{
  "index": "2006-B-6",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG",
    "NT"
  ],
  "difficulty": "",
  "question": "Let $k$ be an integer greater than 1. Suppose $a_0 > 0$, and define\n\\[\na_{n+1} = a_n + \\frac{1}{\\sqrt[k]{a_n}}\n\\]\nfor $n > 0$. Evaluate\n\\[\n\\lim_{n \\to \\infty} \\frac{a_n^{k+1}}{n^k}.\n\\]\n\\end{itemize}\n\n\\end{document}",
  "solution": "\\textbf{First solution:}\nWe start with some easy\nupper and lower bounds on $a_n$.\nWe write $O(f(n))$ and $\\Omega(f(n))$ for functions $g(n)$ such that\n$f(n)/g(n)$ and $g(n)/f(n)$, respectively, are bounded above.\nSince $a_n$ is a nondecreasing sequence, $a_{n+1}-a_n$ is bounded above,\nso $a_n = O(n)$. That means $a_n^{-1/k} = \\Omega(n^{-1/k})$, so\n\\[\na_n = \\Omega \\left( \\sum_{i=1}^n i^{-1/k} \\right)\n= \\Omega(n^{(k-1)/k}).\n\\]\nIn fact, all we will need is that $a_n \\to \\infty$ as $n \\to \\infty$.\n\nBy Taylor's theorem with remainder, for $1 < m < 2$ and $x>0$,\n\\[\n|(1+x)^m - 1 - mx| \\leq \\frac{m(m-1)}{2}x^2.\n\\]\nTaking $m = (k+1)/k$ and $x = a_{n+1}/a_n = 1 + a_n^{-(k+1)/k}$, we obtain\n\\[\n\\left| a_{n+1}^{(k+1)/k} - a_n^{(k+1)/k} - \\frac{k+1}{k} \\right|\n\\leq \\frac{k+1}{2k^2} a_n^{-(k+1)/k}.\n\\]\nIn particular,\n\\[\n\\lim_{n \\to \\infty} a_{n+1}^{(k+1)/k} - a_n^{(k+1)/k} = \\frac{k+1}{k}.\n\\]\n\nIn general, if $x_n$ is a sequence with $\\lim_{n \\to \\infty} x_n = c$, then\nalso\n\\[\n\\lim_{n \\to \\infty} \\frac{1}{n} \\sum_{i=1}^n x_i = c\n\\]\nby Cesaro's lemma. Explicitly, for any $\\epsilon > 0$, we can find $N$ such that\n$|x_n - c| \\leq \\epsilon/2$ for $n \\geq N$, and then\n\\[\n\\left| c - \\frac{1}{n} \\sum_{i=1}^n x_i \\right|\n\\leq \\frac{n-N}{n} \\frac{\\epsilon}{2} + \\frac{N}{n} \\left| \\sum_{i=1}^N (c-x_i) \\right|;\n\\]\nfor $n$ large, the right side is smaller than $\\epsilon$.\n\nIn our case, we deduce that\n\\[\n\\lim_{n \\to \\infty} \\frac{a_n^{(k+1)/k}}{n} = \\frac{k+1}{k}\n\\]\nand so\n\\[\n\\lim_{n \\to \\infty} \\frac{a_n^{k+1}}{n^k} = \\left(\\frac{k+1}{k} \\right)^k,\n\\]\nas desired.\n\n\\textbf{Remark:}\nThe use of Cesaro's lemma above is the special case $b_n = n$\nof the \\emph{Cesaro-Stolz\ntheorem}: if $a_n,b_n$ are sequences such that $b_n$ is positive,\nstrictly increasing, and unbounded, and\n\\[\n\\lim_{n \\to \\infty} \\frac{a_{n+1} - a_n}{b_{n+1} - b_n} = L,\n\\]\nthen\n\\[\n\\lim_{n \\to \\infty} \\frac{a_n}{b_n} = L.\n\\]\n\n\\textbf{Second solution:}\nIn this solution, rather than applying Taylor's theorem with remainder\nto $(1+x)^m$ for $1 < m < 2$ and $x > 0$, we only apply convexity to deduce\nthat $(1+x)^m \\geq 1 + mx$. This gives\n\\[\na_{n+1}^{(k+1)/k} - a_n^{(k+1)/k} \\geq \\frac{k+1}{k},\n\\]\nand so\n\\[\na_n^{(k+1)/k} \\geq \\frac{k+1}{k} n + c\n\\]\nfor some $c \\in \\RR$. In particular,\n\\[\n\\liminf_{n \\to \\infty} \\frac{a_n^{(k+1)/k}}{n} \\geq \\frac{k+1}{k}\n\\]\nand so\n\\[\n\\liminf_{n \\to \\infty} \\frac{a_n}{n^{k/(k+1)}} \\geq \\left(\\frac{k+1}{k} \\right)^{k/(k+1)}.\n\\]\nBut turning this around, the fact that\n\\begin{align*}\n&a_{n+1} - a_n \\\\\n&= a_n^{-1/k} \\\\\n&\\leq \\left(\\frac{k+1}{k} \\right)^{-1/(k+1)} n^{-1/(k+1)}\n(1 + o(1)),\n\\end{align*}\nwhere $o(1)$ denotes a function tending to 0 as $n \\to \\infty$,\nyields\n\\begin{align*}\n&a_n \\\\\n&\\leq\n\\left(\\frac{k+1}{k} \\right)^{-1/(k+1)} \\sum_{i=1}^n i^{-1/(k+1)} (1 + o(1)) \\\\\n&= \\frac{k+1}{k} \\left(\\frac{k+1}{k} \\right)^{-1/(k+1)} n^{k/(k+1)}(1 + o(1)) \\\\\n&= \\left( \\frac{k+1}{k} \\right)^{k/(k+1)} n^{k/(k+1)}(1 + o(1)),\n\\end{align*}\nso\n\\[\n\\limsup_{n \\to \\infty} \\frac{a_n}{n^{k/(k+1)}} \\leq \\left( \\frac{k+1}{k}\n\\right)^{k/(k+1)}\n\\]\nand this completes the proof.\n\n\\textbf{Third solution:}\nWe argue that $a_n \\to \\infty$ as in the first solution.\nWrite $b_n = a_n - L n^{k/(k+1)}$, for a value of $L$ to be determined\nlater.\nWe have\n\\begin{align*}\n&b_{n+1} \\\\\n &= b_n + a_n^{-1/k} - L ((n+1)^{k/(k+1)} - n^{k/(k+1)}) \\\\\n&= e_1 + e_2,\n\\end{align*}\nwhere\n\\begin{align*}\ne_1 &= b_n + a_n^{-1/k} - L^{-1/k} n^{-1/(k+1)} \\\\\ne_2 &= L ((n+1)^{k/(k+1)} - n^{k/(k+1)}) \\\\\n&\\quad - L^{-1/k} n^{-1/(k+1)}.\n\\end{align*}\nWe first estimate $e_1$.\nFor $-1 < m < 0$, by the convexity of $(1+x)^m$\nand $(1+x)^{1-m}$, we have\n\\begin{align*}\n1 + mx &\\leq (1+x)^m \\\\\n&\\leq 1 + mx (1+x)^{m-1}.\n\\end{align*}\nHence\n\\begin{align*}\n-\\frac{1}{k} L^{-(k+1)/k} n^{-1} b_n &\\leq e_1 - b_n \\\\\n&\\leq\n-\\frac{1}{k} b_n a_n^{-(k+1)/k}.\n\\end{align*}\nNote that both bounds have sign opposite to $b_n$; moreover,\nby the bound $a_n = \\Omega(n^{(k-1)/k})$, both bounds have absolutely\nvalue strictly less than that of $b_n$ for $n$ sufficiently large. Consequently,\nfor $n$ large,\n\\[\n|e_1| \\leq |b_n|.\n\\]\nWe now work on $e_2$.\nBy Taylor's theorem\nwith remainder applied to $(1+x)^m$ for $x > 0$ and $0 < m < 1$,\n\\begin{align*}\n1+mx &\\geq (1+x)^m \\\\\n&\\geq 1 + mx + \\frac{m(m-1)}{2} x^2.\n\\end{align*}\nThe ``main term'' of $L ((n+1)^{k/(k+1)} - n^{k/(k+1)})$\nis $L \\frac{k}{k+1} n^{-1/(k+1)}$. To make this coincide with\n$L^{-1/k} n^{-1/(k+1)}$, we take\n\\[\nL = \\left( \\frac{k+1}{k} \\right)^{k/(k+1)}.\n\\]\nWe then find that\n\\[\n|e_2| = O(n^{-2}),\n\\]\nand because $b_{n+1} = e_1 + e_2$, we have\n$|b_{n+1}| \\leq |b_n| + |e_2|$. Hence\n\\[\n|b_n| = O\\left (\\sum_{i=1}^n i^{-2} \\right) = O(1),\n\\]\nand so\n\\[\n\\lim_{n \\to \\infty} \\frac{a_n^{k+1}}{n^k} = L^{k+1} = \\left( \\frac{k+1}{k} \\right)^k.\n\\]\n\n\\textbf{Remark:}\nThe case $k=2$ appeared on the 2004 Romanian Olympiad (district level).\n\n\\textbf{Remark:}\nOne can make a similar argument for any sequence given by\n$a_{n+1} = a_n + f(a_n)$, when $f$ is a \\emph{decreasing} function.\n\n\\textbf{Remark:}\nRichard Stanley suggests a heuristic for determining the asymptotic\nbehavior of sequences of this type: replace the given recursion\n\\[\na_{n+1} - a_n = a_n^{-1/k}\n\\]\nby the differential equation\n\\[\ny' = y^{-1/k}\n\\]\nand determine the asymptotics of the latter.\n\\end{itemize}\n\n\n\\end{document}",
  "vars": [
    "n",
    "a_n",
    "a_n+1",
    "b_n",
    "b_n+1",
    "x_n",
    "x",
    "i",
    "f",
    "g",
    "y",
    "e_1",
    "e_2",
    "a_i"
  ],
  "params": [
    "k",
    "a_0",
    "c",
    "m",
    "N",
    "L"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "n": "indexvar",
        "a_n": "termseq",
        "a_n+1": "nextterm",
        "b_n": "auxterm",
        "b_n+1": "auxtermnext",
        "x_n": "genericseq",
        "x": "variablex",
        "f": "funcfvar",
        "g": "funcgvar",
        "y": "variabley",
        "e_1": "errorone",
        "e_2": "errortwo",
        "a_i": "termiteri",
        "k": "powerparm",
        "a_0": "initiala",
        "c": "limitval",
        "m": "exponent",
        "N": "boundidx",
        "L": "scalefac"
      },
      "question": "Let $powerparm$ be an integer greater than 1. Suppose $initiala > 0$, and define\n\\[\nnextterm = termseq + \\frac{1}{\\sqrt[powerparm]{termseq}}\n\\]\nfor $indexvar > 0$. Evaluate\n\\[\n\\lim_{indexvar \\to \\infty} \\frac{termseq^{powerparm+1}}{indexvar^{powerparm}}.\n\\]",
      "solution": "\\textbf{First solution:}\nWe start with some easy\nupper and lower bounds on $termseq$.\nWe write $O(funcfvar(indexvar))$ and $\\Omega(funcfvar(indexvar))$ for functions $funcgvar(indexvar)$ such that\n$funcfvar(indexvar)/funcgvar(indexvar)$ and $funcgvar(indexvar)/funcfvar(indexvar)$, respectively, are bounded above.\nSince $termseq$ is a nondecreasing sequence, $nextterm - termseq$ is bounded above,\nso $termseq = O(indexvar)$. That means $termseq^{-1/powerparm} = \\Omega(indexvar^{-1/powerparm})$, so\n\\[\ntermseq = \\Omega \\left( \\sum_{i=1}^{indexvar} i^{-1/powerparm} \\right)\n= \\Omega(indexvar^{(powerparm-1)/powerparm}).\n\\]\nIn fact, all we will need is that $termseq \\to \\infty$ as $indexvar \\to \\infty$.\n\nBy Taylor's theorem with remainder, for $1 < exponent < 2$ and $variablex>0$,\n\\[\n|(1+variablex)^{exponent} - 1 - exponent\\,variablex| \\leq \\frac{exponent (exponent-1)}{2}variablex^2.\n\\]\nTaking $exponent = (powerparm+1)/powerparm$ and $variablex = nextterm / termseq = 1 + termseq^{-(powerparm+1)/powerparm}$, we obtain\n\\[\n\\left| nextterm^{(powerparm+1)/powerparm} - termseq^{(powerparm+1)/powerparm} - \\frac{powerparm+1}{powerparm} \\right|\n\\leq \\frac{powerparm+1}{2powerparm^2} termseq^{-(powerparm+1)/powerparm}.\n\\]\nIn particular,\n\\[\n\\lim_{indexvar \\to \\infty} nextterm^{(powerparm+1)/powerparm} - termseq^{(powerparm+1)/powerparm} = \\frac{powerparm+1}{powerparm}.\n\\]\n\nIn general, if $genericseq$ is a sequence with $\\lim_{indexvar \\to \\infty} genericseq = limitval$, then\nalso\n\\[\n\\lim_{indexvar \\to \\infty} \\frac{1}{indexvar} \\sum_{i=1}^{indexvar} genericseq = limitval\n\\]\nby Cesaro's lemma. Explicitly, for any $\\epsilon > 0$, we can find $boundidx$ such that\n$|genericseq - limitval| \\leq \\epsilon/2$ for $indexvar \\geq boundidx$, and then\n\\[\n\\left| limitval - \\frac{1}{indexvar} \\sum_{i=1}^{indexvar} genericseq \\right|\n\\leq \\frac{indexvar-boundidx}{indexvar} \\frac{\\epsilon}{2} + \\frac{boundidx}{indexvar} \\left| \\sum_{i=1}^{boundidx} (limitval-genericseq) \\right|;\n\\]\nfor $indexvar$ large, the right side is smaller than $\\epsilon$.\n\nIn our case, we deduce that\n\\[\n\\lim_{indexvar \\to \\infty} \\frac{termseq^{(powerparm+1)/powerparm}}{indexvar} = \\frac{powerparm+1}{powerparm}\n\\]\nand so\n\\[\n\\lim_{indexvar \\to \\infty} \\frac{termseq^{powerparm+1}}{indexvar^{powerparm}} = \\left(\\frac{powerparm+1}{powerparm} \\right)^{powerparm},\n\\]\nas desired.\n\n\\textbf{Remark:}\nThe use of Cesaro's lemma above is the special case $auxterm = indexvar$\nof the \\emph{Cesaro-Stolz\ntheorem}: if $auxterm, indexvar$ are sequences such that $indexvar$ is positive,\nstrictly increasing, and unbounded, and\n\\[\n\\lim_{indexvar \\to \\infty} \\frac{auxtermnext - auxterm}{(indexvar+1) - indexvar} = limitval,\n\\]\nthen\n\\[\n\\lim_{indexvar \\to \\infty} \\frac{auxterm}{indexvar} = limitval.\n\\]\n\n\\textbf{Second solution:}\nIn this solution, rather than applying Taylor's theorem with remainder\nto $(1+variablex)^{exponent}$ for $1 < exponent < 2$ and $variablex > 0$, we only apply convexity to deduce\nthat $(1+variablex)^{exponent} \\geq 1 + exponent\\,variablex$. This gives\n\\[\nnextterm^{(powerparm+1)/powerparm} - termseq^{(powerparm+1)/powerparm} \\geq \\frac{powerparm+1}{powerparm},\n\\]\nand so\n\\[\ntermseq^{(powerparm+1)/powerparm} \\geq \\frac{powerparm+1}{powerparm} indexvar + limitval\n\\]\nfor some $limitval \\in \\RR$. In particular,\n\\[\n\\liminf_{indexvar \\to \\infty} \\frac{termseq^{(powerparm+1)/powerparm}}{indexvar} \\geq \\frac{powerparm+1}{powerparm}\n\\]\nand so\n\\[\n\\liminf_{indexvar \\to \\infty} \\frac{termseq}{indexvar^{powerparm/(powerparm+1)}} \\geq \\left(\\frac{powerparm+1}{powerparm} \\right)^{powerparm/(powerparm+1)}.\n\\]\nBut turning this around, the fact that\n\\begin{align*}\n&nextterm - termseq \\\\\n&= termseq^{-1/powerparm} \\\\\n&\\leq \\left(\\frac{powerparm+1}{powerparm} \\right)^{-1/(powerparm+1)} indexvar^{-1/(powerparm+1)}\n(1 + o(1)),\n\\end{align*}\nwhere $o(1)$ denotes a function tending to 0 as $indexvar \\to \\infty$,\nyields\n\\begin{align*}\n&termseq \\\\\n&\\leq\n\\left(\\frac{powerparm+1}{powerparm} \\right)^{-1/(powerparm+1)} \\sum_{i=1}^{indexvar} i^{-1/(powerparm+1)} (1 + o(1)) \\\\\n&= \\frac{powerparm+1}{powerparm} \\left(\\frac{powerparm+1}{powerparm} \\right)^{-1/(powerparm+1)} indexvar^{powerparm/(powerparm+1)}(1 + o(1)) \\\\\n&= \\left( \\frac{powerparm+1}{powerparm} \\right)^{powerparm/(powerparm+1)} indexvar^{powerparm/(powerparm+1)}(1 + o(1)),\n\\end{align*}\nso\n\\[\n\\limsup_{indexvar \\to \\infty} \\frac{termseq}{indexvar^{powerparm/(powerparm+1)}} \\leq \\left( \\frac{powerparm+1}{powerparm}\n\\right)^{powerparm/(powerparm+1)}\n\\]\nand this completes the proof.\n\n\\textbf{Third solution:}\nWe argue that $termseq \\to \\infty$ as in the first solution.\nWrite $auxterm = termseq - scalefac indexvar^{powerparm/(powerparm+1)}$, for a value of $scalefac$ to be determined\nlater.\nWe have\n\\begin{align*}\n&auxtermnext \\\\\n &= auxterm + termseq^{-1/powerparm} - scalefac ((indexvar+1)^{powerparm/(powerparm+1)} - indexvar^{powerparm/(powerparm+1)}) \\\\\n&= errorone + errortwo,\n\\end{align*}\nwhere\n\\begin{align*}\nerrorone &= auxterm + termseq^{-1/powerparm} - scalefac^{-1/powerparm} indexvar^{-1/(powerparm+1)} \\\\\nerrortwo &= scalefac ((indexvar+1)^{powerparm/(powerparm+1)} - indexvar^{powerparm/(powerparm+1)}) \\\\\n&\\quad - scalefac^{-1/powerparm} indexvar^{-1/(powerparm+1)}.\n\\end{align*}\nWe first estimate $errorone$.\nFor $-1 < exponent < 0$, by the convexity of $(1+variablex)^{exponent}$\nand $(1+variablex)^{1-exponent}$, we have\n\\begin{align*}\n1 + exponent\\,variablex &\\leq (1+variablex)^{exponent} \\\\\n&\\leq 1 + exponent\\,variablex (1+variablex)^{exponent-1}.\n\\end{align*}\nHence\n\\begin{align*}\n-\\frac{1}{powerparm} scalefac^{-(powerparm+1)/powerparm} indexvar^{-1} auxterm &\\leq errorone - auxterm \\\\\n&\\leq\n-\\frac{1}{powerparm} auxterm termseq^{-(powerparm+1)/powerparm}.\n\\end{align*}\nNote that both bounds have sign opposite to $auxterm$; moreover,\nby the bound $termseq = \\Omega(indexvar^{(powerparm-1)/powerparm})$, both bounds have absolute\nvalue strictly less than that of $auxterm$ for $indexvar$ sufficiently large. Consequently,\nfor $indexvar$ large,\n\\[\n|errorone| \\leq |auxterm|.\n\\]\nWe now work on $errortwo$.\nBy Taylor's theorem\nwith remainder applied to $(1+variablex)^{exponent}$ for $variablex > 0$ and $0 < exponent < 1$,\n\\begin{align*}\n1+exponent\\,variablex &\\geq (1+variablex)^{exponent} \\\\\n&\\geq 1 + exponent\\,variablex + \\frac{exponent(exponent-1)}{2} variablex^2.\n\\end{align*}\nThe ``main term'' of $scalefac ((indexvar+1)^{powerparm/(powerparm+1)} - indexvar^{powerparm/(powerparm+1)})$\nis $scalefac \\frac{powerparm}{powerparm+1} indexvar^{-1/(powerparm+1)}$. To make this coincide with\n$scalefac^{-1/powerparm} indexvar^{-1/(powerparm+1)}$, we take\n\\[\nscalefac = \\left( \\frac{powerparm+1}{powerparm} \\right)^{powerparm/(powerparm+1)}.\n\\]\nWe then find that\n\\[\n|errortwo| = O(indexvar^{-2}),\n\\]\nand because $auxtermnext = errorone + errortwo$, we have\n$|auxtermnext| \\leq |auxterm| + |errortwo|$. Hence\n\\[\n|auxterm| = O\\left (\\sum_{i=1}^{indexvar} i^{-2} \\right) = O(1),\n\\]\nand so\n\\[\n\\lim_{indexvar \\to \\infty} \\frac{termseq^{powerparm+1}}{indexvar^{powerparm}} = scalefac^{powerparm+1} = \\left( \\frac{powerparm+1}{powerparm} \\right)^{powerparm}.\n\\]\n\n\\textbf{Remark:}\nThe case $powerparm=2$ appeared on the 2004 Romanian Olympiad (district level).\n\n\\textbf{Remark:}\nOne can make a similar argument for any sequence given by\n$nextterm = termseq + funcfvar(termseq)$, when $funcfvar$ is a \\emph{decreasing} function.\n\n\\textbf{Remark:}\nRichard Stanley suggests a heuristic for determining the asymptotic\nbehavior of sequences of this type: replace the given recursion\n\\[\nnextterm - termseq = termseq^{-1/powerparm}\n\\]\nby the differential equation\n\\[\nvariabley' = variabley^{-1/powerparm}\n\\]\nand determine the asymptotics of the latter."
    },
    "descriptive_long_confusing": {
      "map": {
        "n": "pinecones",
        "a_n": "overcast",
        "a_n+1": "shipwreck",
        "b_n": "treadmill",
        "b_n+1": "watermill",
        "x_n": "dragonfly",
        "x": "archangel",
        "i": "i",
        "f": "blackbird",
        "g": "huckleber",
        "y": "sailplane",
        "e_1": "horseshoe",
        "e_2": "buttercup",
        "a_i": "scarecrow",
        "k": "bluegrass",
        "a_0": "boardwalk",
        "c": "patchwork",
        "m": "candlewick",
        "N": "goldcrest",
        "L": "cornstalk"
      },
      "question": "Let $bluegrass$ be an integer greater than 1. Suppose $boardwalk > 0$, and define\n\\[\nshipwreck = overcast + \\frac{1}{\\sqrt[bluegrass]{overcast}}\n\\]\nfor $pinecones > 0$. Evaluate\n\\[\n\\lim_{pinecones \\to \\infty} \\frac{overcast^{bluegrass+1}}{pinecones^{bluegrass}}.\n\\]\n\\end{itemize}\n\n\\end{document}",
      "solution": "\\textbf{First solution:}\nWe start with some easy\nupper and lower bounds on $overcast$.\nWe write $O(blackbird(pinecones))$ and $\\Omega(blackbird(pinecones))$ for functions $huckleber(pinecones)$ such that\n$blackbird(pinecones)/huckleber(pinecones)$ and $huckleber(pinecones)/blackbird(pinecones)$, respectively, are bounded above.\nSince $overcast$ is a nondecreasing sequence, $shipwreck-overcast$ is bounded above,\nso $overcast = O(pinecones)$. That means $overcast^{-1/bluegrass} = \\Omega(pinecones^{-1/bluegrass})$, so\n\\[\novercast = \\Omega \\left( \\sum_{i=1}^{pinecones} i^{-1/bluegrass} \\right)\n= \\Omega(pinecones^{(bluegrass-1)/bluegrass}).\n\\]\nIn fact, all we will need is that $overcast \\to \\infty$ as $pinecones \\to \\infty$.\n\nBy Taylor's theorem with remainder, for $1 < candlewick < 2$ and $archangel>0$,\n\\[|(1+archangel)^{candlewick} - 1 - candlewick\\,archangel| \\leq \\frac{candlewick(candlewick-1)}{2}archangel^2.\\]\nTaking $candlewick = (bluegrass+1)/bluegrass$ and $archangel = shipwreck/overcast = 1 + overcast^{-(bluegrass+1)/bluegrass}$, we obtain\n\\[\n\\left| shipwreck^{(bluegrass+1)/bluegrass} - overcast^{(bluegrass+1)/bluegrass} - \\frac{bluegrass+1}{bluegrass} \\right|\n\\leq \\frac{bluegrass+1}{2\\,bluegrass^2} \\, overcast^{-(bluegrass+1)/bluegrass}.\n\\]\nIn particular,\n\\[\n\\lim_{pinecones \\to \\infty} shipwreck^{(bluegrass+1)/bluegrass} - overcast^{(bluegrass+1)/bluegrass} = \\frac{bluegrass+1}{bluegrass}.\n\\]\n\nIn general, if $dragonfly$ is a sequence with $\\lim_{pinecones \\to \\infty} dragonfly = patchwork$, then\nalso\n\\[\n\\lim_{pinecones \\to \\infty} \\frac{1}{pinecones} \\sum_{i=1}^{pinecones} dragonfly = patchwork\n\\]\nby Cesaro's lemma. Explicitly, for any $\\epsilon > 0$, we can find $goldcrest$ such that\n$|dragonfly - patchwork| \\leq \\epsilon/2$ for $pinecones \\geq goldcrest$, and then\n\\[\n\\left| patchwork - \\frac{1}{pinecones} \\sum_{i=1}^{pinecones} dragonfly \\right|\n\\leq \\frac{pinecones-goldcrest}{pinecones} \\frac{\\epsilon}{2} + \\frac{goldcrest}{pinecones} \\left| \\sum_{i=1}^{goldcrest} (patchwork-dragonfly) \\right|;\n\\]\nfor $pinecones$ large, the right side is smaller than $\\epsilon$.\n\nIn our case, we deduce that\n\\[\n\\lim_{pinecones \\to \\infty} \\frac{overcast^{(bluegrass+1)/bluegrass}}{pinecones} = \\frac{bluegrass+1}{bluegrass}\n\\]\nand so\n\\[\n\\lim_{pinecones \\to \\infty} \\frac{overcast^{bluegrass+1}}{pinecones^{bluegrass}} = \\left(\\frac{bluegrass+1}{bluegrass} \\right)^{bluegrass},\n\\]\nas desired.\n\n\\textbf{Remark:}\nThe use of Cesaro's lemma above is the special case $treadmill = pinecones$\nof the \\emph{Cesaro-Stolz\ntheorem}: if $overcast,\\,treadmill$ are sequences such that $treadmill$ is positive,\nstrictly increasing, and unbounded, and\n\\[\n\\lim_{pinecones \\to \\infty} \\frac{shipwreck - overcast}{watermill - treadmill} = cornstalk,\n\\]\nthen\n\\[\n\\lim_{pinecones \\to \\infty} \\frac{overcast}{treadmill} = cornstalk.\n\\]\n\n\\textbf{Second solution:}\nIn this solution, rather than applying Taylor's theorem with remainder\nto $(1+archangel)^{candlewick}$ for $1 < candlewick < 2$ and $archangel > 0$, we only apply convexity to deduce\nthat $(1+archangel)^{candlewick} \\geq 1 + candlewick\\,archangel$. This gives\n\\[\nshipwreck^{(bluegrass+1)/bluegrass} - overcast^{(bluegrass+1)/bluegrass} \\geq \\frac{bluegrass+1}{bluegrass},\n\\]\nand so\n\\[\novercast^{(bluegrass+1)/bluegrass} \\geq \\frac{bluegrass+1}{bluegrass} \\, pinecones + patchwork\n\\]\nfor some $patchwork \\in \\RR$. In particular,\n\\[\n\\liminf_{pinecones \\to \\infty} \\frac{overcast^{(bluegrass+1)/bluegrass}}{pinecones} \\geq \\frac{bluegrass+1}{bluegrass}\n\\]\nand so\n\\[\n\\liminf_{pinecones \\to \\infty} \\frac{overcast}{pinecones^{bluegrass/(bluegrass+1)}} \\geq \\left(\\frac{bluegrass+1}{bluegrass} \\right)^{bluegrass/(bluegrass+1)}.\n\\]\nBut turning this around, the fact that\n\\begin{align*}\n&shipwreck - overcast \\\\\n&= overcast^{-1/bluegrass} \\\\\n&\\leq \\left(\\frac{bluegrass+1}{bluegrass} \\right)^{-1/(bluegrass+1)} \\, pinecones^{-1/(bluegrass+1)}\n(1 + o(1)),\n\\end{align*}\nwhere $o(1)$ denotes a function tending to 0 as $pinecones \\to \\infty$,\nyields\n\\begin{align*}\n&overcast \\\\\n&\\leq\n\\left(\\frac{bluegrass+1}{bluegrass} \\right)^{-1/(bluegrass+1)} \\sum_{i=1}^{pinecones} i^{-1/(bluegrass+1)} (1 + o(1)) \\\\\n&= \\frac{bluegrass+1}{bluegrass} \\left(\\frac{bluegrass+1}{bluegrass} \\right)^{-1/(bluegrass+1)} pinecones^{bluegrass/(bluegrass+1)}(1 + o(1)) \\\\\n&= \\left( \\frac{bluegrass+1}{bluegrass} \\right)^{bluegrass/(bluegrass+1)} pinecones^{bluegrass/(bluegrass+1)}(1 + o(1)),\n\\end{align*}\nso\n\\[\n\\limsup_{pinecones \\to \\infty} \\frac{overcast}{pinecones^{bluegrass/(bluegrass+1)}} \\leq \\left( \\frac{bluegrass+1}{bluegrass}\n\\right)^{bluegrass/(bluegrass+1)}\n\\]\nand this completes the proof.\n\n\\textbf{Third solution:}\nWe argue that $overcast \\to \\infty$ as in the first solution.\nWrite $treadmill = overcast - cornstalk \\, pinecones^{bluegrass/(bluegrass+1)}$, for a value of $cornstalk$ to be determined\nlater.\nWe have\n\\begin{align*}\n&watermill \\\\\n &= treadmill + overcast^{-1/bluegrass} - cornstalk \\bigl( (pinecones+1)^{bluegrass/(bluegrass+1)} - pinecones^{bluegrass/(bluegrass+1)} \\bigr) \\\\\n&= horseshoe + buttercup,\n\\end{align*}\nwhere\n\\begin{align*}\nhorseshoe &= treadmill + overcast^{-1/bluegrass} - cornstalk^{-1/bluegrass} \\, pinecones^{-1/(bluegrass+1)} \\\\\nbuttercup &= cornstalk \\bigl( (pinecones+1)^{bluegrass/(bluegrass+1)} - pinecones^{bluegrass/(bluegrass+1)} \\bigr) \\\\\n&\\quad - cornstalk^{-1/bluegrass} \\, pinecones^{-1/(bluegrass+1)}.\n\\end{align*}\nWe first estimate $horseshoe$.\nFor $-1 < candlewick < 0$, by the convexity of $(1+archangel)^{candlewick}$\nand $(1+archangel)^{1-candlewick}$, we have\n\\begin{align*}\n1 + candlewick\\,archangel &\\leq (1+archangel)^{candlewick} \\\\\n&\\leq 1 + candlewick\\,archangel \\, (1+archangel)^{candlewick-1}.\n\\end{align*}\nHence\n\\begin{align*}\n-\\frac{1}{bluegrass} \\, cornstalk^{-(bluegrass+1)/bluegrass} \\, pinecones^{-1} \\, treadmill &\\leq horseshoe - treadmill \\\\\n&\\leq\n-\\frac{1}{bluegrass} \\, treadmill \\, overcast^{-(bluegrass+1)/bluegrass}.\n\\end{align*}\nNote that both bounds have sign opposite to $treadmill$; moreover,\nby the bound $overcast = \\Omega(pinecones^{(bluegrass-1)/bluegrass})$, both bounds have absolute\nvalue strictly less than that of $treadmill$ for $pinecones$ sufficiently large. Consequently,\nfor $pinecones$ large,\n\\[\n|horseshoe| \\leq |treadmill|.\n\\]\nWe now work on $buttercup$.\nBy Taylor's theorem\nwith remainder applied to $(1+archangel)^{candlewick}$ for $archangel > 0$ and $0 < candlewick < 1$,\n\\begin{align*}\n1+candlewick\\,archangel &\\geq (1+archangel)^{candlewick} \\\\\n&\\geq 1 + candlewick\\,archangel + \\frac{candlewick(candlewick-1)}{2} \\, archangel^2.\n\\end{align*}\nThe ``main term'' of $cornstalk \\bigl( (pinecones+1)^{bluegrass/(bluegrass+1)} - pinecones^{bluegrass/(bluegrass+1)} \\bigr)$\nis $cornstalk \\frac{bluegrass}{bluegrass+1} \\, pinecones^{-1/(bluegrass+1)}$. To make this coincide with\n$cornstalk^{-1/bluegrass} \\, pinecones^{-1/(bluegrass+1)}$, we take\n\\[\ncornstalk = \\left( \\frac{bluegrass+1}{bluegrass} \\right)^{bluegrass/(bluegrass+1)}.\n\\]\nWe then find that\n\\[\n|buttercup| = O(pinecones^{-2}),\n\\]\nand because $watermill = horseshoe + buttercup$, we have\n$|watermill| \\leq |treadmill| + |buttercup|$. Hence\n\\[\n|treadmill| = O\\left (\\sum_{i=1}^{pinecones} i^{-2} \\right) = O(1),\n\\]\nand so\n\\[\n\\lim_{pinecones \\to \\infty} \\frac{overcast^{bluegrass+1}}{pinecones^{bluegrass}} = cornstalk^{bluegrass+1} = \\left( \\frac{bluegrass+1}{bluegrass} \\right)^{bluegrass}.\n\\]\n\n\\textbf{Remark:}\nThe case $bluegrass=2$ appeared on the 2004 Romanian Olympiad (district level).\n\n\\textbf{Remark:}\nOne can make a similar argument for any sequence given by\n$shipwreck = overcast + blackbird(overcast)$, when $blackbird$ is a \\emph{decreasing} function.\n\n\\textbf{Remark:}\nRichard Stanley suggests a heuristic for determining the asymptotic\nbehavior of sequences of this type: replace the given recursion\n\\[\nshipwreck - overcast = overcast^{-1/bluegrass}\n\\]\nby the differential equation\n\\[\nsailplane' = sailplane^{-1/bluegrass}\n\\]\nand determine the asymptotics of the latter.\n\\end{itemize}\n\n\\end{document}"
    },
    "descriptive_long_misleading": {
      "map": {
        "n": "humongousindex",
        "a_n": "stagnantvalue",
        "a_n+1": "retreatvalue",
        "b_n": "quiescentbase",
        "b_n+1": "quiescentbasenext",
        "x_n": "fixedpointset",
        "x": "constantalpha",
        "i": "infinityidx",
        "f": "rigidmap",
        "g": "staticfunc",
        "y": "steadyvar",
        "e_1": "calmerrorone",
        "e_2": "calmerrortwo",
        "a_i": "laggardterm",
        "k": "softconst",
        "a_0": "finalanchor",
        "c": "variability",
        "m": "meekexponent",
        "N": "tinycutoff",
        "L": "movingvalue"
      },
      "question": "Let $softconst$ be an integer greater than 1. Suppose $finalanchor > 0$, and define\n\\[\nretreatvalue = stagnantvalue + \\frac{1}{\\sqrt[softconst]{stagnantvalue}}\n\\]\nfor $humongousindex > 0$. Evaluate\n\\[\n\\lim_{humongousindex \\to \\infty} \\frac{stagnantvalue^{softconst+1}}{humongousindex^{softconst}}.\n\\]",
      "solution": "\\textbf{First solution:}\nWe start with some easy\nupper and lower bounds on $stagnantvalue$.\nWe write $O(rigidmap(humongousindex))$ and $\\Omega(rigidmap(humongousindex))$ for functions $staticfunc(humongousindex)$ such that\n$rigidmap(humongousindex)/staticfunc(humongousindex)$ and $staticfunc(humongousindex)/rigidmap(humongousindex)$, respectively, are bounded above.\nSince $stagnantvalue$ is a nondecreasing sequence, $retreatvalue-stagnantvalue$ is bounded above,\nso $stagnantvalue = O(humongousindex)$. That means $stagnantvalue^{-1/softconst} = \\Omega(humongousindex^{-1/softconst})$, so\n\\[\nstagnantvalue = \\Omega \\left( \\sum_{infinityidx=1}^{humongousindex} infinityidx^{-1/softconst} \\right)\n= \\Omega(humongousindex^{(softconst-1)/softconst}).\n\\]\nIn fact, all we will need is that $stagnantvalue \\to \\infty$ as $humongousindex \\to \\infty$.\n\nBy Taylor's theorem with remainder, for $1 < meekexponent < 2$ and $constantalpha>0$,\n\\[\n|(1+constantalpha)^{meekexponent} - 1 - meekexponent constantalpha| \\leq \\frac{meekexponent(meekexponent-1)}{2}constantalpha^2.\n\\]\nTaking $meekexponent = (softconst+1)/softconst$ and $constantalpha = retreatvalue/stagnantvalue = 1 + stagnantvalue^{-(softconst+1)/softconst}$, we obtain\n\\[\n\\left| retreatvalue^{(softconst+1)/softconst} - stagnantvalue^{(softconst+1)/softconst} - \\frac{softconst+1}{softconst} \\right|\n\\leq \\frac{softconst+1}{2softconst^2} stagnantvalue^{-(softconst+1)/softconst}.\n\\]\nIn particular,\n\\[\n\\lim_{humongousindex \\to \\infty} retreatvalue^{(softconst+1)/softconst} - stagnantvalue^{(softconst+1)/softconst} = \\frac{softconst+1}{softconst}.\n\\]\n\nIn general, if $fixedpointset$ is a sequence with $\\lim_{humongousindex \\to \\infty} fixedpointset = variability$, then also\n\\[\n\\lim_{humongousindex \\to \\infty} \\frac{1}{humongousindex} \\sum_{infinityidx=1}^{humongousindex} fixedpointset = variability\n\\]\nby Cesaro's lemma. Explicitly, for any $\\epsilon > 0$, we can find $tinycutoff$ such that\n$|fixedpointset - variability| \\leq \\epsilon/2$ for $humongousindex \\geq tinycutoff$, and then\n\\[\n\\left| variability - \\frac{1}{humongousindex} \\sum_{infinityidx=1}^{humongousindex} fixedpointset \\right|\n\\leq \\frac{humongousindex-tinycutoff}{humongousindex} \\frac{\\epsilon}{2} + \\frac{tinycutoff}{humongousindex} \\left| \\sum_{infinityidx=1}^{tinycutoff} (variability-fixedpointset) \\right|;\n\\]\nfor $humongousindex$ large, the right side is smaller than $\\epsilon$.\n\nIn our case, we deduce that\n\\[\n\\lim_{humongousindex \\to \\infty} \\frac{stagnantvalue^{(softconst+1)/softconst}}{humongousindex} = \\frac{softconst+1}{softconst}\n\\]\nand so\n\\[\n\\lim_{humongousindex \\to \\infty} \\frac{stagnantvalue^{softconst+1}}{humongousindex^{softconst}} = \\left(\\frac{softconst+1}{softconst} \\right)^{softconst},\n\\]\nas desired.\n\n\\textbf{Remark:}\nThe use of Cesaro's lemma above is the special case $quiescentbase = humongousindex$\nof the \\emph{Cesaro-Stolz theorem}: if $stagnantvalue,quiescentbase$ are sequences such that $quiescentbase$ is positive,\nstrictly increasing, and unbounded, and\n\\[\n\\lim_{humongousindex \\to \\infty} \\frac{retreatvalue - stagnantvalue}{quiescentbasenext - quiescentbase} = movingvalue,\n\\]\nthen\n\\[\n\\lim_{humongousindex \\to \\infty} \\frac{stagnantvalue}{quiescentbase} = movingvalue.\n\\]\n\n\\textbf{Second solution:}\nIn this solution, rather than applying Taylor's theorem with remainder\nto $(1+constantalpha)^{meekexponent}$ for $1 < meekexponent < 2$ and $constantalpha > 0$, we only apply convexity to deduce\nthat $(1+constantalpha)^{meekexponent} \\geq 1 + meekexponent constantalpha$. This gives\n\\[\nretreatvalue^{(softconst+1)/softconst} - stagnantvalue^{(softconst+1)/softconst} \\geq \\frac{softconst+1}{softconst},\n\\]\nand so\n\\[\nstagnantvalue^{(softconst+1)/softconst} \\geq \\frac{softconst+1}{softconst} humongousindex + variability\n\\]\nfor some $variability \\in \\RR$. In particular,\n\\[\n\\liminf_{humongousindex \\to \\infty} \\frac{stagnantvalue^{(softconst+1)/softconst}}{humongousindex} \\geq \\frac{softconst+1}{softconst}\n\\]\nand so\n\\[\n\\liminf_{humongousindex \\to \\infty} \\frac{stagnantvalue}{humongousindex^{softconst/(softconst+1)}} \\geq \\left(\\frac{softconst+1}{softconst} \\right)^{softconst/(softconst+1)}.\n\\]\nBut turning this around, the fact that\n\\begin{align*}\n&retreatvalue - stagnantvalue \\\\\n&= stagnantvalue^{-1/softconst} \\\\\n&\\leq \\left(\\frac{softconst+1}{softconst} \\right)^{-1/(softconst+1)} humongousindex^{-1/(softconst+1)}\n(1 + o(1)),\n\\end{align*}\nwhere $o(1)$ denotes a function tending to 0 as $humongousindex \\to \\infty$,\nyields\n\\begin{align*}\n&stagnantvalue \\\\\n&\\leq\n\\left(\\frac{softconst+1}{softconst} \\right)^{-1/(softconst+1)} \\sum_{infinityidx=1}^{humongousindex} infinityidx^{-1/(softconst+1)} (1 + o(1)) \\\\\n&= \\frac{softconst+1}{softconst} \\left(\\frac{softconst+1}{softconst} \\right)^{-1/(softconst+1)} humongousindex^{softconst/(softconst+1)}(1 + o(1)) \\\\\n&= \\left( \\frac{softconst+1}{softconst} \\right)^{softconst/(softconst+1)} humongousindex^{softconst/(softconst+1)}(1 + o(1)),\n\\end{align*}\nso\n\\[\n\\limsup_{humongousindex \\to \\infty} \\frac{stagnantvalue}{humongousindex^{softconst/(softconst+1)}} \\leq \\left( \\frac{softconst+1}{softconst}\n\\right)^{softconst/(softconst+1)}\n\\]\nand this completes the proof.\n\n\\textbf{Third solution:}\nWe argue that $stagnantvalue \\to \\infty$ as in the first solution.\nWrite $quiescentbase = stagnantvalue - movingvalue humongousindex^{softconst/(softconst+1)}$, for a value of $movingvalue$ to be determined\nlater.\nWe have\n\\begin{align*}\n&quiescentbasenext \\\\\n &= quiescentbase + stagnantvalue^{-1/softconst} - movingvalue ((humongousindex+1)^{softconst/(softconst+1)} - humongousindex^{softconst/(softconst+1)}) \\\\\n&= calmerrorone + calmerrortwo,\n\\end{align*}\nwhere\n\\begin{align*}\ncalmerrorone &= quiescentbase + stagnantvalue^{-1/softconst} - movingvalue^{-1/softconst} humongousindex^{-1/(softconst+1)} \\\\\ncalmerrortwo &= movingvalue ((humongousindex+1)^{softconst/(softconst+1)} - humongousindex^{softconst/(softconst+1)}) \\\\\n&\\quad - movingvalue^{-1/softconst} humongousindex^{-1/(softconst+1)}.\n\\end{align*}\nWe first estimate $calmerrorone$.\nFor $-1 < meekexponent < 0$, by the convexity of $(1+constantalpha)^{meekexponent}$\nand $(1+constantalpha)^{1-meekexponent}$, we have\n\\begin{align*}\n1 + meekexponent constantalpha &\\leq (1+constantalpha)^{meekexponent} \\\\\n&\\leq 1 + meekexponent constantalpha (1+constantalpha)^{meekexponent-1}.\n\\end{align*}\nHence\n\\begin{align*}\n-\\frac{1}{softconst} movingvalue^{-(softconst+1)/softconst} humongousindex^{-1} quiescentbase &\\leq calmerrorone - quiescentbase \\\\\n&\\leq\n-\\frac{1}{softconst} quiescentbase stagnantvalue^{-(softconst+1)/softconst}.\n\\end{align*}\nNote that both bounds have sign opposite to $quiescentbase$; moreover,\nby the bound $stagnantvalue = \\Omega(humongousindex^{(softconst-1)/softconst})$, both bounds have absolute\nvalue strictly less than that of $quiescentbase$ for $humongousindex$ sufficiently large. Consequently,\nfor $humongousindex$ large,\n\\[\n|calmerrorone| \\leq |quiescentbase|.\n\\]\n\nWe now work on $calmerrortwo$.\nBy Taylor's theorem\nwith remainder applied to $(1+constantalpha)^{meekexponent}$ for $constantalpha > 0$ and $0 < meekexponent < 1$,\n\\begin{align*}\n1+meekexponent constantalpha &\\geq (1+constantalpha)^{meekexponent} \\\\\n&\\geq 1 + meekexponent constantalpha + \\frac{meekexponent(meekexponent-1)}{2} constantalpha^2.\n\\end{align*}\nThe ``main term'' of $movingvalue ((humongousindex+1)^{softconst/(softconst+1)} - humongousindex^{softconst/(softconst+1)})$\nis $movingvalue \\frac{softconst}{softconst+1} humongousindex^{-1/(softconst+1)}$. To make this coincide with\n$movingvalue^{-1/softconst} humongousindex^{-1/(softconst+1)}$, we take\n\\[\nmovingvalue = \\left( \\frac{softconst+1}{softconst} \\right)^{softconst/(softconst+1)}.\n\\]\nWe then find that\n\\[\n|calmerrortwo| = O(humongousindex^{-2}),\n\\]\nand because $quiescentbasenext = calmerrorone + calmerrortwo$, we have\n$|quiescentbasenext| \\leq |quiescentbase| + |calmerrortwo|$. Hence\n\\[\n|quiescentbase| = O\\left (\\sum_{infinityidx=1}^{humongousindex} infinityidx^{-2} \\right) = O(1),\n\\]\nand so\n\\[\n\\lim_{humongousindex \\to \\infty} \\frac{stagnantvalue^{softconst+1}}{humongousindex^{softconst}} = movingvalue^{softconst+1} = \\left( \\frac{softconst+1}{softconst} \\right)^{softconst}.\n\\]\n\n\\textbf{Remark:}\nThe case $softconst=2$ appeared on the 2004 Romanian Olympiad (district level).\n\n\\textbf{Remark:}\nOne can make a similar argument for any sequence given by\n$retreatvalue = stagnantvalue + rigidmap(stagnantvalue)$, when $rigidmap$ is a \\emph{decreasing} function.\n\n\\textbf{Remark:}\nRichard Stanley suggests a heuristic for determining the asymptotic\nbehavior of sequences of this type: replace the given recursion\n\\[\nretreatvalue - stagnantvalue = stagnantvalue^{-1/softconst}\n\\]\nby the differential equation\n\\[\nsteadyvar' = steadyvar^{-1/softconst}\n\\]\nand determine the asymptotics of the latter."
    },
    "garbled_string": {
      "map": {
        "n": "qzxwvtnp",
        "a_n": "hjgrksla",
        "a_n+1": "mldvpsqe",
        "b_n": "cyrthnma",
        "b_n+1": "zvfklgeo",
        "x_n": "wbrtcoai",
        "x": "pdnyslqe",
        "i": "nmxgrtav",
        "f": "qlzbwcks",
        "g": "tpsrdqva",
        "y": "kcfpldgw",
        "e_1": "rszvwkue",
        "e_2": "vhqslmrb",
        "a_i": "fdwoxlei",
        "k": "uhljvcos",
        "a_0": "jbkthqew",
        "c": "rgmqnshe",
        "m": "hdnrplsi",
        "N": "vqslbmtp",
        "L": "szdkqjow"
      },
      "question": "Let $uhljvcos$ be an integer greater than 1. Suppose $jbkthqew > 0$, and define\n\\[\nmldvpsqe = hjgrksla + \\frac{1}{\\sqrt[uhljvcos]{hjgrksla}}\n\\]\nfor $qzxwvtnp > 0$. Evaluate\n\\[\n\\lim_{qzxwvtnp \\to \\infty} \\frac{hjgrksla^{uhljvcos+1}}{qzxwvtnp^{uhljvcos}}.\n\\]\n\\end{itemize}\n\n\\end{document}",
      "solution": "\\textbf{First solution:}\nWe start with some easy\nupper and lower bounds on $hjgrksla$.\nWe write $O(qlzbwcks(qzxwvtnp))$ and $\\Omega(qlzbwcks(qzxwvtnp))$ for functions $tpsrdqva(qzxwvtnp)$ such that\n$qlzbwcks(qzxwvtnp)/tpsrdqva(qzxwvtnp)$ and $tpsrdqva(qzxwvtnp)/qlzbwcks(qzxwvtnp)$, respectively, are bounded above.\nSince $hjgrksla$ is a nondecreasing sequence, $mldvpsqe-hjgrksla$ is bounded above,\nso $hjgrksla = O(qzxwvtnp)$. That means $hjgrksla^{-1/uhljvcos} = \\Omega(qzxwvtnp^{-1/uhljvcos})$, so\n\\[\nhjgrksla = \\Omega \\left( \\sum_{nmxgrtav=1}^{qzxwvtnp} nmxgrtav^{-1/uhljvcos} \\right)\n= \\Omega(qzxwvtnp^{(uhljvcos-1)/uhljvcos}).\n\\]\nIn fact, all we will need is that $hjgrksla \\to \\infty$ as $qzxwvtnp \\to \\infty$.\n\nBy Taylor's theorem with remainder, for $1 < hdnrplsi < 2$ and $pdnyslqe>0$,\n\\[\n|(1+pdnyslqe)^{hdnrplsi} - 1 - hdnrplsi\\,pdnyslqe| \\leq \\frac{hdnrplsi(hdnrplsi-1)}{2}pdnyslqe^2.\n\\]\nTaking $hdnrplsi = (uhljvcos+1)/uhljvcos$ and $pdnyslqe = mldvpsqe/hjgrksla = 1 + hjgrksla^{-(uhljvcos+1)/uhljvcos}$, we obtain\n\\[\n\\left| mldvpsqe^{(uhljvcos+1)/uhljvcos} - hjgrksla^{(uhljvcos+1)/uhljvcos} - \\frac{uhljvcos+1}{uhljvcos} \\right|\n\\leq \\frac{uhljvcos+1}{2\\,uhljvcos^2} \\, hjgrksla^{-(uhljvcos+1)/uhljvcos}.\n\\]\nIn particular,\n\\[\n\\lim_{qzxwvtnp \\to \\infty} mldvpsqe^{(uhljvcos+1)/uhljvcos} - hjgrksla^{(uhljvcos+1)/uhljvcos} = \\frac{uhljvcos+1}{uhljvcos}.\n\\]\n\nIn general, if $wbrtcoai$ is a sequence with $\\lim_{qzxwvtnp \\to \\infty} wbrtcoai = rgmqnshe$, then also\n\\[\n\\lim_{qzxwvtnp \\to \\infty} \\frac{1}{qzxwvtnp} \\sum_{nmxgrtav=1}^{qzxwvtnp} wbrtcoai = rgmqnshe\n\\]\nby Cesaro's lemma. Explicitly, for any $\\epsilon > 0$, we can find $vqslbmtp$ such that\n$|wbrtcoai - rgmqnshe| \\leq \\epsilon/2$ for $qzxwvtnp \\geq vqslbmtp$, and then\n\\[\n\\left| rgmqnshe - \\frac{1}{qzxwvtnp} \\sum_{nmxgrtav=1}^{qzxwvtnp} x_{nmxgrtav} \\right|\n\\leq \\frac{qzxwvtnp-vqslbmtp}{qzxwvtnp} \\frac{\\epsilon}{2} + \\frac{vqslbmtp}{qzxwvtnp} \\left| \\sum_{nmxgrtav=1}^{vqslbmtp} (rgmqnshe-x_{nmxgrtav}) \\right|;\n\\]\nfor $qzxwvtnp$ large, the right side is smaller than $\\epsilon$.\n\nIn our case, we deduce that\n\\[\n\\lim_{qzxwvtnp \\to \\infty} \\frac{hjgrksla^{(uhljvcos+1)/uhljvcos}}{qzxwvtnp} = \\frac{uhljvcos+1}{uhljvcos}\n\\]\nand so\n\\[\n\\lim_{qzxwvtnp \\to \\infty} \\frac{hjgrksla^{uhljvcos+1}}{qzxwvtnp^{uhljvcos}} = \\left(\\frac{uhljvcos+1}{uhljvcos} \\right)^{uhljvcos},\n\\]\nas desired.\n\n\\textbf{Remark:}\nThe use of Cesaro's lemma above is the special case $cyrthnma = qzxwvtnp$\nof the \\emph{Cesaro-Stolz\ntheorem}: if $hjgrksla,cyrthnma$ are sequences such that $cyrthnma$ is positive,\nstrictly increasing, and unbounded, and\n\\[\n\\lim_{qzxwvtnp \\to \\infty} \\frac{mldvpsqe - hjgrksla}{zvfklgeo - cyrthnma} = szdkqjow,\n\\]\nthen\n\\[\n\\lim_{qzxwvtnp \\to \\infty} \\frac{hjgrksla}{cyrthnma} = szdkqjow.\n\\]\n\n\\textbf{Second solution:}\nIn this solution, rather than applying Taylor's theorem with remainder\nto $(1+pdnyslqe)^{hdnrplsi}$ for $1 < hdnrplsi < 2$ and $pdnyslqe > 0$, we only apply convexity to deduce\nthat $(1+pdnyslqe)^{hdnrplsi} \\geq 1 + hdnrplsi pdnyslqe$. This gives\n\\[\nmldvpsqe^{(uhljvcos+1)/uhljvcos} - hjgrksla^{(uhljvcos+1)/uhljvcos} \\geq \\frac{uhljvcos+1}{uhljvcos},\n\\]\nand so\n\\[\nhjgrksla^{(uhljvcos+1)/uhljvcos} \\geq \\frac{uhljvcos+1}{uhljvcos} \\, qzxwvtnp + rgmqnshe\n\\]\nfor some $rgmqnshe \\in \\RR$. In particular,\n\\[\n\\liminf_{qzxwvtnp \\to \\infty} \\frac{hjgrksla^{(uhljvcos+1)/uhljvcos}}{qzxwvtnp} \\geq \\frac{uhljvcos+1}{uhljvcos}\n\\]\nand so\n\\[\n\\liminf_{qzxwvtnp \\to \\infty} \\frac{hjgrksla}{qzxwvtnp^{uhljvcos/(uhljvcos+1)}} \\geq \\left(\\frac{uhljvcos+1}{uhljvcos} \\right)^{uhljvcos/(uhljvcos+1)}.\n\\]\nBut turning this around, the fact that\n\\begin{align*}\n& mldvpsqe - hjgrksla \\\\\n&= hjgrksla^{-1/uhljvcos} \\\\\n&\\leq \\left(\\frac{uhljvcos+1}{uhljvcos} \\right)^{-1/(uhljvcos+1)} qzxwvtnp^{-1/(uhljvcos+1)}\n(1 + o(1)),\n\\end{align*}\nwhere $o(1)$ denotes a function tending to 0 as $qzxwvtnp \\to \\infty$,\nyields\n\\begin{align*}\n& hjgrksla \\\\\n&\\leq\n\\left(\\frac{uhljvcos+1}{uhljvcos} \\right)^{-1/(uhljvcos+1)} \\sum_{nmxgrtav=1}^{qzxwvtnp} nmxgrtav^{-1/(uhljvcos+1)} (1 + o(1)) \\\\\n&= \\frac{uhljvcos+1}{uhljvcos} \\left(\\frac{uhljvcos+1}{uhljvcos} \\right)^{-1/(uhljvcos+1)} qzxwvtnp^{uhljvcos/(uhljvcos+1)}(1 + o(1)) \\\\\n&= \\left( \\frac{uhljvcos+1}{uhljvcos} \\right)^{uhljvcos/(uhljvcos+1)} qzxwvtnp^{uhljvcos/(uhljvcos+1)}(1 + o(1)),\n\\end{align*}\nso\n\\[\n\\limsup_{qzxwvtnp \\to \\infty} \\frac{hjgrksla}{qzxwvtnp^{uhljvcos/(uhljvcos+1)}} \\leq \\left( \\frac{uhljvcos+1}{uhljvcos}\n\\right)^{uhljvcos/(uhljvcos+1)}\n\\]\nand this completes the proof.\n\n\\textbf{Third solution:}\nWe argue that $hjgrksla \\to \\infty$ as in the first solution.\nWrite $cyrthnma = hjgrksla - szdkqjow \\, qzxwvtnp^{uhljvcos/(uhljvcos+1)}$, for a value of $szdkqjow$ to be determined\nlater.\nWe have\n\\begin{align*}\n& zvfklgeo \\\\\n &= cyrthnma + hjgrksla^{-1/uhljvcos} - szdkqjow \\left((qzxwvtnp+1)^{uhljvcos/(uhljvcos+1)} - qzxwvtnp^{uhljvcos/(uhljvcos+1)}\\right) \\\\\n&= rszvwkue + vhqslmrb,\n\\end{align*}\nwhere\n\\begin{align*}\nrszvwkue &= cyrthnma + hjgrksla^{-1/uhljvcos} - szdkqjow^{-1/uhljvcos} qzxwvtnp^{-1/(uhljvcos+1)} \\\\\nvhqslmrb &= szdkqjow \\left((qzxwvtnp+1)^{uhljvcos/(uhljvcos+1)} - qzxwvtnp^{uhljvcos/(uhljvcos+1)}\\right) \\\\\n&\\quad - szdkqjow^{-1/uhljvcos} qzxwvtnp^{-1/(uhljvcos+1)}.\n\\end{align*}\nWe first estimate $rszvwkue$.\nFor $-1 < hdnrplsi < 0$, by the convexity of $(1+pdnyslqe)^{hdnrplsi}$\nand $(1+pdnyslqe)^{1-hdnrplsi}$, we have\n\\begin{align*}\n1 + hdnrplsi pdnyslqe &\\leq (1+pdnyslqe)^{hdnrplsi} \\\\\n&\\leq 1 + hdnrplsi pdnyslqe (1+pdnyslqe)^{hdnrplsi-1}.\n\\end{align*}\nHence\n\\begin{align*}\n-\\frac{1}{uhljvcos} szdkqjow^{-(uhljvcos+1)/uhljvcos} qzxwvtnp^{-1} cyrthnma &\\leq rszvwkue - cyrthnma \\\\\n&\\leq\n-\\frac{1}{uhljvcos} cyrthnma \\, hjgrksla^{-(uhljvcos+1)/uhljvcos}.\n\\end{align*}\nNote that both bounds have sign opposite to $cyrthnma$; moreover,\nby the bound $hjgrksla = \\Omega(qzxwvtnp^{(uhljvcos-1)/uhljvcos})$, both bounds have absolute\nvalue strictly less than that of $cyrthnma$ for $qzxwvtnp$ sufficiently large. Consequently,\nfor $qzxwvtnp$ large,\n\\[\n|rszvwkue| \\leq |cyrthnma|.\n\\]\nWe now work on $vhqslmrb$.\nBy Taylor's theorem\nwith remainder applied to $(1+pdnyslqe)^{hdnrplsi}$ for $pdnyslqe > 0$ and $0 < hdnrplsi < 1$,\n\\begin{align*}\n1+hdnrplsi pdnyslqe &\\geq (1+pdnyslqe)^{hdnrplsi} \\\\\n&\\geq 1 + hdnrplsi pdnyslqe + \\frac{hdnrplsi(hdnrplsi-1)}{2} pdnyslqe^2.\n\\end{align*}\nThe ``main term'' of $szdkqjow ((qzxwvtnp+1)^{uhljvcos/(uhljvcos+1)} - qzxwvtnp^{uhljvcos/(uhljvcos+1)})$\nis $szdkqjow \\frac{uhljvcos}{uhljvcos+1} qzxwvtnp^{-1/(uhljvcos+1)}$. To make this coincide with\n$szdkqjow^{-1/uhljvcos} qzxwvtnp^{-1/(uhljvcos+1)}$, we take\n\\[\nszdkqjow = \\left( \\frac{uhljvcos+1}{uhljvcos} \\right)^{uhljvcos/(uhljvcos+1)}.\n\\]\nWe then find that\n\\[\n|vhqslmrb| = O(qzxwvtnp^{-2}),\n\\]\nand because $zvfklgeo = rszvwkue + vhqslmrb$, we have\n$|zvfklgeo| \\leq |cyrthnma| + |vhqslmrb|$. Hence\n\\[\n|cyrthnma| = O\\left (\\sum_{nmxgrtav=1}^{qzxwvtnp} nmxgrtav^{-2} \\right) = O(1),\n\\]\nand so\n\\[\n\\lim_{qzxwvtnp \\to \\infty} \\frac{hjgrksla^{uhljvcos+1}}{qzxwvtnp^{uhljvcos}} = szdkqjow^{uhljvcos+1} = \\left( \\frac{uhljvcos+1}{uhljvcos} \\right)^{uhljvcos}.\n\\]\n\n\\textbf{Remark:}\nThe case $uhljvcos=2$ appeared on the 2004 Romanian Olympiad (district level).\n\n\\textbf{Remark:}\nOne can make a similar argument for any sequence given by\n$mldvpsqe = hjgrksla + qlzbwcks(hjgrksla)$, when $qlzbwcks$ is a \\emph{decreasing} function.\n\n\\textbf{Remark:}\nRichard Stanley suggests a heuristic for determining the asymptotic\nbehavior of sequences of this type: replace the given recursion\n\\[\nmldvpsqe - hjgrksla = hjgrksla^{-1/uhljvcos}\n\\]\nby the differential equation\n\\[\nkcfpldgw' = kcfpldgw^{-1/uhljvcos}\n\\]\nand determine the asymptotics of the latter.\n\\end{itemize}\n\n\\end{document}"
    },
    "kernel_variant": {
      "question": "────────────────────────────────────────────────────\nFix two real parameters α > 1 and c > 0.  For every real β > 0 start with  \nx₀ = 3 and define the sequence (xₙ)ₙ≥0 recursively by  \n\n  x_{n+1} = x_n + c·x_n^{-1/α} + n^{-β}  (n ≥ 0).       (★)\n\na)  Prove that (xₙ) is strictly increasing and diverges to +∞.  \n\nb)  Evaluate  \n\n  L(β) = lim_{n→∞} x_n^{α+1}/n^{α},                        (♢)\n\ndetermine for which β the limit is finite and positive, and describe\nits behaviour in the remaining cases.",
      "solution": "────────────────────────────────────────────────────\nThroughout we abbreviate  \n\n m := (α+1)/α  (so 1 < m < 2).\n\nStep 1 – Monotonicity and divergence.  \nSince both summands in (★) are positive, x_{n+1} > x_n for every n, so\n(xₙ) is increasing.  If it were bounded, its increments\nc·x_n^{-1/α} + n^{-β} would eventually be bounded below by a fixed\npositive number, contradicting boundedness.  Hence xₙ → +∞.\n\nStep 2 – A rough growth window.  \nBecause x_{n+1} – x_n ≥ n^{-β}, we have the lower bound  \n\n x_n ≥ Σ_{k=0}^{n-1} k^{-β} = Ω(n^{1-β})           (1)\n\n(when β ≠ 1 we use the integral test; the logarithmic case β = 1 is\neven larger than n^{1-β} and causes no difficulty).  On the other hand,\n(★) gives x_{n+1} – x_n ≤ c x_n^{-1/α}, so telescoping and comparing\nwith the differential equation y′ = c·y^{-1/α} yield  \n\n x_n = O(n^{α/(α+1)}).                              (2)\n\nConsequently xₙ^{-1/α} = O(n^{-1/(α+1)}).\n\nStep 3 – A first–order expansion for the main term.  \nPut aₙ := x_n^m.  Because m > 1 we may apply Taylor with remainder to  \n\n x_{n+1}^m = (x_n+Δ_n)^m, Δ_n := c·x_n^{-1/α}+n^{-β}.     (3)\n\nWriting y_n := Δ_n/x_n we have |y_n| ≤ constant·n^{-1-ε} for some ε>0\nby (2), so  \n\n |(1+y_n)^m – 1 – m y_n| ≤ C y_n².                 (4)\n\nMultiplying (4) by x_n^m and using (3) gives  \n\n |a_{n+1} – a_n – m x_n^{m-1}Δ_n| ≤ C′x_n^{m-2}Δ_n² = o(1). (5)\n\nHence  \n\n a_{n+1} – a_n = m x_n^{m-1}Δ_n + o(1).             (6)\n\nBut x_n^{m-1} = x_n^{(α+1)/α – 1} = x_n^{-1/α}, so\n\n a_{n+1} – a_n = m[ c·x_n^{-1/α}+n^{-β} ]·x_n^{-1/α} + o(1)\n     = m c + m x_n^{-1}n^{-β} + o(1).              (7)\n\nRecall x_n ≈ n^{α/(α+1)}, whence x_n^{-1}n^{-β} behaves like\nn^{1-(β+α/(α+1))}.  Three different regimes appear.\n\n—————————————————————————————\nREGIME I:  β > 1/(α+1).  \n—————————————————————————————\nThen the exponent 1 – (β+α/(α+1)) is negative, so the middle term in (7)\ntends to 0.  Thus  \n\n a_{n+1} – a_n → m c.\n\nSince bₙ := n is strictly increasing and unbounded, the Cesàro–Stolz\ntheorem applies:  \n\n lim_{n→∞} a_n/n = m c.                             (8)\n\nBecause a_n = x_n^m, dividing by n and raising to the α-th power gives  \n\n L(β) = ( m c )^{α} = [ c(α+1)/α ]^{α}.            (9)\n\nThe limit is finite and positive.\n\n—————————————————————————————\nREGIME II:  β = 1/(α+1).  \n—————————————————————————————\nNow the exponent mentioned above equals 0, so the term\nm x_n^{-1}n^{-β} in (7) tends to m K^m, where\n\n K^m := lim_{n→∞} x_n^m/n                       (10)\n\n(if the limit exists; we shall see that it does).\nTaking limits in (7) therefore yields\n\n lim_{n→∞}(a_{n+1}−a_n) = m c + m K^{m}.           (11)\n\nUsing Cesàro–Stolz once more gives\n\n K^{m}=lim_{n→∞}a_n/n = m c+m K^{m}.               (12)\n\nRearranging, (1 – m)K^m = m c, i.e.\n\n K^{m} = (m c)/(m−1) = [(α+1)/α] c /(1/α) = (α+1)c. (13)\n\nHence K = [ (α+1)c ]^{α/(α+1)} and therefore\n\n L(β)=K^{α+1} = [ (α+1)c ]^{α}.     (Limit finite, new constant.)  (14)\n\n—————————————————————————————\nREGIME III:  0 < β < 1/(α+1).  \n—————————————————————————————\nHere the exponent 1 – (β+α/(α+1)) in (7) is positive, so\na_{n+1} – a_n → ∞.  Consequently a_n/n → ∞ and\n\n L(β)=∞.                                          (15)\n\nStep 4 – Collecting the three cases.  \nPutting the conclusions (9), (14) and (15) together we obtain\n\n • If β > 1/(α+1): L(β) = [ c(α+1)/α ]^{α};  \n • If β = 1/(α+1): L(β) = [ (α+1)c ]^{α};  \n • If 0 < β < 1/(α+1): L(β) = +∞.\n\n(When β = 0 the sequence grows faster still, so the last line remains\nvalid.)\n\nThis completes the required analysis. □",
      "_replacement_note": {
        "replaced_at": "2025-07-05T22:17:12.029053",
        "reason": "Original kernel variant was too easy compared to the original problem"
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}