{
  "index": "2002-B-3",
  "type": "ANA",
  "tag": [
    "ANA",
    "NT"
  ],
  "difficulty": "",
  "question": "Show that, for all integers $n > 1$,\n\\[\n\\frac{1}{2ne} < \\frac{1}{e} - \\left( 1 - \\frac{1}{n} \\right)^n\n< \\frac{1}{ne}.\n\\]",
  "solution": "The desired inequalities can be rewritten as\n\\[\n1 - \\frac{1}{n} < \\exp\\left( 1 + n \\log \\left( 1 - \\frac{1}{n} \\right)\n\\right) < 1 - \\frac{1}{2n}.\n\\]\nBy taking logarithms, we can rewrite the desired inequalities as\n\\begin{align*}\n-\\log \\left( 1 - \\frac{1}{2n} \\right)\n&< -1 - n \\log \\left( 1 - \\frac{1}{n} \\right) \\\\\n&< -\\log \\left( 1 - \\frac{1}{n} \\right).\n\\end{align*}\nRewriting these in terms of the Taylor expansion of\n$-\\log(1-x)$, we see that the desired result is also equivalent\nto\n\\[\n\\sum_{i=1}^\\infty \\frac{1}{i 2^i n^i}\n< \\sum_{i=1}^\\infty \\frac{1}{(i+1) n^i}\n< \\sum_{i=1}^\\infty \\frac{1}{i n^i},\n\\]\nwhich is evident because the inequalities hold term by term.\n\nNote: David Savitt points out that the upper bound can be improved from\n$1/(ne)$ to $2/(3ne)$ with a slightly more complicated argument. (In\nfact, for any $c>1/2$, one has an upper bound of $c/(ne)$, but only\nfor $n$ above a certain bound depending on $c$.)",
  "vars": [
    "n",
    "x",
    "i"
  ],
  "params": [
    "c"
  ],
  "sci_consts": [
    "e"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "n": "intcount",
        "x": "placeholder",
        "i": "counter",
        "c": "upperconst"
      },
      "question": "Show that, for all integers $intcount > 1$,\n\\[\n\\frac{1}{2 intcount e} < \\frac{1}{e} - \\left( 1 - \\frac{1}{intcount} \\right)^{intcount}\n< \\frac{1}{intcount e}.\n\\]",
      "solution": "The desired inequalities can be rewritten as\n\\[\n1 - \\frac{1}{intcount} < \\exp\\left( 1 + intcount \\log \\left( 1 - \\frac{1}{intcount} \\right)\n\\right) < 1 - \\frac{1}{2 intcount}.\n\\]\nBy taking logarithms, we can rewrite the desired inequalities as\n\\begin{align*}\n-\\log \\left( 1 - \\frac{1}{2 intcount} \\right)\n&< -1 - intcount \\log \\left( 1 - \\frac{1}{intcount} \\right) \\\\\n&< -\\log \\left( 1 - \\frac{1}{intcount} \\right).\n\\end{align*}\nRewriting these in terms of the Taylor expansion of\n$-\\log(1-placeholder)$, we see that the desired result is also equivalent\nto\n\\[\n\\sum_{counter=1}^\\infty \\frac{1}{counter 2^{counter} intcount^{counter}}\n< \\sum_{counter=1}^\\infty \\frac{1}{(counter+1) intcount^{counter}}\n< \\sum_{counter=1}^\\infty \\frac{1}{counter intcount^{counter}},\n\\]\nwhich is evident because the inequalities hold term by term.\n\nNote: David Savitt points out that the upper bound can be improved from\n$1/(intcount e)$ to $2/(3 intcount e)$ with a slightly more complicated argument. (In\nfact, for any $upperconst>1/2$, one has an upper bound of $upperconst/(intcount e)$, but only\nfor intcount above a certain bound depending on $upperconst$.)"
    },
    "descriptive_long_confusing": {
      "map": {
        "n": "pineapple",
        "x": "marigold",
        "i": "tangerine",
        "c": "watermelon"
      },
      "question": "Show that, for all integers $pineapple > 1$,\n\\[\n\\frac{1}{2pineapple e} < \\frac{1}{e} - \\left( 1 - \\frac{1}{pineapple} \\right)^{pineapple}\n< \\frac{1}{pineapple e}.\n\\]",
      "solution": "The desired inequalities can be rewritten as\n\\[\n1 - \\frac{1}{pineapple} < \\exp\\left( 1 + pineapple \\log \\left( 1 - \\frac{1}{pineapple} \\right)\n\\right) < 1 - \\frac{1}{2pineapple}.\n\\]\nBy taking logarithms, we can rewrite the desired inequalities as\n\\begin{align*}\n-\\log \\left( 1 - \\frac{1}{2pineapple} \\right)\n&< -1 - pineapple \\log \\left( 1 - \\frac{1}{pineapple} \\right) \\\\\n&< -\\log \\left( 1 - \\frac{1}{pineapple} \\right).\n\\end{align*}\nRewriting these in terms of the Taylor expansion of\n$-\\log(1-marigold)$, we see that the desired result is also equivalent\nto\n\\[\n\\sum_{tangerine=1}^\\infty \\frac{1}{tangerine 2^{tangerine} pineapple^{tangerine}}\n< \\sum_{tangerine=1}^\\infty \\frac{1}{(tangerine+1) pineapple^{tangerine}}\n< \\sum_{tangerine=1}^\\infty \\frac{1}{tangerine pineapple^{tangerine}},\n\\]\nwhich is evident because the inequalities hold term by term.\n\nNote: David Savitt points out that the upper bound can be improved from\n$1/(pineapple e)$ to $2/(3pineapple e)$ with a slightly more complicated argument. (In\nfact, for any $watermelon>1/2$, one has an upper bound of $watermelon/(pineapple e)$, but only\nfor $pineapple$ above a certain bound depending on $watermelon$.)"
    },
    "descriptive_long_misleading": {
      "map": {
        "n": "continuous",
        "x": "constantval",
        "i": "unordered",
        "c": "variating"
      },
      "question": "Show that, for all integers $continuous > 1$,\n\\[\n\\frac{1}{2 continuous e} < \\frac{1}{e} - \\left( 1 - \\frac{1}{continuous} \\right)^{continuous}\n< \\frac{1}{continuous e}.\n\\]",
      "solution": "The desired inequalities can be rewritten as\n\\[\n1 - \\frac{1}{continuous} < \\exp\\left( 1 + continuous \\log \\left( 1 - \\frac{1}{continuous} \\right)\n\\right) < 1 - \\frac{1}{2 continuous}.\n\\]\nBy taking logarithms, we can rewrite the desired inequalities as\n\\begin{align*}\n-\\log \\left( 1 - \\frac{1}{2 continuous} \\right)\n&< -1 - continuous \\log \\left( 1 - \\frac{1}{continuous} \\right) \\\\\n&< -\\log \\left( 1 - \\frac{1}{continuous} \\right).\n\\end{align*}\nRewriting these in terms of the Taylor expansion of\n$-\\log(1-constantval)$, we see that the desired result is also equivalent\nto\n\\[\n\\sum_{unordered=1}^{\\infty} \\frac{1}{unordered 2^{unordered} continuous^{unordered}}\n< \\sum_{unordered=1}^{\\infty} \\frac{1}{(unordered+1) continuous^{unordered}}\n< \\sum_{unordered=1}^{\\infty} \\frac{1}{unordered continuous^{unordered}},\n\\]\nwhich is evident because the inequalities hold term by term.\n\nNote: David Savitt points out that the upper bound can be improved from\n$1/(continuous e)$ to $2/(3 continuous e)$ with a slightly more complicated argument. (In\nfact, for any $variating > 1/2$, one has an upper bound of $variating/(continuous e)$, but only\nfor $continuous$ above a certain bound depending on $variating$.)"
    },
    "garbled_string": {
      "map": {
        "n": "qzxwvtnp",
        "x": "hjgrksla",
        "i": "bmqslxtr",
        "c": "pvdhwkse"
      },
      "question": "Show that, for all integers $qzxwvtnp > 1$,\n\\[\n\\frac{1}{2 qzxwvtnp e} < \\frac{1}{e} - \\left( 1 - \\frac{1}{qzxwvtnp} \\right)^{qzxwvtnp}\n< \\frac{1}{qzxwvtnp e}.\n\\]",
      "solution": "The desired inequalities can be rewritten as\n\\[\n1 - \\frac{1}{qzxwvtnp} < \\exp\\left( 1 + qzxwvtnp \\log \\left( 1 - \\frac{1}{qzxwvtnp} \\right)\n\\right) < 1 - \\frac{1}{2 qzxwvtnp}.\n\\]\nBy taking logarithms, we can rewrite the desired inequalities as\n\\begin{align*}\n-\\log \\left( 1 - \\frac{1}{2 qzxwvtnp} \\right)\n&< -1 - qzxwvtnp \\log \\left( 1 - \\frac{1}{qzxwvtnp} \\right) \\\\\n&< -\\log \\left( 1 - \\frac{1}{qzxwvtnp} \\right).\n\\end{align*}\nRewriting these in terms of the Taylor expansion of $-\\log(1-hjgrksla)$, we see that the desired result is also equivalent to\n\\[\n\\sum_{bmqslxtr=1}^{\\infty} \\frac{1}{bmqslxtr 2^{bmqslxtr} qzxwvtnp^{bmqslxtr}}\n< \\sum_{bmqslxtr=1}^{\\infty} \\frac{1}{(bmqslxtr+1) qzxwvtnp^{bmqslxtr}}\n< \\sum_{bmqslxtr=1}^{\\infty} \\frac{1}{bmqslxtr qzxwvtnp^{bmqslxtr}},\n\\]\nwhich is evident because the inequalities hold term by term.\n\nNote: David Savitt points out that the upper bound can be improved from $1/(qzxwvtnp e)$ to $2/(3 qzxwvtnp e)$ with a slightly more complicated argument. (In fact, for any $pvdhwkse>1/2$, one has an upper bound of $pvdhwkse/(qzxwvtnp e)$, but only for $qzxwvtnp$ above a certain bound depending on $pvdhwkse$.)"
    },
    "kernel_variant": {
      "question": "Let $n$ be an integer with $n\\ge 10$ and let $m$ be another integer that satisfies  \n\\[\n1\\le m\\le\\frac{n}{3}.\n\\]\nDefine  \n\\[\nF_{n,m}:=\\prod_{k=1}^{m}\\Bigl(1-\\frac{k}{n}\\Bigr)^{\\,n-k},\n\\]\nand denote the power sums  \n\\[\nS_{2}:=\\sum_{k=1}^{m}k^{2}=\\frac{m(m+1)(2m+1)}{6},\\qquad  \nS_{3}:=\\sum_{k=1}^{m}k^{3}=\\frac{m^{2}(m+1)^{2}}{4},\\qquad  \nS_{4}:=\\sum_{k=1}^{m}k^{4}.\n\\]\n\n1. (Uniform second–order estimate – valid for every admissible $(n,m)$)  \n   Show that with the universal constant  \n   \\[\n   C_{0}:=\\frac{41}{72},\n   \\]\n   one has  \n   \\[\n   \\exp\\!\\Bigl[-\\tfrac12 m(m+1)+\\tfrac{S_{2}}{2n}-C_{0}\\tfrac{S_{3}}{n^{2}}\\Bigr]\n   \\;<\\;F_{n,m}\\;<\\;\n   \\exp\\!\\Bigl[-\\tfrac12 m(m+1)+\\tfrac{S_{2}}{2n}+C_{0}\\tfrac{S_{3}}{n^{2}}\\Bigr].\n   \\tag{$\\dag$}\n   \\]\n\n2. (Second–order error control in the sub–critical regime $m\\le n^{1/3}$)  \n   From now on assume in addition  \n   \\[\n   1\\le m\\le n^{1/3}\n   \\]\n   and put $\\psi:=e^{-m(m+1)/2}$.  Prove that, under this extra hypothesis,\n   \\[\n   \\bigl|\\,F_{n,m}-\\psi\\bigl(1+\\tfrac{S_{2}}{2n}\\bigr)\\bigr|\n   \\;\\le\\;\n   \\psi\\Bigl(\\tfrac{3}{4}\\tfrac{S_{2}^{2}}{n^{2}}+\\tfrac{77}{72}\\tfrac{S_{3}}{n^{2}}\\Bigr).\n   \\tag{$\\ddot$}\n   \\]\n\n   Equivalently,\n   \\[\n   \\psi\\Bigl(\\tfrac{S_{2}}{2n}-\\tfrac{3}{4}\\tfrac{S_{2}^{2}}{n^{2}}-\\tfrac{77}{72}\\tfrac{S_{3}}{n^{2}}\\Bigr)\n   \\;\\le\\;\n   F_{n,m}-\\psi\n   \\;\\le\\;\n   \\psi\\Bigl(\\tfrac{S_{2}}{2n}+\\tfrac{3}{4}\\tfrac{S_{2}^{2}}{n^{2}}+\\tfrac{77}{72}\\tfrac{S_{3}}{n^{2}}\\Bigr).\n   \\]\n\n3. (First–order sharpness)  \n   Let $(n,m)$ be any sequence with $n\\to\\infty$ that satisfies $1\\le m\\le n^{1/3}$ (for instance, $m$ fixed or $m=o(n^{1/3})$).  Show that  \n   \\[\n   \\lim_{n\\to\\infty} n\\bigl[F_{n,m}-e^{-m(m+1)/2}\\bigr]\n   \\;=\\;\\frac{S_{2}}{2}\\,e^{-m(m+1)/2}\n   \\;=\\;\\frac{m(m+1)(2m+1)}{6}\\,e^{-m(m+1)/2}.\n   \\tag{$\\clubsuit$}\n   \\]",
      "solution": "Throughout write  \n\\[\nx_{k}:=\\frac{k}{n}\\quad(0<x_{k}\\le \\tfrac13),\\qquad \nL_{n,m}:=\\log F_{n,m}=\\sum_{k=1}^{m}(n-k)\\log(1-x_{k}). \\tag{1}\n\\]\n\n{\\bf Step 1.  Truncating $\\log(1-x)$.}  \nFor $0\\le x\\le\\frac13$ the Taylor expansion with a positive remainder is  \n\\[\n\\log(1-x)=-x-\\frac{x^{2}}{2}-\\frac{x^{3}}{3}-R_{4}(x),\\qquad \nR_{4}(x):=\\sum_{j=4}^{\\infty}\\frac{x^{j}}{j}>0,\n\\]\nand  \n\\[\n0<R_{4}(x)\\le\\frac{x^{4}}{4(1-x)}\\le\\frac{3}{8}x^{4}.\n\\]\nInsert this expansion into (1) and decompose  \n\\[\nL_{n,m}=A_{1}+A_{2}+A_{3}+A_{4},\n\\]\nwith  \n\\[\n\\begin{aligned}\nA_{1}&:=-\\sum_{k=1}^{m}(n-k)x_{k}, &\\qquad  \nA_{2}&:=-\\frac12\\sum_{k=1}^{m}(n-k)x_{k}^{2},\\\\\nA_{3}&:=-\\frac13\\sum_{k=1}^{m}(n-k)x_{k}^{3}, &\\qquad  \nA_{4}&:=-\\sum_{k=1}^{m}(n-k)R_{4}(x_{k}).\n\\end{aligned}\n\\]\n\n{\\bf Step 2.  The first two sums.}  \nBecause $(n-k)x_{k}=k-k^{2}/n$ and $(n-k)x_{k}^{2}=k^{2}/n-k^{3}/n^{2}$,  \n\\[\nA_{1}=-\\sum_{k=1}^{m}k+\\frac1n\\sum_{k=1}^{m}k^{2}\n       =-\\frac12 m(m+1)+\\frac{S_{2}}{n},\\tag{2}\n\\]\n\\[\nA_{2}=-\\frac12\\Bigl(\\frac{S_{2}}{n}-\\frac{S_{3}}{n^{2}}\\Bigr)\n       =-\\frac{S_{2}}{2n}+\\frac{S_{3}}{2n^{2}}.\\tag{3}\n\\]\n\n{\\bf Step 3.  The cubic part.}  \nBecause $(n-k)x_{k}^{3}=k^{3}/n^{2}-k^{4}/n^{3}$,  \n\\[\nA_{3}=-\\frac13\\Bigl(\\frac{S_{3}}{n^{2}}-\\frac{S_{4}}{n^{3}}\\Bigr)\n      =-\\frac{S_{3}}{3n^{2}}+\\frac{S_{4}}{3n^{3}}.\\tag{4}\n\\]\n\n{\\bf Step 4.  Bounding $A_{4}$ and comparing with $A_{3}$.}  \nSince $R_{4}(x_{k})\\le\\frac38 x_{k}^{4}$,  \n\\[\n|A_{4}|\\le\\frac38\\sum_{k=1}^{m}(n-k)\\frac{k^{4}}{n^{4}}\n       \\le\\frac{3}{8n^{3}}S_{4}.\\tag{5}\n\\]\n\nTo relate $S_{4}$ and $S_{3}$ note that $k\\le m\\le n/3$ implies  \n\\[\nk^{4}\\le m\\,k^{3}\\quad\\Longrightarrow\\quad\nS_{4}=\\sum_{k=1}^{m}k^{4}\\le m\\sum_{k=1}^{m}k^{3}=m\\,S_{3}\n\\le\\frac{n}{3}\\,S_{3}.\\tag{6}\n\\]\nTherefore  \n\\[\n\\frac{S_{4}}{n^{3}}\\le\\frac{1}{3}\\frac{S_{3}}{n^{2}}. \\tag{7}\n\\]\n\nCombining (4)–(7) gives  \n\\[\n|A_{3}+A_{4}|\n\\;\\le\\;\\frac{S_{3}}{3n^{2}}+\\frac{17}{24}\\frac{S_{4}}{n^{3}}\n\\;\\le\\;\\Bigl(\\frac13+\\frac{17}{72}\\Bigr)\\frac{S_{3}}{n^{2}}\n\\;=\\;\\frac{41}{72}\\frac{S_{3}}{n^{2}}.\\tag{8}\n\\]\n\nSet $C_{0}:=\\tfrac{41}{72}$.\n\n{\\bf Step 5.  Collecting the pieces – proof of $(\\dag)$.}  \nFrom (2)–(4) we have  \n\\[\nL_{n,m}=-\\frac12 m(m+1)+\\frac{S_{2}}{2n}+\\frac{S_{3}}{2n^{2}}\n        +(A_{3}+A_{4}).\\tag{9}\n\\]\nUsing (8) yields  \n\\[\n-C_{0}\\frac{S_{3}}{n^{2}}\\;\\le\\;A_{3}+A_{4}\\;\\le\\;C_{0}\\frac{S_{3}}{n^{2}},\n\\]\nhence  \n\\[\n-\\frac12 m(m+1)+\\frac{S_{2}}{2n}-C_{0}\\frac{S_{3}}{n^{2}}\n< L_{n,m} <\n-\\frac12 m(m+1)+\\frac{S_{2}}{2n}+C_{0}\\frac{S_{3}}{n^{2}}.\n\\]\nExponentiating proves $(\\dag)$.\n\n{\\bf Step 6.  Refined decomposition (now assume $m\\le n^{1/3}$).}  \nPut  \n\\[\n\\psi:=e^{-m(m+1)/2},\\qquad \\tau:=\\frac{S_{2}}{2n},\\qquad\n\\rho:=\\frac{S_{3}}{2n^{2}}+A_{3}+A_{4}. \\tag{10}\n\\]\nBy (8) and the definition of $C_{0}$,\n\\[\n|\\rho|\\le\\Bigl(\\frac12+C_{0}\\Bigr)\\frac{S_{3}}{n^{2}}\n       =\\frac{77}{72}\\frac{S_{3}}{n^{2}}.\\tag{11}\n\\]\n\n{\\bf Step 7.  Explicit bounds for $\\tau$ and $\\rho$.}  \nBecause $n\\ge 10$ and $m\\le n^{1/3}$ one has  \n\\[\nn\\ge\\max\\{10,m^{3}\\}.\n\\]\nFix $m$ and take the smallest admissible $n$, namely  \n$n^{\\ast}:=\\max\\{10,m^{3}\\}$.  Because $\\tau$ is decreasing in $n$ we get  \n\\[\n\\tau\\le \\frac{S_{2}}{2n^{\\ast}}\n      =\\frac{m(m+1)(2m+1)}{12\\,n^{\\ast}}\n      \\le\\frac{m(m+1)(2m+1)}{12\\,m^{3}}\n      =\\frac{1}{6}+\\frac{1}{4m}+\\frac{1}{12m^{2}}\n      \\le\\frac{7}{27}<0.26.\\tag{12}\n\\]\nConsequently\n\\[\n\\tau<\\frac{7}{27},\\qquad\\text{hence }e^{\\tau}\\le e^{0.26}<1.3.\n\\]\n\n\\emph{Improved bound for $\\dfrac{|\\rho|}{\\tau}$.}  \nFrom (11) we have  \n\\[\n\\frac{|\\rho|}{\\tau}\\le\\frac{77}{36}\\frac{S_{3}}{S_{2}n}. \n\\]\nBecause $S_{3}/S_{2}\\le\\frac34(m+1)$ and $m+1\\le n^{1/3}+1$, we use the elementary inequality\n\\[\nn^{1/3}+1\\le\\frac32\\,n^{1/3}\\qquad(n\\ge 10) \\tag{13}\n\\]\n(the right–hand side exceeds the left because $n^{1/3}\\ge 2$).  Hence\n\\[\n\\frac{|\\rho|}{\\tau}\\le\\frac{77}{36}\\cdot\\frac34\\cdot\n      \\frac{3}{2}\\,n^{-2/3}\n     =\\frac{77}{32}\\,n^{-2/3}<0.52,\\qquad(n\\ge 10). \\tag{14}\n\\]\nThus\n\\[\n|\\rho|\\le 0.52\\,\\tau,\\qquad\n|\\tau+\\rho|\\le 1.52\\,\\tau\\le 1.52\\cdot\\frac{7}{27}\n           =\\frac{106.4}{27}<0.4<\\log 2.\\tag{15}\n\\]\n\n{\\bf Step 8.  Taylor expansion of $e^{\\tau+\\rho}$ with a corrected remainder bound.}  \nBecause $|\\tau+\\rho|<0.4<\\log 2$ we have $e^{|\\tau+\\rho|}<2$.  \nFor the third–order remainder set  \n\\[\nR:=\\frac{(\\tau+\\rho)^{3}}{6}e^{|\\tau+\\rho|},\n\\]\nso\n\\[\n|R|\\le\\frac{2}{6}\\,|\\,\\tau+\\rho\\,|^{3}\n     \\le\\frac13|\\,\\tau+\\rho\\,|^{2}. \n\\]\nWrite  \n\\[\ne^{\\tau+\\rho}=1+(\\tau+\\rho)+\\frac{(\\tau+\\rho)^{2}}{2}+R\n            =1+\\tau+E,\\qquad \nE:=\\rho+\\frac{(\\tau+\\rho)^{2}}{2}+R. \\tag{16}\n\\]\n\nUsing $|\\rho|\\le 0.52\\,\\tau$ and $|\\tau+\\rho|\\le 1.52\\,\\tau$ one gets  \n\\[\n\\frac{(\\tau+\\rho)^{2}}{2}\\le\\frac{(1.52)^{2}}{2}\\,\\tau^{2}<1.2\\,\\tau^{2},\\qquad\n|R|\\le\\frac13(1.52)^{2}\\tau^{2}<0.8\\,\\tau^{2}.\n\\]\nTherefore  \n\\[\n|E|\\le|\\rho|+1.2\\,\\tau^{2}+0.8\\,\\tau^{2}\n      \\le|\\rho|+\\frac{3}{2}\\,\\tau^{2}. \\tag{17}\n\\]\n\n{\\bf Step 9.  Completion of the proof of $(\\ddot)$.}  \nCombining (11) and (17) gives  \n\\[\n|E|\\le\\frac{77}{72}\\frac{S_{3}}{n^{2}}+\\frac{3}{4}\\frac{S_{2}^{2}}{n^{2}}.\n\\]\nSince $F_{n,m}=\\psi\\,e^{\\tau+\\rho}=\\psi(1+\\tau+E)$,  \n\\[\n\\bigl|\\,F_{n,m}-\\psi(1+\\tau)\\bigr|\n=\\psi\\,|E|\n\\le\\psi\\Bigl(\\frac{3}{4}\\frac{S_{2}^{2}}{n^{2}}+\\frac{77}{72}\\frac{S_{3}}{n^{2}}\\Bigr),\n\\]\nwhich is exactly $(\\ddot)$.\n\n{\\bf Step 10.  Proof of the limit $(\\clubsuit)$.}  \nLet $n\\to\\infty$ along any sequence satisfying $m\\le n^{1/3}$.  \nMultiplying the bound in Step 9 by $n$ gives  \n\\[\nn\\psi|E|\n\\;\\le\\;\\psi\\left(\\frac{77}{72}\\frac{S_{3}}{n}+\\frac{3}{4}\\frac{S_{2}^{2}}{n}\\right).\n\\tag{18}\n\\]\n\nBounding the two summands separately:\n\n(i) If $m$ is bounded (in particular fixed) then   \n$S_{3}=O(1)$ and $S_{2}^{2}=O(1)$, hence the right–hand side of (18) is \n$O\\!\\bigl(\\psi/n\\bigr)\\to 0$.\n\n(ii) If $m\\to\\infty$ but $m\\le n^{1/3}$, then  \n$S_{3}=O(m^{4})$ and $S_{2}^{2}=O(m^{6})$.  \nBecause $n\\ge m^{3}$,  \n\\[\n\\frac{S_{3}}{n}=O(m),\\qquad\n\\frac{S_{2}^{2}}{n}=O(m^{3}). \\tag{19}\n\\]\nMoreover $m\\le n^{1/3}$ implies $m\\to\\infty$ together with $n$, whence  \n\\[\n\\psi=e^{-m(m+1)/2}\\le e^{-m^{2}/2}.\n\\]\nThus  \n\\[\nn\\psi|E|\\;\\ll\\;e^{-m^{2}/2}\\,(m+m^{3})\\;\\xrightarrow[n\\to\\infty]{}\\;0,\n\\]\nbecause $e^{-m^{2}/2}$ dominates every polynomial in $m$.\n\nHence in all cases $n\\psi E\\to 0$.  Returning to\n\\[\nn\\bigl[F_{n,m}-\\psi\\bigr]=n\\psi\\tau+n\\psi E,\n\\]\nwe have $n\\psi E\\to 0$ while \n$n\\psi\\tau=n\\psi\\dfrac{S_{2}}{2n}\n=\\dfrac{S_{2}}{2}\\,\\psi$.  Therefore\n\\[\n\\lim_{n\\to\\infty} n\\bigl[F_{n,m}-e^{-m(m+1)/2}\\bigr]\n=\\frac{S_{2}}{2}\\,e^{-m(m+1)/2},\n\\]\nwhich is exactly $(\\clubsuit)$. \\hfill$\\square$",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.776153",
        "was_fixed": false,
        "difficulty_analysis": "• Dimension & Variables:  The original inequality involves a single factor (1–c/n)ⁿ.  The enhanced version introduces a product over m distinct factors whose exponents themselves vary (n–k).  Two discrete variables (n,m) now interact non-trivially.\n\n• Deeper Expansion:  A full third-order Taylor expansion with a rigorously controlled remainder is required, as opposed to the first-order expansion sufficient for the original.\n\n• Summation Theory:  Precise closed forms for Σk, Σk², Σk³, Σk⁴ are exploited, demanding fluency with Faulhaber polynomials and error bounding of several nested sums.\n\n• Two-parameter Asymptotics:  The estimate must remain uniform for all m≤n/3, and a sharp 1/n-order asymptotic constant is proved (♣).  Establishing such sharpness is far beyond the scope of the original exercise.\n\n• Multiple Concepts Combined:  The solution interweaves Taylor analysis, combinatorial summations, exponential inequalities, and limit arguments.  Each component alone is elementary; their coordinated use in four separate error layers markedly raises the conceptual load.\n\nConsequently the problem is significantly more intricate than both the original and the current kernel variant, requiring subtler bounding techniques and a much longer chain of reasoning."
      }
    },
    "original_kernel_variant": {
      "question": "Let $n$ be an integer with $n\\ge 10$ and let $m$ be another integer that satisfies  \n\\[\n1\\;\\le\\;m\\;\\le\\;\\frac{n}{3}.\n\\]\nDefine  \n\\[\nF_{n,m}:=\\prod_{k=1}^{m}\\Bigl(1-\\frac{k}{n}\\Bigr)^{\\,n-k},\n\\qquad\\qquad(\\star)\n\\]\nand denote the power sums  \n\\[\nS_{2}:=\\sum_{k=1}^{m}k^{2}=\\frac{m(m+1)(2m+1)}{6},\\qquad  \nS_{3}:=\\sum_{k=1}^{m}k^{3}=\\frac{m^{2}(m+1)^{2}}{4},\\qquad  \nS_{4}:=\\sum_{k=1}^{m}k^{4}.\n\\]\n\n1. (Uniform second–order estimate – valid for every admissible $(n,m)$)  \n   Show that with the universal constant  \n   \\[\n   C_{0}:=\\frac{41}{72}\\qquad(\\ddag)\n   \\]\n   one has  \n   \\[\n   \\exp\\!\\Bigl[-\\tfrac12 m(m+1)+\\tfrac{S_{2}}{2n}-C_{0}\\tfrac{S_{3}}{n^{2}}\\Bigr]\n   \\;<\\;F_{n,m}\\;<\\;\n   \\exp\\!\\Bigl[-\\tfrac12 m(m+1)+\\tfrac{S_{2}}{2n}+C_{0}\\tfrac{S_{3}}{n^{2}}\\Bigr].\n   \\tag{\\dag}\n   \\]\n\n2. (Second–order error control in the sub–critical regime $m\\le n^{1/3}$)  \n   From now on assume in addition  \n   \\[\n   1\\;\\le\\;m\\;\\le\\;n^{1/3}\\qquad(\\star\\star)\n   \\]\n   and put $\\psi:=e^{-m(m+1)/2}$.  Prove that, under $(\\star\\star)$,\n   \\[\n   \\bigl|\\,F_{n,m}-\\psi\\bigl(1+\\tfrac{S_{2}}{2n}\\bigr)\\bigr|\n   \\;\\le\\;\n   \\psi\\Bigl(\\tfrac{3}{4}\\tfrac{S_{2}^{2}}{n^{2}}+\\tfrac{77}{72}\\tfrac{S_{3}}{n^{2}}\\Bigr).\n   \\tag{\\ddot}\n   \\]\n\n   Equivalently\n   \\[\n   \\psi\\Bigl(\\tfrac{S_{2}}{2n}-\\tfrac{3}{4}\\tfrac{S_{2}^{2}}{n^{2}}-\\tfrac{77}{72}\\tfrac{S_{3}}{n^{2}}\\Bigr)\n   \\;\\le\\;\n   F_{n,m}-\\psi\n   \\;\\le\\;\n   \\psi\\Bigl(\\tfrac{S_{2}}{2n}+\\tfrac{3}{4}\\tfrac{S_{2}^{2}}{n^{2}}+\\tfrac{77}{72}\\tfrac{S_{3}}{n^{2}}\\Bigr).\n   \\]\n\n3. (First–order sharpness)  \n   Let $(n,m)$ be any sequence with $n\\to\\infty$ that satisfies $(\\star\\star)$ (for instance, $m$ fixed or $m=o(n^{1/3})$).  Show that  \n   \\[\n   \\lim_{n\\to\\infty} n\\bigl[F_{n,m}-e^{-m(m+1)/2}\\bigr]\n   \\;=\\;\\frac{S_{2}}{2}\\,e^{-m(m+1)/2}\n   \\;=\\;\\frac{m(m+1)(2m+1)}{6}\\,e^{-m(m+1)/2}.\n   \\tag{\\clubsuit}\n   \\]",
      "solution": "Throughout write  \n\\[\nx_{k}:=\\frac{k}{n}\\quad(0<x_{k}\\le \\tfrac13),\\qquad \nL_{n,m}:=\\log F_{n,m}=\\sum_{k=1}^{m}(n-k)\\log(1-x_{k}). \\tag{1}\n\\]\n\nStep 1.  Truncating $\\log(1-x)$.  \nFor $0\\le x\\le\\frac13$ the Taylor expansion with positive remainder is  \n\\[\n\\log(1-x)=-x-\\frac{x^{2}}{2}-\\frac{x^{3}}{3}-R_{4}(x),\\qquad \nR_{4}(x):=\\sum_{j=4}^{\\infty}\\frac{x^{j}}{j}>0,\n\\]\nand  \n\\[\n0<R_{4}(x)\\le\\frac{x^{4}}{4(1-x)}\\le\\frac{3}{8}x^{4}.\n\\]\nInsert this expansion into (1) and decompose  \n\\[\nL_{n,m}=A_{1}+A_{2}+A_{3}+A_{4},\n\\]\nwith  \n\\[\n\\begin{aligned}\nA_{1}&:=-\\sum_{k=1}^{m}(n-k)x_{k}, &\\qquad  \nA_{2}&:=-\\frac12\\sum_{k=1}^{m}(n-k)x_{k}^{2},\\\\[2pt]\nA_{3}&:=-\\frac13\\sum_{k=1}^{m}(n-k)x_{k}^{3}, &\\qquad  \nA_{4}&:=-\\sum_{k=1}^{m}(n-k)R_{4}(x_{k}).\n\\end{aligned}\n\\]\n\nStep 2.  The first two sums.  \nBecause $(n-k)x_{k}=k-k^{2}/n$ and $(n-k)x_{k}^{2}=k^{2}/n-k^{3}/n^{2}$,  \n\\[\nA_{1}=-\\sum_{k=1}^{m}k+\\frac1n\\sum_{k=1}^{m}k^{2}\n       =-\\frac12 m(m+1)+\\frac{S_{2}}{n},\\tag{2}\n\\]\n\\[\nA_{2}=-\\frac12\\Bigl(\\frac{S_{2}}{n}-\\frac{S_{3}}{n^{2}}\\Bigr)\n       =-\\frac{S_{2}}{2n}+\\frac{S_{3}}{2n^{2}}.\\tag{3}\n\\]\n\nStep 3.  The cubic part.  \nBecause $(n-k)x_{k}^{3}=k^{3}/n^{2}-k^{4}/n^{3}$,  \n\\[\nA_{3}=-\\frac13\\Bigl(\\frac{S_{3}}{n^{2}}-\\frac{S_{4}}{n^{3}}\\Bigr)\n      =-\\frac{S_{3}}{3n^{2}}+\\frac{S_{4}}{3n^{3}}.\\tag{4}\n\\]\n\nStep 4.  Bounding $A_{4}$ and comparing with $A_{3}$.  \nSince $R_{4}(x_{k})\\le\\frac38 x_{k}^{4}$,  \n\\[\n|A_{4}|\\le\\frac38\\sum_{k=1}^{m}(n-k)\\frac{k^{4}}{n^{4}}\n       \\le\\frac{3}{8n^{3}}S_{4}.\\tag{5}\n\\]\n\nTo relate $S_{4}$ and $S_{3}$ note that $k\\le m\\le n/3$ implies  \n\\[\nk^{4}\\le m\\,k^{3}\\quad\\Longrightarrow\\quad\nS_{4}=\\sum_{k=1}^{m}k^{4}\\le m\\sum_{k=1}^{m}k^{3}=m\\,S_{3}\n\\le\\frac{n}{3}\\,S_{3}.\\tag{6}\n\\]\nTherefore  \n\\[\n\\frac{S_{4}}{n^{3}}\\le\\frac{1}{3}\\frac{S_{3}}{n^{2}}. \\tag{7}\n\\]\n\nCombining (4)–(7) gives  \n\\[\n|A_{3}+A_{4}|\n\\;\\le\\;\\frac{S_{3}}{3n^{2}}+\\frac{17}{24}\\frac{S_{4}}{n^{3}}\n\\;\\le\\;\\Bigl(\\frac13+\\frac{17}{72}\\Bigr)\\frac{S_{3}}{n^{2}}\n\\;=\\;\\frac{41}{72}\\frac{S_{3}}{n^{2}}.\\tag{8}\n\\]\n\nSet $C_{0}:=\\tfrac{41}{72}$.\n\nStep 5.  Collecting the pieces – proof of $(\\dag)$.  \nFrom (2)–(4) we have  \n\\[\nL_{n,m}=-\\frac12 m(m+1)+\\frac{S_{2}}{2n}+\\frac{S_{3}}{2n^{2}}\n        +(A_{3}+A_{4}).\\tag{9}\n\\]\nUsing (8) yields  \n\\[\n-C_{0}\\frac{S_{3}}{n^{2}}\\;\\le\\;A_{3}+A_{4}\\;\\le\\;C_{0}\\frac{S_{3}}{n^{2}},\n\\]\nhence  \n\\[\n-\\frac12 m(m+1)+\\frac{S_{2}}{2n}-C_{0}\\frac{S_{3}}{n^{2}}\n< L_{n,m} <\n-\\frac12 m(m+1)+\\frac{S_{2}}{2n}+C_{0}\\frac{S_{3}}{n^{2}}.\n\\]\nExponentiating proves $(\\dag)$.\n\nStep 6.  Refined decomposition (assume $(\\star\\star)$).  \nPut  \n\\[\n\\psi:=e^{-m(m+1)/2},\\qquad \\tau:=\\frac{S_{2}}{2n},\\qquad\n\\rho:=\\frac{S_{3}}{2n^{2}}+A_{3}+A_{4}. \\tag{10}\n\\]\nBy (8) and the definition of $C_{0}$,\n\\[\n|\\rho|\\le\\Bigl(\\frac12+C_{0}\\Bigr)\\frac{S_{3}}{n^{2}}\n       =\\frac{77}{72}\\frac{S_{3}}{n^{2}}.\\tag{11}\n\\]\n\nStep 7.  Relating $\\rho$ and $\\tau$.  \nCompute  \n\\[\n\\frac{|\\rho|}{\\tau}\\le\\frac{77}{72}\\frac{S_{3}}{n^{2}}\\cdot\\frac{2n}{S_{2}}\n                    =\\frac{77}{36}\\frac{S_{3}}{S_{2}n}. \\tag{12}\n\\]\nSince $S_{3}/S_{2}=\\dfrac{3}{2}\\dfrac{m(m+1)}{2m+1}\\le\\frac34(m+1)$,  \n\\[\n\\frac{|\\rho|}{\\tau}\\le\\frac{77}{48}\\frac{m+1}{n}.\\tag{13}\n\\]\n\nUnder $m\\le n^{1/3}$ and $n\\ge 10$ one has $(m+1)/n\\le 0.32$, hence  \n\\[\n\\frac{|\\rho|}{\\tau}\\le 0.52,\\qquad\n|\\rho|\\le 0.52\\,\\tau. \\tag{14}\n\\]\nConsequently $|\\,\\tau+\\rho\\,|\\le 1.52\\,\\tau\\le 0.255<\\frac13$ for every $n\\ge 10$.\n\nStep 8.  Taylor expansion of $e^{\\tau+\\rho}$ with a corrected remainder bound.  \nBecause $|\\,\\tau+\\rho\\,|<\\frac13$ we use\n\\[\nR:=\\frac{(\\tau+\\rho)^{3}}{6}e^{|\\tau+\\rho|}\n\\]\nfor the third–order remainder and note $e^{|\\tau+\\rho|}\\le e^{1/3}<2$.  \nHence  \n\\[\n|R|\\le\\frac{2}{6}\\,|\\,\\tau+\\rho\\,|^{3}\\le\\frac13|\\,\\tau+\\rho\\,|^{2}.\n\\]\nWrite  \n\\[\ne^{\\tau+\\rho}=1+(\\tau+\\rho)+\\frac{(\\tau+\\rho)^{2}}{2}+R\n            =1+\\tau+E,\\qquad \nE:=\\rho+\\frac{(\\tau+\\rho)^{2}}{2}+R. \\tag{15}\n\\]\n\nUsing $|\\,\\rho\\,|\\le 0.52\\,\\tau$ and $|\\,\\tau+\\rho\\,|\\le 1.52\\,\\tau$ one gets  \n\\[\n\\frac{(\\tau+\\rho)^{2}}{2}\\le\\frac{(1.52)^{2}}{2}\\,\\tau^{2}<1.2\\,\\tau^{2},\\qquad\n|R|\\le\\frac13(1.52)^{2}\\tau^{2}<0.8\\,\\tau^{2}.\n\\]\nTherefore  \n\\[\n|E|\\le|\\rho|+1.2\\,\\tau^{2}+0.8\\,\\tau^{2}\n      \\le|\\rho|+3\\,\\tau^{2}. \\tag{16}\n\\]\n\nStep 9.  From (16) and (11) we have  \n\\[\n|E|\\le\\frac{77}{72}\\frac{S_{3}}{n^{2}}+3\\Bigl(\\frac{S_{2}}{2n}\\Bigr)^{2}\n     =\\frac{77}{72}\\frac{S_{3}}{n^{2}}+\\frac{3}{4}\\frac{S_{2}^{2}}{n^{2}}.\n\\]\nSince $F_{n,m}=\\psi\\,e^{\\tau+\\rho}=\\psi(1+\\tau+E)$,  \n\\[\n\\bigl|\\,F_{n,m}-\\psi(1+\\tau)\\bigr|\n=\\psi\\,|E|\n\\le\\psi\\Bigl(\\frac{3}{4}\\frac{S_{2}^{2}}{n^{2}}+\\frac{77}{72}\\frac{S_{3}}{n^{2}}\\Bigr),\n\\]\nwhich is exactly the desired inequality $(\\ddot)$.\n\nStep 10.  Proof of the limit $(\\clubsuit)$.  \nLet $n\\to\\infty$ along any sequence satisfying $(\\star\\star)$.  \nMultiplying the bound in Step 9 by $n$ gives  \n\\[\nn\\psi|E|\n\\;\\le\\;\\psi\\left(\\frac{77}{72}\\frac{S_{3}}{n}+\\frac{3}{4}\\frac{S_{2}^{2}}{n}\\right).\n\\tag{17}\n\\]\n\nBounding the two summands separately:\n\n(i) If $m$ is bounded (in particular fixed) then   \n$S_{3}=O(1)$ and $S_{2}^{2}=O(1)$, hence the right–hand side of (17) is \n$O\\!\\bigl(\\psi/n\\bigr)\\to 0$.\n\n(ii) If $m\\to\\infty$ but $m\\le n^{1/3}$, then  \n$S_{3}=O(m^{4})$ and $S_{2}^{2}=O(m^{6})$.  \nBecause $n\\ge m^{3}$,  \n\\[\n\\frac{S_{3}}{n}=O(m),\\qquad\n\\frac{S_{2}^{2}}{n}=O(m^{3}). \\tag{18}\n\\]\nMoreover $m\\le n^{1/3}$ implies $m\\to\\infty$ together with $n$, whence  \n\\[\n\\psi=e^{-m(m+1)/2}\\le e^{-m^{2}/2}.\n\\]\nThus  \n\\[\nn\\psi|E|\\;\\ll\\;e^{-m^{2}/2}\\,(m+m^{3})\\;\\xrightarrow[n\\to\\infty]{}\\;0,\n\\]\nbecause $e^{-m^{2}/2}$ dominates every polynomial in $m$.\n\nHence in all cases $n\\psi E\\to 0$.  Returning to\n\\[\nn\\bigl[F_{n,m}-\\psi\\bigr]=n\\psi\\tau+n\\psi E,\n\\]\nwe have $n\\psi E\\to 0$ while \n$n\\psi\\tau=n\\psi\\dfrac{S_{2}}{2n}\n=\\dfrac{S_{2}}{2}\\,\\psi$.  Therefore\n\\[\n\\lim_{n\\to\\infty} n\\bigl[F_{n,m}-e^{-m(m+1)/2}\\bigr]\n=\\frac{S_{2}}{2}\\,e^{-m(m+1)/2},\n\\]\nwhich is exactly $(\\clubsuit)$. \\hfill$\\square$",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.594414",
        "was_fixed": false,
        "difficulty_analysis": "• Dimension & Variables:  The original inequality involves a single factor (1–c/n)ⁿ.  The enhanced version introduces a product over m distinct factors whose exponents themselves vary (n–k).  Two discrete variables (n,m) now interact non-trivially.\n\n• Deeper Expansion:  A full third-order Taylor expansion with a rigorously controlled remainder is required, as opposed to the first-order expansion sufficient for the original.\n\n• Summation Theory:  Precise closed forms for Σk, Σk², Σk³, Σk⁴ are exploited, demanding fluency with Faulhaber polynomials and error bounding of several nested sums.\n\n• Two-parameter Asymptotics:  The estimate must remain uniform for all m≤n/3, and a sharp 1/n-order asymptotic constant is proved (♣).  Establishing such sharpness is far beyond the scope of the original exercise.\n\n• Multiple Concepts Combined:  The solution interweaves Taylor analysis, combinatorial summations, exponential inequalities, and limit arguments.  Each component alone is elementary; their coordinated use in four separate error layers markedly raises the conceptual load.\n\nConsequently the problem is significantly more intricate than both the original and the current kernel variant, requiring subtler bounding techniques and a much longer chain of reasoning."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}