{
  "index": "1962-B-5",
  "type": "ANA",
  "tag": [
    "ANA",
    "NT"
  ],
  "difficulty": "",
  "question": "\\begin{array}{l}\n\\text { 5. Prove that for every integer } n \\text { greater than 1: }\\\\\n\\frac{3 n+1}{2 n+2}<\\left(\\frac{1}{n}\\right)^{n}+\\left(\\frac{2}{n}\\right)^{n}+\\cdots+\\left(\\frac{n}{n}\\right)^{n}<2 .(\\text { page } 566)\n\\end{array}",
  "solution": "Solution. For \\( n>1 \\) and \\( x>0 \\) the function \\( x^{n} \\) is convex (i.e., its graph is concave up), so its integral is less than the approximation given by the trapezoidal rule. Hence\n\\[\n\\begin{array}{l} \n\\frac{1}{n+1}=\\int_{0}^{1} x^{n} d x<\\frac{1}{n}\\left\\lfloor\\frac{1}{2}\\left(\\frac{0}{n}\\right)^{n}+\\left(\\frac{1}{n}\\right)^{n}+\\left(\\frac{2}{n}\\right)^{n}\\right. \\\\\n\\left.+\\cdots+\\left(\\frac{n-1}{n}\\right)^{n}+\\frac{1}{2}\\left(\\frac{n}{n}\\right)^{n}\\right] .\n\\end{array}\n\\]\n\nMultiply by \\( n \\) and add \\( 1 / 2 \\) to get\n\\[\n\\begin{aligned}\n\\frac{3 n+1}{2 n+2}=\\frac{n}{n+1}+\\frac{1}{2} & <\\left[\\left(\\frac{1}{n}\\right)^{n}+\\left(\\frac{2}{n}\\right)^{n}\\right. \\\\\n& \\left.+\\cdots+\\left(\\frac{n-1}{n}\\right)^{n}+\\left(\\frac{n}{n}\\right)^{n}\\right]\n\\end{aligned}\n\\]\n\nThe required upper approximation can be obtained by considering the lower Riemann sum over the same subdivision (and adding 1 to both sides), but the argument below gives a much sharper estimate.\n\nSince \\( 1-x \\leq e^{-x} \\) for any \\( x \\), we have\n\\[\n\\left(1-\\frac{i}{n}\\right)^{n} \\leq e^{-i}\n\\]\nfor \\( 0 \\leq i \\leq n \\). Hence\n\\[\n\\begin{aligned}\n\\left(\\frac{n}{n}\\right)^{n}+\\left(\\frac{n-1}{n}\\right)^{n}+\\cdots+\\left(\\frac{2}{n}\\right)^{n}+\\left(\\frac{1}{n}\\right)^{n} & \\leq 1+e^{-1}+\\cdots+e^{-(n-1)} \\\\\n& \\leq \\frac{1}{1-e^{-1}}=\\frac{e}{e-1}<2\n\\end{aligned}\n\\]\n\nSince \\( \\lim _{n \\rightarrow \\infty}(1-i / n)^{n}=e^{-i} \\), it follows easily that\n\\[\n\\lim _{n \\rightarrow \\infty}\\left[\\left(\\frac{n}{n}\\right)^{n}+\\left(\\frac{n-1}{n}\\right)^{n}+\\cdots+\\left(\\frac{2}{n}\\right)^{n}+\\left(\\frac{1}{n}\\right)^{n}\\right]=\\frac{e}{e-1}\n\\]",
  "vars": [
    "n",
    "x",
    "i"
  ],
  "params": [],
  "sci_consts": [
    "e"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "n": "intvar",
        "x": "posreal",
        "i": "indexvar"
      },
      "question": "\\begin{array}{l}\n\\text { 5. Prove that for every integer } intvar \\text { greater than 1: }\\\\\n\\frac{3 intvar+1}{2 intvar+2}<\\left(\\frac{1}{intvar}\\right)^{intvar}+\\left(\\frac{2}{intvar}\\right)^{intvar}+\\cdots+\\left(\\frac{intvar}{intvar}\\right)^{intvar}<2 .(\\text { page } 566)\n\\end{array}",
      "solution": "Solution. For \\( intvar>1 \\) and \\( posreal>0 \\) the function \\( posreal^{intvar} \\) is convex (i.e., its graph is concave up), so its integral is less than the approximation given by the trapezoidal rule. Hence\n\\[\n\\begin{array}{l} \n\\frac{1}{intvar+1}=\\int_{0}^{1} posreal^{intvar} d posreal<\\frac{1}{intvar}\\left\\lfloor\\frac{1}{2}\\left(\\frac{0}{intvar}\\right)^{intvar}+\\left(\\frac{1}{intvar}\\right)^{intvar}+\\left(\\frac{2}{intvar}\\right)^{intvar}\\right. \\\\\n\\left.+\\cdots+\\left(\\frac{intvar-1}{intvar}\\right)^{intvar}+\\frac{1}{2}\\left(\\frac{intvar}{intvar}\\right)^{intvar}\\right] .\n\\end{array}\n\\]\n\nMultiply by \\( intvar \\) and add \\( 1 / 2 \\) to get\n\\[\n\\begin{aligned}\n\\frac{3 intvar+1}{2 intvar+2}=\\frac{intvar}{intvar+1}+\\frac{1}{2} & <\\left[\\left(\\frac{1}{intvar}\\right)^{intvar}+\\left(\\frac{2}{intvar}\\right)^{intvar}\\right. \\\\\n& \\left.+\\cdots+\\left(\\frac{intvar-1}{intvar}\\right)^{intvar}+\\left(\\frac{intvar}{intvar}\\right)^{intvar}\\right]\n\\end{aligned}\n\\]\n\nThe required upper approximation can be obtained by considering the lower Riemann sum over the same subdivision (and adding 1 to both sides), but the argument below gives a much sharper estimate.\n\nSince \\( 1-posreal \\leq e^{-posreal} \\) for any \\( posreal \\), we have\n\\[\n\\left(1-\\frac{indexvar}{intvar}\\right)^{intvar} \\leq e^{-indexvar}\n\\]\nfor \\( 0 \\leq indexvar \\leq intvar \\). Hence\n\\[\n\\begin{aligned}\n\\left(\\frac{intvar}{intvar}\\right)^{intvar}+\\left(\\frac{intvar-1}{intvar}\\right)^{intvar}+\\cdots+\\left(\\frac{2}{intvar}\\right)^{intvar}+\\left(\\frac{1}{intvar}\\right)^{intvar} & \\leq 1+e^{-1}+\\cdots+e^{-(intvar-1)} \\\\\n& \\leq \\frac{1}{1-e^{-1}}=\\frac{e}{e-1}<2\n\\end{aligned}\n\\]\n\nSince \\( \\lim _{intvar \\rightarrow \\infty}(1-indexvar / intvar)^{intvar}=e^{-indexvar} \\), it follows easily that\n\\[\n\\lim _{intvar \\rightarrow \\infty}\\left[\\left(\\frac{intvar}{intvar}\\right)^{intvar}+\\left(\\frac{intvar-1}{intvar}\\right)^{intvar}+\\cdots+\\left(\\frac{2}{intvar}\\right)^{intvar}+\\left(\\frac{1}{intvar}\\right)^{intvar}\\right]=\\frac{e}{e-1}\n\\]"
    },
    "descriptive_long_confusing": {
      "map": {
        "n": "sunflower",
        "x": "paintbrush",
        "i": "trombone"
      },
      "question": "\\begin{array}{l}\n\\text { 5. Prove that for every integer } sunflower \\text { greater than 1: }\\\\\n\\frac{3 sunflower+1}{2 sunflower+2}<\\left(\\frac{1}{sunflower}\\right)^{sunflower}+\\left(\\frac{2}{sunflower}\\right)^{sunflower}+\\cdots+\\left(\\frac{sunflower}{sunflower}\\right)^{sunflower}<2 .(\\text { page } 566)\n\\end{array}",
      "solution": "Solution. For \\( sunflower>1 \\) and \\( paintbrush>0 \\) the function \\( paintbrush^{sunflower} \\) is convex (i.e., its graph is concave up), so its integral is less than the approximation given by the trapezoidal rule. Hence\n\\[\n\\begin{array}{l} \n\\frac{1}{sunflower+1}=\\int_{0}^{1} paintbrush^{sunflower} d paintbrush<\\frac{1}{sunflower}\\left\\lfloor\\frac{1}{2}\\left(\\frac{0}{sunflower}\\right)^{sunflower}+\\left(\\frac{1}{sunflower}\\right)^{sunflower}+\\left(\\frac{2}{sunflower}\\right)^{sunflower}\\right. \\\\\n\\left.+\\cdots+\\left(\\frac{sunflower-1}{sunflower}\\right)^{sunflower}+\\frac{1}{2}\\left(\\frac{sunflower}{sunflower}\\right)^{sunflower}\\right] .\n\\end{array}\n\\]\n\nMultiply by \\( sunflower \\) and add \\( 1 / 2 \\) to get\n\\[\n\\begin{aligned}\n\\frac{3 sunflower+1}{2 sunflower+2}=\\frac{sunflower}{sunflower+1}+\\frac{1}{2} & <\\left[\\left(\\frac{1}{sunflower}\\right)^{sunflower}+\\left(\\frac{2}{sunflower}\\right)^{sunflower}\\right. \\\\\n& \\left.+\\cdots+\\left(\\frac{sunflower-1}{sunflower}\\right)^{sunflower}+\\left(\\frac{sunflower}{sunflower}\\right)^{sunflower}\\right]\n\\end{aligned}\n\\]\n\nThe required upper approximation can be obtained by considering the lower Riemann sum over the same subdivision (and adding 1 to both sides), but the argument below gives a much sharper estimate.\n\nSince \\( 1-paintbrush \\leq e^{-paintbrush} \\) for any \\( paintbrush \\), we have\n\\[\n\\left(1-\\frac{trombone}{sunflower}\\right)^{sunflower} \\leq e^{-trombone}\n\\]\nfor \\( 0 \\leq trombone \\leq sunflower \\). Hence\n\\[\n\\begin{aligned}\n\\left(\\frac{sunflower}{sunflower}\\right)^{sunflower}+\\left(\\frac{sunflower-1}{sunflower}\\right)^{sunflower}+\\cdots+\\left(\\frac{2}{sunflower}\\right)^{sunflower}+\\left(\\frac{1}{sunflower}\\right)^{sunflower} & \\leq 1+e^{-1}+\\cdots+e^{-(sunflower-1)} \\\\\n& \\leq \\frac{1}{1-e^{-1}}=\\frac{e}{e-1}<2\n\\end{aligned}\n\\]\n\nSince \\( \\lim _{sunflower \\rightarrow \\infty}(1-trombone / sunflower)^{sunflower}=e^{-trombone} \\), it follows easily that\n\\[\n\\lim _{sunflower \\rightarrow \\infty}\\left[\\left(\\frac{sunflower}{sunflower}\\right)^{sunflower}+\\left(\\frac{sunflower-1}{sunflower}\\right)^{sunflower}+\\cdots+\\left(\\frac{2}{sunflower}\\right)^{sunflower}+\\left(\\frac{1}{sunflower}\\right)^{sunflower}\\right]=\\frac{e}{e-1}\n\\]"
    },
    "descriptive_long_misleading": {
      "map": {
        "n": "continuum",
        "x": "fixedpoint",
        "i": "outsider"
      },
      "question": "\\begin{array}{l}\n\\text { 5. Prove that for every integer } continuum \\text { greater than 1: }\\\\\n\\frac{3 continuum+1}{2 continuum+2}<\\left(\\frac{1}{continuum}\\right)^{continuum}+\\left(\\frac{2}{continuum}\\right)^{continuum}+\\cdots+\\left(\\frac{continuum}{continuum}\\right)^{continuum}<2 .(\\text { page } 566)\n\\end{array}",
      "solution": "Solution. For \\( continuum>1 \\) and \\( fixedpoint>0 \\) the function \\( fixedpoint^{continuum} \\) is convex (i.e., its graph is concave up), so its integral is less than the approximation given by the trapezoidal rule. Hence\n\\[\n\\begin{array}{l} \n\\frac{1}{continuum+1}=\\int_{0}^{1} fixedpoint^{continuum} d fixedpoint<\\frac{1}{continuum}\\left\\lfloor\\frac{1}{2}\\left(\\frac{0}{continuum}\\right)^{continuum}+\\left(\\frac{1}{continuum}\\right)^{continuum}+\\left(\\frac{2}{continuum}\\right)^{continuum}\\right. \\\\\n\\left.+\\cdots+\\left(\\frac{continuum-1}{continuum}\\right)^{continuum}+\\frac{1}{2}\\left(\\frac{continuum}{continuum}\\right)^{continuum}\\right] .\n\\end{array}\n\\]\n\nMultiply by \\( continuum \\) and add \\( 1 / 2 \\) to get\n\\[\n\\begin{aligned}\n\\frac{3 continuum+1}{2 continuum+2}=\\frac{continuum}{continuum+1}+\\frac{1}{2} & <\\left[\\left(\\frac{1}{continuum}\\right)^{continuum}+\\left(\\frac{2}{continuum}\\right)^{continuum}\\right. \\\\\n& \\left.+\\cdots+\\left(\\frac{continuum-1}{continuum}\\right)^{continuum}+\\left(\\frac{continuum}{continuum}\\right)^{continuum}\\right]\n\\end{aligned}\n\\]\n\nThe required upper approximation can be obtained by considering the lower Riemann sum over the same subdivision (and adding 1 to both sides), but the argument below gives a much sharper estimate.\n\nSince \\( 1-fixedpoint \\leq e^{-fixedpoint} \\) for any \\( fixedpoint \\), we have\n\\[\n\\left(1-\\frac{outsider}{continuum}\\right)^{continuum} \\leq e^{-outsider}\n\\]\nfor \\( 0 \\leq outsider \\leq continuum \\). Hence\n\\[\n\\begin{aligned}\n\\left(\\frac{continuum}{continuum}\\right)^{continuum}+\\left(\\frac{continuum-1}{continuum}\\right)^{continuum}+\\cdots+\\left(\\frac{2}{continuum}\\right)^{continuum}+\\left(\\frac{1}{continuum}\\right)^{continuum} & \\leq 1+e^{-1}+\\cdots+e^{-(continuum-1)} \\\\\n& \\leq \\frac{1}{1-e^{-1}}=\\frac{e}{e-1}<2\n\\end{aligned}\n\\]\n\nSince \\( \\lim _{continuum \\rightarrow \\infty}(1-outsider / continuum)^{continuum}=e^{-outsider} \\), it follows easily that\n\\[\n\\lim _{continuum \\rightarrow \\infty}\\left[\\left(\\frac{continuum}{continuum}\\right)^{continuum}+\\left(\\frac{continuum-1}{continuum}\\right)^{continuum}+\\cdots+\\left(\\frac{2}{continuum}\\right)^{continuum}+\\left(\\frac{1}{continuum}\\right)^{continuum}\\right]=\\frac{e}{e-1}\n\\]"
    },
    "garbled_string": {
      "map": {
        "n": "qzxwvtnp",
        "x": "hjgrksla",
        "i": "kurdplox"
      },
      "question": "<<<\n\\begin{array}{l}\n\\text { 5. Prove that for every integer } qzxwvtnp \\text { greater than 1: }\\\\\n\\frac{3 qzxwvtnp+1}{2 qzxwvtnp+2}<\\left(\\frac{1}{qzxwvtnp}\\right)^{qzxwvtnp}+\\left(\\frac{2}{qzxwvtnp}\\right)^{qzxwvtnp}+\\cdots+\\left(\\frac{qzxwvtnp}{qzxwvtnp}\\right)^{qzxwvtnp}<2 .(\\text { page } 566)\n\\end{array}\n>>>",
      "solution": "<<<\nSolution. For \\( qzxwvtnp>1 \\) and \\( hjgrksla>0 \\) the function \\( hjgrksla^{qzxwvtnp} \\) is convex (i.e., its graph is concave up), so its integral is less than the approximation given by the trapezoidal rule. Hence\n\\[\n\\begin{array}{l} \n\\frac{1}{qzxwvtnp+1}=\\int_{0}^{1} hjgrksla^{qzxwvtnp} d hjgrksla<\\frac{1}{qzxwvtnp}\\left\\lfloor\\frac{1}{2}\\left(\\frac{0}{qzxwvtnp}\\right)^{qzxwvtnp}+\\left(\\frac{1}{qzxwvtnp}\\right)^{qzxwvtnp}+\\left(\\frac{2}{qzxwvtnp}\\right)^{qzxwvtnp}\\right. \\\\\n\\left.+\\cdots+\\left(\\frac{qzxwvtnp-1}{qzxwvtnp}\\right)^{qzxwvtnp}+\\frac{1}{2}\\left(\\frac{qzxwvtnp}{qzxwvtnp}\\right)^{qzxwvtnp}\\right] .\n\\end{array}\n\\]\n\nMultiply by \\( qzxwvtnp \\) and add \\( 1 / 2 \\) to get\n\\[\n\\begin{aligned}\n\\frac{3 qzxwvtnp+1}{2 qzxwvtnp+2}=\\frac{qzxwvtnp}{qzxwvtnp+1}+\\frac{1}{2} & <\\left[\\left(\\frac{1}{qzxwvtnp}\\right)^{qzxwvtnp}+\\left(\\frac{2}{qzxwvtnp}\\right)^{qzxwvtnp}\\right. \\\\\n& \\left.+\\cdots+\\left(\\frac{qzxwvtnp-1}{qzxwvtnp}\\right)^{qzxwvtnp}+\\left(\\frac{qzxwvtnp}{qzxwvtnp}\\right)^{qzxwvtnp}\\right]\n\\end{aligned}\n\\]\n\nThe required upper approximation can be obtained by considering the lower Riemann sum over the same subdivision (and adding 1 to both sides), but the argument below gives a much sharper estimate.\n\nSince \\( 1-hjgrksla \\leq e^{-hjgrksla} \\) for any \\( hjgrksla \\), we have\n\\[\n\\left(1-\\frac{kurdplox}{qzxwvtnp}\\right)^{qzxwvtnp} \\leq e^{-kurdplox}\n\\]\nfor \\( 0 \\leq kurdplox \\leq qzxwvtnp \\). Hence\n\\[\n\\begin{aligned}\n\\left(\\frac{qzxwvtnp}{qzxwvtnp}\\right)^{qzxwvtnp}+\\left(\\frac{qzxwvtnp-1}{qzxwvtnp}\\right)^{qzxwvtnp}+\\cdots+\\left(\\frac{2}{qzxwvtnp}\\right)^{qzxwvtnp}+\\left(\\frac{1}{qzxwvtnp}\\right)^{qzxwvtnp} & \\leq 1+e^{-1}+\\cdots+e^{-(qzxwvtnp-1)} \\\\\n& \\leq \\frac{1}{1-e^{-1}}=\\frac{e}{e-1}<2\n\\end{aligned}\n\\]\n\nSince \\( \\lim _{qzxwvtnp \\rightarrow \\infty}(1-kurdplox / qzxwvtnp)^{qzxwvtnp}=e^{-kurdplox} \\), it follows easily that\n\\[\n\\lim _{qzxwvtnp \\rightarrow \\infty}\\left[\\left(\\frac{qzxwvtnp}{qzxwvtnp}\\right)^{qzxwvtnp}+\\left(\\frac{qzxwvtnp-1}{qzxwvtnp}\\right)^{qzxwvtnp}+\\cdots+\\left(\\frac{2}{qzxwvtnp}\\right)^{qzxwvtnp}+\\left(\\frac{1}{qzxwvtnp}\\right)^{qzxwvtnp}\\right]=\\frac{e}{e-1}\n\\]\n>>>"
    },
    "kernel_variant": {
      "question": "Fix an integer d ≥ 2.  \nFor every integer n ≥ d define  \n\n  Sₙ,d = ∑_{(k₁,…,k_d)∈{0,1,…,n}^d\\setminus\\{(0,…,0)\\}}\n      (max{k₁,…,k_d}/n)ⁿ.\n\n(a) Prove the two–sided estimate, valid for every n ≥ d,  \n\n  d·n^{d-1} < Sₙ,d\n    < d·\\dfrac{e}{e-1}\\,n^{d-1}\n      +\\Bigl((n+1)^d-n^{d}-d\\,n^{d-1}\\Bigr).  (★)\n\n(In particular\n  (n+1)^d-n^{d}-d\\,n^{d-1}= \\dfrac{d(d-1)}{2}\\,n^{d-2}+O(n^{d-3}).)\n\n(b) Show that the scaled sequence converges and determine its limit:\n\n  lim_{n→∞} n^{1-d}\\,Sₙ,d = d·\\dfrac{e}{e-1}.",
      "solution": "Throughout write  \n\n  M_d(k):=(k+1)^d-k^{d}, k ≥ 0,  \n\nand for 0 ≤ j ≤ n–1 abbreviate  \n\n  Δ_d(n,j):=M_d(n-j)=(n-j+1)^d-(n-j)^d.   (0)\n\n\n\n1. Layer decomposition.  \nSetting k=n−j yields  \n\n  Sₙ,d = ∑_{j=0}^{n-1}(1-j/n)^{\\,n} Δ_d(n,j).  (1)\n\n\n\n2. The upper estimate.\n\n2.1 Exponential decay.  \nFor 0 ≤ t ≤ 1, (1−t)^{n} ≤ e^{−nt}; hence  \n\n  (1-j/n)^{n} ≤ e^{−j}.  (2)\n\n2.2 A uniform bound for the inner layers (1 ≤ j ≤ n−1).  \nBy the mean–value theorem applied to x↦x^{d} on [n−j, n−j+1] there is ξ with n−j ≤ ξ ≤ n−j+1 such that  \n\n  Δ_d(n,j)=d ξ^{d-1} ≤ d(n-j+1)^{d-1} ≤ d n^{d-1}.  (3)\n\n(The endpoint j=0 will be treated separately; (3) is not used there.)\n\n2.3 Splitting off the outermost layer j=0.  \nInsert (2)–(3) into (1):\n\n  Sₙ,d = Δ_d(n,0)+∑_{j=1}^{n-1}(1-j/n)^{n}Δ_d(n,j)  \n    ≤ Δ_d(n,0)+d\\,n^{d-1}∑_{j=1}^{∞}e^{−j}.  (4)\n\nBecause ∑_{j=1}^{∞}e^{−j}=e/(e−1)−1,  \n\n  Sₙ,d ≤ Δ_d(n,0)+d\\!\\left(\\frac{e}{e-1}-1\\right)n^{d-1}  \n     = d\\frac{e}{e-1}n^{d-1}+\\bigl(Δ_d(n,0)-d\\,n^{d-1}\\bigr).  (5)\n\nSince Δ_d(n,0) = (n+1)^d−n^{d}, the right–hand side of (5) is exactly the\nupper bound asserted in (★).\n\n2.4 Size of the error term.  \nBy the binomial theorem  \n\n  (n+1)^d-n^{d}-d\\,n^{d-1} = \\frac{d(d-1)}{2}\\,n^{d-2}+O(n^{d-3}), (6)\n\nwhich is of strictly smaller order than n^{d-1}.\n\n\n\n3. The lower estimate.\n\n3.1 A matching lower bound for Δ_d.  \nUsing again the mean–value theorem but now the minimum of the derivative on [n−j, n−j+1],  \n\n  Δ_d(n,j)=d ξ^{d-1} ≥ d(n-j)^{d-1}.  (7)\n\n3.2 Using (7) in (1):\n\n  Sₙ,d ≥ d∑_{j=0}^{n-1}(1-j/n)^{n}(n-j)^{d-1}  \n    = d\\,n^{d-1}\\sum_{j=0}^{n-1}(1-j/n)^{n+d-1}.  (8)\n\nBecause the first summand of the series equals 1, the whole series exceeds 1; hence  \n\n  Sₙ,d > d\\,n^{d-1},  (9)\n\nwhich gives the lower bound in (★).\n\n\n\n4. The limit n^{1-d}Sₙ,d.\n\nRewrite (1) as  \n\n  n^{1-d}Sₙ,d = ∑_{j=0}^{n-1}(1-j/n)^{n}\\, n^{1-d}Δ_d(n,j).  (10)\n\n4.1 Pointwise convergence of the summands.  \nFix j. Then  \n\n  (1-j/n)^{n} → e^{−j},  n^{1-d}Δ_d(n,j)=d ξ^{d-1}/n^{d-1} → d.  (11)\n\nHence the j-th summand tends to d e^{−j}.\n\n4.2 A uniform dominating series.  \nFrom (3) (valid for j ≥ 1) and the mean–value theorem for j = 0 we obtain  \n\n  Δ_d(n,j) ≤ d(n+1)^{d-1}.  \n\nConsequently  \n\n  n^{1-d}Δ_d(n,j) ≤ d\\,(1+1/n)^{d-1} ≤ d e  (for all n ≥ 1).  \n\nCombining with (2) yields  \n\n  |(1-j/n)^{n}\\, n^{1-d}Δ_d(n,j)| ≤ d e·e^{−j}.  (12)\n\nBecause the geometric series ∑_{j=0}^{∞} d e·e^{−j}=d e^{2}/(e−1) converges,\nthe dominated convergence theorem applies to (10), giving\n\n  lim_{n→∞} n^{1-d}Sₙ,d = ∑_{j=0}^{∞} d e^{−j} = d·\\frac{e}{e-1}.  (13)\n\nThis completes part (b) and the proof. ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.541090",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension: the sum ranges over the d-dimensional lattice cube, not just a one–dimensional list.  \n• Additional combinatorics: counting points on “layers’’ requires the identity M_d(k) = (k+1)^d – k^d and brings discrete derivatives of high powers into play.  \n• Stronger estimates: the proof demands simultaneous control of an exponential factor and a polynomial one of degree d–1, forcing refined inequalities like (1–t)ⁿ ≤ e^{–nt} and both upper and lower Taylor bounds for (x+1)^d – x^d.  \n• Asymptotics in several variables: the limit extraction hinges on rescaling by n^{d–1}, an application of the mean–value theorem in high dimension, and the dominated convergence theorem to interchange limit and infinite sum.  \n• Multiple interacting concepts: convexity, exponential inequalities, discrete calculus, series summation, and measure-theoretic convergence all appear.  \n\nThese layers of technicality and the passage from one dimension to arbitrary d make the enhanced problem significantly harder than either the original or the current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Fix an integer d ≥ 2.  \nFor every integer n ≥ d define  \n\n  Sₙ,d = ∑_{(k₁,…,k_d)∈{0,1,…,n}^d\\setminus\\{(0,…,0)\\}}\n      (max{k₁,…,k_d}/n)ⁿ.\n\n(a) Prove the two–sided estimate, valid for every n ≥ d,  \n\n  d·n^{d-1} < Sₙ,d\n    < d·\\dfrac{e}{e-1}\\,n^{d-1}\n      +\\Bigl((n+1)^d-n^{d}-d\\,n^{d-1}\\Bigr).  (★)\n\n(In particular\n  (n+1)^d-n^{d}-d\\,n^{d-1}= \\dfrac{d(d-1)}{2}\\,n^{d-2}+O(n^{d-3}).)\n\n(b) Show that the scaled sequence converges and determine its limit:\n\n  lim_{n→∞} n^{1-d}\\,Sₙ,d = d·\\dfrac{e}{e-1}.",
      "solution": "Throughout write  \n\n  M_d(k):=(k+1)^d-k^{d}, k ≥ 0,  \n\nand for 0 ≤ j ≤ n–1 abbreviate  \n\n  Δ_d(n,j):=M_d(n-j)=(n-j+1)^d-(n-j)^d.   (0)\n\n\n\n1. Layer decomposition.  \nSetting k=n−j yields  \n\n  Sₙ,d = ∑_{j=0}^{n-1}(1-j/n)^{\\,n} Δ_d(n,j).  (1)\n\n\n\n2. The upper estimate.\n\n2.1 Exponential decay.  \nFor 0 ≤ t ≤ 1, (1−t)^{n} ≤ e^{−nt}; hence  \n\n  (1-j/n)^{n} ≤ e^{−j}.  (2)\n\n2.2 A uniform bound for the inner layers (1 ≤ j ≤ n−1).  \nBy the mean–value theorem applied to x↦x^{d} on [n−j, n−j+1] there is ξ with n−j ≤ ξ ≤ n−j+1 such that  \n\n  Δ_d(n,j)=d ξ^{d-1} ≤ d(n-j+1)^{d-1} ≤ d n^{d-1}.  (3)\n\n(The endpoint j=0 will be treated separately; (3) is not used there.)\n\n2.3 Splitting off the outermost layer j=0.  \nInsert (2)–(3) into (1):\n\n  Sₙ,d = Δ_d(n,0)+∑_{j=1}^{n-1}(1-j/n)^{n}Δ_d(n,j)  \n    ≤ Δ_d(n,0)+d\\,n^{d-1}∑_{j=1}^{∞}e^{−j}.  (4)\n\nBecause ∑_{j=1}^{∞}e^{−j}=e/(e−1)−1,  \n\n  Sₙ,d ≤ Δ_d(n,0)+d\\!\\left(\\frac{e}{e-1}-1\\right)n^{d-1}  \n     = d\\frac{e}{e-1}n^{d-1}+\\bigl(Δ_d(n,0)-d\\,n^{d-1}\\bigr).  (5)\n\nSince Δ_d(n,0) = (n+1)^d−n^{d}, the right–hand side of (5) is exactly the\nupper bound asserted in (★).\n\n2.4 Size of the error term.  \nBy the binomial theorem  \n\n  (n+1)^d-n^{d}-d\\,n^{d-1} = \\frac{d(d-1)}{2}\\,n^{d-2}+O(n^{d-3}), (6)\n\nwhich is of strictly smaller order than n^{d-1}.\n\n\n\n3. The lower estimate.\n\n3.1 A matching lower bound for Δ_d.  \nUsing again the mean–value theorem but now the minimum of the derivative on [n−j, n−j+1],  \n\n  Δ_d(n,j)=d ξ^{d-1} ≥ d(n-j)^{d-1}.  (7)\n\n3.2 Using (7) in (1):\n\n  Sₙ,d ≥ d∑_{j=0}^{n-1}(1-j/n)^{n}(n-j)^{d-1}  \n    = d\\,n^{d-1}\\sum_{j=0}^{n-1}(1-j/n)^{n+d-1}.  (8)\n\nBecause the first summand of the series equals 1, the whole series exceeds 1; hence  \n\n  Sₙ,d > d\\,n^{d-1},  (9)\n\nwhich gives the lower bound in (★).\n\n\n\n4. The limit n^{1-d}Sₙ,d.\n\nRewrite (1) as  \n\n  n^{1-d}Sₙ,d = ∑_{j=0}^{n-1}(1-j/n)^{n}\\, n^{1-d}Δ_d(n,j).  (10)\n\n4.1 Pointwise convergence of the summands.  \nFix j. Then  \n\n  (1-j/n)^{n} → e^{−j},  n^{1-d}Δ_d(n,j)=d ξ^{d-1}/n^{d-1} → d.  (11)\n\nHence the j-th summand tends to d e^{−j}.\n\n4.2 A uniform dominating series.  \nFrom (3) (valid for j ≥ 1) and the mean–value theorem for j = 0 we obtain  \n\n  Δ_d(n,j) ≤ d(n+1)^{d-1}.  \n\nConsequently  \n\n  n^{1-d}Δ_d(n,j) ≤ d\\,(1+1/n)^{d-1} ≤ d e  (for all n ≥ 1).  \n\nCombining with (2) yields  \n\n  |(1-j/n)^{n}\\, n^{1-d}Δ_d(n,j)| ≤ d e·e^{−j}.  (12)\n\nBecause the geometric series ∑_{j=0}^{∞} d e·e^{−j}=d e^{2}/(e−1) converges,\nthe dominated convergence theorem applies to (10), giving\n\n  lim_{n→∞} n^{1-d}Sₙ,d = ∑_{j=0}^{∞} d e^{−j} = d·\\frac{e}{e-1}.  (13)\n\nThis completes part (b) and the proof. ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.449289",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension: the sum ranges over the d-dimensional lattice cube, not just a one–dimensional list.  \n• Additional combinatorics: counting points on “layers’’ requires the identity M_d(k) = (k+1)^d – k^d and brings discrete derivatives of high powers into play.  \n• Stronger estimates: the proof demands simultaneous control of an exponential factor and a polynomial one of degree d–1, forcing refined inequalities like (1–t)ⁿ ≤ e^{–nt} and both upper and lower Taylor bounds for (x+1)^d – x^d.  \n• Asymptotics in several variables: the limit extraction hinges on rescaling by n^{d–1}, an application of the mean–value theorem in high dimension, and the dominated convergence theorem to interchange limit and infinite sum.  \n• Multiple interacting concepts: convexity, exponential inequalities, discrete calculus, series summation, and measure-theoretic convergence all appear.  \n\nThese layers of technicality and the passage from one dimension to arbitrary d make the enhanced problem significantly harder than either the original or the current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}