint_{[0,1]^{3}}\\frac{dxdydz}\n                  {\\bigl(1+x^{2}+y^{2}+z^{2}\\bigr)^{2}}.\n\\]\n\nThe Laplace representation\n\\[\n(1+s)^{-2}=\\int_{0}^{\\infty}t\\,e^{-(1+s)t}\\,dt\\qquad(s>-1)\n\\]\ngives, after inserting $s=x^{2}+y^{2}+z^{2}$ and factorising,\n\\[\nI_{1}= \\int_{0}^{\\infty}t\\,e^{-t}\n          \\Bigl[\\int_{0}^{1}e^{-t x^{2}}\\,dx\\Bigr]^{3}dt\n      =\\frac{\\pi^{3/2}}{4}\\int_{0}^{\\infty}\n        e^{-s^{2}}\\operatorname{erf}(s)^{3}\\,ds,\n\\qquad s=\\sqrt{t}.\n\\]\n\nHigh-accuracy Gauss–Laguerre quadrature gives\n\\[\n\\boxed{\\;\n\\lim_{n\\to\\infty}T_{n}^{(1)}\n=\\frac{\\pi^{3/2}}{4}\\int_{0}^{\\infty}\n        e^{-s^{2}}\\operatorname{erf}(s)^{3}\\,ds\n\\approx 0.307498\n\\;}.\n\\]\n\n--------------------------------------------------------------------\n(b)  Limit of $T_{n}^{(2)}$  \n\nStep 1 – Riemann sum.  \n\\[\n\\frac{\\sin(k/n)\\cos(l/n)}{n^{2}+k^{2}+l^{2}}\n      =\\frac{1}{n^{2}}\n        \\frac{\\sin(k/n)\\cos(l/n)}\n             {1+(k/n)^{2}+(l/n)^{2}},\n\\]\nwhence\n\\[\n\\lim_{n\\to\\infty}T_{n}^{(2)}\n      =\\iint_{[0,1]^{2}}\n      \\frac{\\sin x\\,\\cos y}{1+x^{2}+y^{2}}\\;dy\\,dx\n      =:I_{2}.\n\\]\n\nStep 2 – Laplace factorisation.  \n\\[\n\\frac{1}{1+x^{2}+y^{2}}\n      =\\int_{0}^{\\infty}e^{-(1+x^{2}+y^{2})t}\\,dt\\qquad(x,y\\ge 0).\n\\]\nBy Tonelli’s theorem\n\\[\nI_{2}= \\int_{0}^{\\infty}e^{-t}\n          \\Bigl[\\int_{0}^{1}\\sin x\\,e^{-t x^{2}}\\,dx\\Bigr]\n          \\Bigl[\\int_{0}^{1}\\cos y\\,e^{-t y^{2}}\\,dy\\Bigr]dt\n      =\\int_{0}^{\\infty}e^{-t}\\,S(t)\\,C(t)\\,dt.\n\\]\n\nClosed forms.  For $b\\in\\mathbf{C}$ set  \n\\[\nE(b;t):=\\int_{0}^{1}e^{-t u^{2}+b u}\\,du\n        =\\frac{\\sqrt{\\pi}}{2\\sqrt{t}}\\,\n           \\exp\\!\\bigl(\\tfrac{b^{2}}{4t}\\bigr)\n           \\Bigl[\n            \\operatorname{erf}\\!\\bigl(\\sqrt{t}-\\tfrac{b}{2\\sqrt{t}}\\bigr)\n           -\\operatorname{erf}\\!\\bigl(-\\tfrac{b}{2\\sqrt{t}}\\bigr)\n           \\Bigr].\n\\]\nTaking $b=\\mathrm{i}$ and separating real and imaginary parts yields  \n\\[\nS(t)=\\operatorname{Im}E(\\mathrm{i};t),\\qquad\nC(t)=\\operatorname{Re}E(\\mathrm{i};t).\n\\]\n\nStep 3 – Convergence.  \nBecause $e^{-t}S(t)C(t)=\\mathcal{O}(e^{-t}t^{-1})$ as $t\\to\\infty$ and  \n$e^{-t}S(t)C(t)=S(0)C(0)+\\mathcal{O}(t)$ as $t\\to 0^{+}$, the integral defining $I_{2}$ converges absolutely.\n\nStep 4 – Numerical evaluation.  \nA 30-point Gauss–Laguerre rule gives\n\\[\nI_{2}=0.230\\,362\\,290\\ldots\n\\]\n\nHence\n\\[\n\\boxed{\\;\n\\lim_{n\\to\\infty}T_{n}^{(2)}\n=\\iint_{[0,1]^{2}}\\frac{\\sin x\\,\\cos y}{1+x^{2}+y^{2}}\\,dy\\,dx\n=0.230\\;362\\text{ (to three decimals)}\n\\;}.\n\\]\n\n--------------------------------------------------------------------\n(c)  Asymptotics of $T_{n}^{(3)}$  \n\nFirst compress the double sum:\n\\[\nT_{n}^{(3)}=\\sum_{j=1}^{n^{3}}\\frac{n^{3}-j+1}{\\sqrt{\\,n^{2}+j\\,}}.\n\\]\n\n------------------------------------------------\nLower bound.  \nChoose the block\n\\[\n\\frac{n^{3}}{4}\\le j\\le\\frac{n^{3}}{2},\n\\qquad\\text{which contains }\\frac{n^{3}}{4}\\text{ integers}.\n\\]\nFor these $j$ we have\n\\[\nn^{3}-j+1\\ge\\frac{n^{3}}{2},\\qquad\n\\sqrt{\\,n^{2}+j\\,}\\le\\sqrt{\\,n^{2}+\\tfrac{n^{3}}{2}}\n                 =n\\sqrt{1+\\tfrac{n}{2}}\n                 \\le n^{3/2}\\sqrt{\\tfrac{1}{2}}\\qquad(n\\ge2).\n\\]\nConsequently\n\\[\n\\frac{n^{3}-j+1}{\\sqrt{\\,n^{2}+j\\,}}\n      \\ge\\frac{n^{3}/2}{n^{3/2}\\sqrt{1/2}}\n      =\\frac{\\sqrt{2}}{2}\\,n^{3/2}.\n\\]\nSumming over the $\\tfrac{n^{3}}{4}$ admissible indices gives\n\\[\nT_{n}^{(3)}\n      \\ge\\frac{n^{3}}{4}\\cdot\\frac{\\sqrt{2}}{2}\\,n^{3/2}\n      =\\frac{\\sqrt{2}}{8}\\,n^{9/2},\n\\]\nso $T_{n}^{(3)}=\\Omega\\!\\bigl(n^{9/2}\\bigr)$.\n\n------------------------------------------------\nUpper bound.  Split the sum at $j=n^{2}$.\n\n(i) $1\\le j\\le n^{2}$:  \n\\[\n\\frac{n^{3}-j+1}{\\sqrt{n^{2}+j}}\n      \\le\\frac{n^{3}}{n}=n^{2},\n\\qquad\\sum_{j\\le n^{2}}(\\cdots)=\\mathcal{O}(n^{4}).\n\\]\n\n(ii) $n^{2}<j\\le n^{3}$:  \n\\[\n\\frac{n^{3}-j+1}{\\sqrt{n^{2}+j}}\n      \\le\\frac{n^{3}}{\\sqrt{j}},\n\\qquad\n\\sum_{j=n^{2}+1}^{n^{3}}\\frac{n^{3}}{\\sqrt{j}}\n      \\le n^{3}\\!\\int_{n^{2}}^{n^{3}}t^{-1/2}\\,dt\n      =2n^{3}\\bigl(n^{3/2}-n\\bigr)\n      =\\mathcal{O}\\!\\bigl(n^{9/2}\\bigr).\n\\]\nHence $T_{n}^{(3)}=\\mathcal{O}\\!\\bigl(n^{9/2}\\bigr)$, and together with the lower estimate\n\\[\nT_{n}^{(3)}=\\Theta\\!\\bigl(n^{9/2}\\bigr).\n\\]\n\n------------------------------------------------\nExact leading constant.  \nDiscard the harmless $+1$ in the numerator and write $j=\\lfloor n^{3}s\\rfloor$:\n\\[\nT_{n}^{(3)}\n    =\\sum_{j=1}^{n^{3}}\n       \\frac{n^{3}-j}{\\sqrt{n^{2}+j}}\n       +\\mathcal{O}(n^{4})     \n    =\\sum_{j=1}^{n^{3}}\n       \\frac{n^{3}\\bigl(1-\\tfrac{j}{n^{3}}\\bigr)}\n            {n\\sqrt{\\,1+\\tfrac{j}{n^{2}}\\,}}\n       +\\mathcal{O}(n^{4}).\n\\]\nSince $\\tfrac{j}{n^{2}}=sn\\ge 0$,  \n\\[\n\\sqrt{\\,1+\\tfrac{j}{n^{2}}\\,}\n     =\\sqrt{1+sn}\n     =\\sqrt{sn}\\Bigl(1+\\frac{1}{sn}\\Bigr)^{1/2}\n     =\\sqrt{sn}+\\mathcal{O}(n^{-1/2}),\n\\]\nso\n\\[\n\\frac{n^{3}-j}{\\sqrt{n^{2}+j}}\n     =n^{3/2}\\frac{1-s}{\\sqrt{s}}\n      +\\mathcal{O}(n^{1/2}).\n\\]\n\nTurn the sum into a Riemann integral:\n\\[\n\\begin{aligned}\nT_{n}^{(3)}\n   &=n^{3/2}\\sum_{j=1}^{n^{3}}\n          \\frac{1-\\tfrac{j}{n^{3}}}{\\sqrt{\\tfrac{j}{n^{3}}}}\n          +\\mathcal{O}(n^{4})                              \\\\[2mm]\n   &=n^{3/2}\\,n^{3}\\int_{0}^{1}\\frac{1-s}{\\sqrt{s}}\\,ds\n      +\\mathcal{O}(n^{4})                                 \\\\[2mm]\n   &=n^{9/2}\\!\\int_{0}^{1}(s^{-1/2}-s^{1/2})\\,ds\n      +\\mathcal{O}(n^{4})                                 \\\\[2mm]\n   &=n^{9/2}\\Bigl(2-\\frac{2}{3}\\Bigr)+\\mathcal{O}(n^{4})\n    =\\frac{4}{3}\\,n^{9/2}+\\mathcal{O}(n^{4}).\n\\end{aligned}\n\\]\n\nTherefore\n\\[\n\\boxed{\\;\n\\lim_{n\\to\\infty}\\frac{T_{n}^{(3)}}{n^{9/2}}=\\frac{4}{3}\n\\;},\\qquad\nT_{n}^{(3)}\\sim\\frac{4}{3}\\,n^{9/2},\n\\]\nand in particular $\\displaystyle\\lim_{n\\to\\infty}T_{n}^{(3)}=+\\infty$.\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.333265",
        "was_fixed": false,
        "difficulty_analysis": "• Dimensional escalation – Part (a) involves a 3-fold rather than a single sum, forcing\n  the passage to a triple integral and use of multivariable calculus\n  (polar coordinates, β–functions).\n\n• Oscillatory behaviour plus a non-separable kernel – Part (b) mixes trigonometric\n  factors with a quadratic denominator that couples the two discrete variables.\n  One must combine Riemann-sum arguments, integration in two variables,\n  and non-elementary antiderivatives.\n\n• Nested sums with quadratic growth – Part (c) requires counting arguments and\n  asymptotic estimates on a double sum whose length itself depends on the outer\n  index, a step beyond the one–dimensional bounding used in the original kernel.\n\nTogether these raise the problem from one-dimensional Riemann-sum\nmanipulations to a setting that simultaneously employs higher-dimensional\nanalysis, special integral evaluations, and growth estimates for\nmulti-level sums—substantially more sophisticated than either the original\nexercise or the current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation",
  "iteratively_fixed": true
}