{
  "index": "1996-B-2",
  "type": "ANA",
  "tag": [
    "ANA",
    "NT"
  ],
  "difficulty": "",
  "question": "Show that for every positive integer $n$,\n\\[\n\\left( \\frac{2n-1}{e} \\right)^{\\frac{2n-1}{2}} < 1 \\cdot 3 \\cdot 5\n\\cdots (2n-1) < \\left( \\frac{2n+1}{e} \\right)^{\\frac{2n+1}{2}}.\n\\]",
  "solution": "By estimating the area under the graph of $\\ln x$ using upper and\nlower rectangles of width 2, we get\n\\begin{align*}\n\\int_1^{2n-1} \\ln x\\,dx &\\leq 2(\\ln(3) + \\cdots + \\ln(2n-1)) \\\\\n&\\leq \\int_3^{2n+1} \\ln x\\,dx.\n\\end{align*}\nSince $\\int \\ln x\\,dx = x \\ln x - x + C$, we have, upon exponentiating\nand taking square roots,\n%\\begin{align*}\n%\\left( \\frac{2n-1}{e} \\right)^{\\frac{2n-1}{2}}\n%< (2n-1)^{\\frac{2n-1}{2}} e^{-n+1}\n%&\\leq 1 \\cdot 3 \\cdots (2n-1) \\\\\n%&\\leq (2n+1)^{\\frac{2n+1}{2}} \\frac{e^{-n+1}}{3^{3/2}}\n%< \\left( \\frac{2n+1}{e} \\right)^{\\frac{2n+1}{2}},\n%\\end{align*}\n\\begin{align*}\n\\left( \\frac{2n-1}{e} \\right)^{\\frac{2n-1}{2}}\n&< (2n-1)^{\\frac{2n-1}{2}} e^{-n+1} \\\\\n& \\leq 1 \\cdot 3 \\cdots (2n-1) \\\\\n& \\leq (2n+1)^{\\frac{2n+1}{2}} \\frac{e^{-n+1}}{3^{3/2}} \\\\\n& < \\left( \\frac{2n+1}{e} \\right)^{\\frac{2n+1}{2}},\n\\end{align*}\nusing the fact that $1 < e < 3$.",
  "vars": [
    "x"
  ],
  "params": [
    "n",
    "C"
  ],
  "sci_consts": [
    "e"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "realvalue",
        "n": "indexvalue",
        "C": "constante"
      },
      "question": "Show that for every positive integer $indexvalue$,\\[\n\\left( \\frac{2indexvalue-1}{e} \\right)^{\\frac{2indexvalue-1}{2}} < 1 \\cdot 3 \\cdot 5\n\\cdots (2indexvalue-1) < \\left( \\frac{2indexvalue+1}{e} \\right)^{\\frac{2indexvalue+1}{2}}.\\]",
      "solution": "By estimating the area under the graph of $\\ln realvalue$ using upper and\nlower rectangles of width 2, we get\n\\begin{align*}\n\\int_1^{2indexvalue-1} \\ln realvalue\\,drealvalue &\\leq 2(\\ln(3) + \\cdots + \\ln(2indexvalue-1)) \\\\\n&\\leq \\int_3^{2indexvalue+1} \\ln realvalue\\,drealvalue.\n\\end{align*}\nSince $\\int \\ln realvalue\\,drealvalue = realvalue \\ln realvalue - realvalue + constante$, we have, upon exponentiating\nand taking square roots,\n%\\begin{align*}\n%\\left( \\frac{2indexvalue-1}{e} \\right)^{\\frac{2indexvalue-1}{2}}\n%< (2indexvalue-1)^{\\frac{2indexvalue-1}{2}} e^{-indexvalue+1}\n%&\\leq 1 \\cdot 3 \\cdots (2indexvalue-1) \\\\\n%&\\leq (2indexvalue+1)^{\\frac{2indexvalue+1}{2}} \\frac{e^{-indexvalue+1}}{3^{3/2}}\n%< \\left( \\frac{2indexvalue+1}{e} \\right)^{\\frac{2indexvalue+1}{2}},\n%\\end{align*}\n\\begin{align*}\n\\left( \\frac{2indexvalue-1}{e} \\right)^{\\frac{2indexvalue-1}{2}}\n&< (2indexvalue-1)^{\\frac{2indexvalue-1}{2}} e^{-indexvalue+1} \\\\\n& \\leq 1 \\cdot 3 \\cdots (2indexvalue-1) \\\\\n& \\leq (2indexvalue+1)^{\\frac{2indexvalue+1}{2}} \\frac{e^{-indexvalue+1}}{3^{3/2}} \\\\\n& < \\left( \\frac{2indexvalue+1}{e} \\right)^{\\frac{2indexvalue+1}{2}},\n\\end{align*}\nusing the fact that $1 < e < 3$."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "sandcastle",
        "n": "algorithm",
        "C": "ballooner"
      },
      "question": "Show that for every positive integer $algorithm$,\n\\[\n\\left( \\frac{2algorithm-1}{e} \\right)^{\\frac{2algorithm-1}{2}} < 1 \\cdot 3 \\cdot 5\n\\cdots (2algorithm-1) < \\left( \\frac{2algorithm+1}{e} \\right)^{\\frac{2algorithm+1}{2}}.\n\\]",
      "solution": "By estimating the area under the graph of $\\ln sandcastle$ using upper and\nlower rectangles of width 2, we get\n\\begin{align*}\n\\int_1^{2algorithm-1} \\ln sandcastle\\,d sandcastle &\\leq 2(\\ln(3) + \\cdots + \\ln(2algorithm-1)) \\\\\n&\\leq \\int_3^{2algorithm+1} \\ln sandcastle\\,d sandcastle.\n\\end{align*}\nSince $\\int \\ln sandcastle\\,d sandcastle = sandcastle \\ln sandcastle - sandcastle + ballooner$, we have, upon exponentiating\nand taking square roots,\n%\\begin{align*}\n%\\left( \\frac{2algorithm-1}{e} \\right)^{\\frac{2algorithm-1}{2}}\n%< (2algorithm-1)^{\\frac{2algorithm-1}{2}} e^{-algorithm+1}\n%&\\leq 1 \\cdot 3 \\cdots (2algorithm-1) \\\\\n%&\\leq (2algorithm+1)^{\\frac{2algorithm+1}{2}} \\frac{e^{-algorithm+1}}{3^{3/2}}\n%< \\left( \\frac{2algorithm+1}{e} \\right)^{\\frac{2algorithm+1}{2}},\n%\\end{align*}\n\\begin{align*}\n\\left( \\frac{2algorithm-1}{e} \\right)^{\\frac{2algorithm-1}{2}}\n&< (2algorithm-1)^{\\frac{2algorithm-1}{2}} e^{-algorithm+1} \\\\\n& \\leq 1 \\cdot 3 \\cdots (2algorithm-1) \\\\\n& \\leq (2algorithm+1)^{\\frac{2algorithm+1}{2}} \\frac{e^{-algorithm+1}}{3^{3/2}} \\\\\n& < \\left( \\frac{2algorithm+1}{e} \\right)^{\\frac{2algorithm+1}{2}},\n\\end{align*}\nusing the fact that $1 < e < 3$. "
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "constantval",
        "n": "negfraction",
        "C": "variable"
      },
      "question": "Show that for every positive integer $negfraction$,\\[\n\\left( \\frac{2negfraction-1}{e} \\right)^{\\frac{2negfraction-1}{2}} < 1 \\cdot 3 \\cdot 5\n\\cdots (2negfraction-1) < \\left( \\frac{2negfraction+1}{e} \\right)^{\\frac{2negfraction+1}{2}}.\\]\n",
      "solution": "By estimating the area under the graph of $\\ln constantval$ using upper and\nlower rectangles of width 2, we get\n\\begin{align*}\n\\int_1^{2negfraction-1} \\ln constantval\\,dconstantval &\\leq 2(\\ln(3) + \\cdots + \\ln(2negfraction-1)) \\\\\n&\\leq \\int_3^{2negfraction+1} \\ln constantval\\,dconstantval.\n\\end{align*}\nSince $\\int \\ln constantval\\,dconstantval = constantval \\ln constantval - constantval + variable$, we have, upon exponentiating\nand taking square roots,\n%\\begin{align*}\n%\\left( \\frac{2negfraction-1}{e} \\right)^{\\frac{2negfraction-1}{2}}\n%< (2negfraction-1)^{\\frac{2negfraction-1}{2}} e^{-negfraction+1}\n%&\\leq 1 \\cdot 3 \\cdots (2negfraction-1) \\\\\n%&\\leq (2negfraction+1)^{\\frac{2negfraction+1}{2}} \\frac{e^{-negfraction+1}}{3^{3/2}}\n%< \\left( \\frac{2negfraction+1}{e} \\right)^{\\frac{2negfraction+1}{2}},\n%\\end{align*}\n\\begin{align*}\n\\left( \\frac{2negfraction-1}{e} \\right)^{\\frac{2negfraction-1}{2}}\n&< (2negfraction-1)^{\\frac{2negfraction-1}{2}} e^{-negfraction+1} \\\\\n& \\leq 1 \\cdot 3 \\cdots (2negfraction-1) \\\\\n& \\leq (2negfraction+1)^{\\frac{2negfraction+1}{2}} \\frac{e^{-negfraction+1}}{3^{3/2}} \\\\\n& < \\left( \\frac{2negfraction+1}{e} \\right)^{\\frac{2negfraction+1}{2}},\n\\end{align*}\nusing the fact that $1 < e < 3$.",
      "errors": []
    },
    "garbled_string": {
      "map": {
        "x": "hjgrksla",
        "n": "qzxwvtnp",
        "C": "pjzrmvdk"
      },
      "question": "Show that for every positive integer $qzxwvtnp$,\n\\[\n\\left( \\frac{2qzxwvtnp-1}{e} \\right)^{\\frac{2qzxwvtnp-1}{2}} < 1 \\cdot 3 \\cdot 5\n\\cdots (2qzxwvtnp-1) < \\left( \\frac{2qzxwvtnp+1}{e} \\right)^{\\frac{2qzxwvtnp+1}{2}}.\n\\]",
      "solution": "By estimating the area under the graph of $\\ln hjgrksla$ using upper and\nlower rectangles of width 2, we get\n\\begin{align*}\n\\int_1^{2qzxwvtnp-1} \\ln hjgrksla\\,dhjgrksla &\\leq 2(\\ln(3) + \\cdots + \\ln(2qzxwvtnp-1)) \\\\\n&\\leq \\int_3^{2qzxwvtnp+1} \\ln hjgrksla\\,dhjgrksla.\n\\end{align*}\nSince $\\int \\ln hjgrksla\\,dhjgrksla = hjgrksla \\ln hjgrksla - hjgrksla + pjzrmvdk$, we have, upon exponentiating\nand taking square roots,\n%\\begin{align*}\n%\\left( \\frac{2qzxwvtnp-1}{e} \\right)^{\\frac{2qzxwvtnp-1}{2}}\n%< (2qzxwvtnp-1)^{\\frac{2qzxwvtnp-1}{2}} e^{-qzxwvtnp+1}\n%&\\leq 1 \\cdot 3 \\cdots (2qzxwvtnp-1) \\\\\n%&\\leq (2qzxwvtnp+1)^{\\frac{2qzxwvtnp+1}{2}} \\frac{e^{-qzxwvtnp+1}}{3^{3/2}}\n%< \\left( \\frac{2qzxwvtnp+1}{e} \\right)^{\\frac{2qzxwvtnp+1}{2}},\n%\\end{align*}\n\\begin{align*}\n\\left( \\frac{2qzxwvtnp-1}{e} \\right)^{\\frac{2qzxwvtnp-1}{2}}\n&< (2qzxwvtnp-1)^{\\frac{2qzxwvtnp-1}{2}} e^{-qzxwvtnp+1} \\\\\n& \\leq 1 \\cdot 3 \\cdots (2qzxwvtnp-1) \\\\\n& \\leq (2qzxwvtnp+1)^{\\frac{2qzxwvtnp+1}{2}} \\frac{e^{-qzxwvtnp+1}}{3^{3/2}} \\\\\n& < \\left( \\frac{2qzxwvtnp+1}{e} \\right)^{\\frac{2qzxwvtnp+1}{2}},\n\\end{align*}\nusing the fact that $1 < e < 3$.",
      "latex": ""
    },
    "kernel_variant": {
      "question": "Let\n\\[\nP_n\\;=\\;1\\cdot4\\cdot7\\cdots(3n-2)=\\prod_{k=0}^{n-1}(3k+1),\\qquad n\\in\\mathbb N.\n\\]\nShow that for every positive integer $n$ one has\n\\[\n\\left(\\frac{3n-2}{e}\\right)^{\\!\\frac{3n-2}{3}}\n\\;<\\;P_n\\;<\\;\\left(\\frac{3n+1}{e}\\right)^{\\!\\frac{3n+1}{3}}.\n\\]",
      "solution": "1. Taking logarithms.  Put S_n = ln P_n = ∑_{k=0}^{n-1} ln(3k + 1).  The desired inequalities\n   ((3n−2)/e)^{(3n−2)/3} < P_n < ((3n+1)/e)^{(3n+1)/3}\nare equivalent to\n   (3n−2)/3·ln((3n−2)/e) < S_n < (3n+1)/3·ln((3n+1)/e).\n\n2. Estimate S_n by Riemann sums for ∫ln x dx with step Δx=3.  Since ln x is increasing, on each interval [3k+1, 3k+4] we have\n   3 ln(3k+1) ≤ ∫_{3k+1}^{3k+4} ln x dx ≤ 3 ln(3k+4).\nLower bound on S_n: summing the upper‐integral inequalities for k=0,…,n−2 gives\n   ∫_{1}^{3n−2} ln x dx ≤ 3 ∑_{k=0}^{n-2} ln(3k+4)\n                     = 3 ∑_{j=1}^{n-1} ln(3j+1)\n                     = 3 S_n.\nUpper bound on S_n: summing the lower‐integral inequalities for k=1,…,n−1 gives\n   3 ∑_{k=1}^{n-1} ln(3k+1)\n   = 3 S_n\n   ≤ ∫_{4}^{3n+1} ln x dx.\nHence\n   ∫_{1}^{3n−2} ln x dx ≤ 3 S_n ≤ ∫_{4}^{3n+1} ln x dx.\n\n3. Evaluate the integrals (since ∫ln x dx = x ln x − x + C):\n   ∫_{1}^{3n−2} ln x dx = (3n−2) ln(3n−2) − (3n−2) + 1,\n   ∫_{4}^{3n+1} ln x dx = (3n+1) ln(3n+1) − (3n+1) − (4 ln 4 − 4).\nDivide by 3:\n   [(3n−2) ln(3n−2) − (3n−2) + 1]/3 ≤ S_n ≤ [(3n+1) ln(3n+1) − (3n+1) − 4 ln 4 + 4]/3.\n\n4. Exponentiation:\n Lower bound:\n   P_n = e^{S_n}\n       ≥ exp([(3n−2) ln(3n−2) − (3n−2) + 1]/3)\n       = e^{1/3}\bigl((3n−2)/e\bigr)^{(3n−2)/3}\n       > \bigl((3n−2)/e\bigr)^{(3n−2)/3}.\n Upper bound:\n   P_n ≤ exp([(3n+1) ln(3n+1) − (3n+1) − 4 ln 4 + 4]/3)\n       = \bigl((3n+1)/e\bigr)^{(3n+1)/3}\bigl(e/4\bigr)^{4/3}.\n\n5. Since e < 4 we have (e/4)^{4/3} < 1, hence\n   P_n < \bigl((3n+1)/e\bigr)^{(3n+1)/3}.\n\nCombining with the lower bound gives the claimed\n   ((3n−2)/e)^{(3n−2)/3} < P_n < ((3n+1)/e)^{(3n+1)/3}.",
      "_meta": {
        "core_steps": [
          "Take natural logarithm so the product becomes a sum of logs",
          "Bound that sum above and below by right- and left-Riemann sums of ln x using rectangles of fixed width 2",
          "Evaluate the bounding integrals via ∫ln x dx = x ln x – x",
          "Exponentiate the inequalities and take the appropriate (square-)root to undo the earlier factor 2",
          "Insert the rough numerical bound 1 < e < 3 to tidy the end constants and obtain the stated double inequality"
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Common gap between successive factors / rectangle width ; determines the later square-root",
            "original": "2"
          },
          "slot2": {
            "description": "Specific constant chosen as an upper bound for e when simplifying the right-hand inequality",
            "original": "3"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}