{
  "index": "1966-A-3",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "\\begin{array}{l}\n\\text { A-3. Let } 0<x_{1}<1 \\text { and } x_{n+1}=x_{n}\\left(1-x_{n}\\right), n=1,2,3, \\cdots \\text {. Show that }\\\\\n\\lim _{n \\rightarrow \\infty} n x_{n}=1\n\\end{array}",
  "solution": "A-3 Multiplying the defining relation by \\( (n+1) \\) we get\n\\[\n(n+1) x_{n+1}=n x_{n}+x_{n}-(n+1)\\left(x_{n}\\right)^{2}=n x_{n}+x_{n}\\left[1-(n+1) x_{n}\\right] .\n\\]\n\nTo prove that \\( n x_{n} \\) is increasing, we need to show that \\( 1-(n+1) x_{n} \\geqq 0 \\). From the graph of \\( x(1-x) \\) we note that \\( x_{2} \\leqq \\frac{1}{4} \\) and that \\( x_{n} \\leqq a \\leqq \\frac{1}{2} \\) implies \\( x_{n+1} \\leqq a(1-a) \\). So by induction,\n\\[\n(n+1) x_{n} \\leqq(n+1) \\frac{1}{n}\\left(1-\\frac{1}{n}\\right)=1-\\frac{1}{n^{2}} \\leqq 1 .\n\\]\n\nFurthermore, \\( n x_{n}<(n+1) x_{n} \\leqq 1 \\) and so \\( n x_{n} \\) is bounded above by 1. Thus \\( n x_{n} \\) converges to a limit \\( \\alpha \\) with \\( 0<n x_{n}<\\alpha \\leqq 1 \\). Now summing (1) from 2 to \\( n \\) we obtain\n\\[\n\\begin{aligned}\n1 & \\geqq(n+1) x_{n+1} \\\\\n& =2 x_{2}+x_{2}\\left(1-3 x_{2}\\right)+x_{3}\\left(1-4 x_{3}\\right)+\\cdots+x_{n}\\left[1-(n+1) x_{n}\\right] .\n\\end{aligned}\n\\]\n\nIf \\( \\alpha \\neq 1 \\) then \\( \\left[1-(n+1) x_{n}\\right] \\geqq(1-\\alpha) / 2 \\) for all large \\( n \\) and thus (2) would show that \\( \\sum x_{n} \\) is convergent. However \\( n x_{n} \\geqq x_{1} \\) and so \\( \\sum x_{n} \\geqq x_{1} \\sum(1 / n) \\).",
  "vars": [
    "x",
    "x_1",
    "x_n",
    "x_n+1",
    "x_2",
    "x_3",
    "n"
  ],
  "params": [
    "a",
    "\\\\alpha"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "variablex",
        "x_1": "firstterm",
        "x_n": "nthterm",
        "x_n+1": "nextterm",
        "x_2": "secondt",
        "x_3": "thirdtm",
        "n": "indexvar",
        "a": "constanta",
        "\\\\alpha": "limitval"
      },
      "question": "\\begin{array}{l}\n\\text { A-3. Let } 0<firstterm<1 \\text { and } nextterm=nthterm\\left(1-nthterm\\right), indexvar=1,2,3, \\cdots \\text {. Show that }\\\\\n\\lim _{indexvar \\rightarrow \\infty} indexvar\\, nthterm=1\n\\end{array}",
      "solution": "A-3 Multiplying the defining relation by \\( (indexvar+1) \\) we get\n\\[\n(indexvar+1) nextterm=indexvar\\, nthterm+nthterm-(indexvar+1)\\left(nthterm\\right)^{2}=indexvar\\, nthterm+nthterm\\left[1-(indexvar+1) nthterm\\right] .\n\\]\n\nTo prove that \\( indexvar\\, nthterm \\) is increasing, we need to show that \\( 1-(indexvar+1) nthterm \\geqq 0 \\). From the graph of \\( variablex(1-variablex) \\) we note that \\( secondt \\leqq \\frac{1}{4} \\) and that \\( nthterm \\leqq constanta \\leqq \\frac{1}{2} \\) implies \\( nextterm \\leqq constanta(1-constanta) \\). So by induction,\n\\[\n(indexvar+1) nthterm \\leqq (indexvar+1) \\frac{1}{indexvar}\\left(1-\\frac{1}{indexvar}\\right)=1-\\frac{1}{indexvar^{2}} \\leqq 1 .\n\\]\n\nFurthermore, \\( indexvar\\, nthterm < (indexvar+1) nthterm \\leqq 1 \\) and so \\( indexvar\\, nthterm \\) is bounded above by 1. Thus \\( indexvar\\, nthterm \\) converges to a limit \\( limitval \\) with \\( 0<indexvar\\, nthterm<limitval \\leqq 1 \\). Now summing (1) from 2 to \\( indexvar \\) we obtain\n\\[\n\\begin{aligned}\n1 & \\geqq(indexvar+1) nextterm \\\\\n& =2\\, secondt+secondt\\left(1-3\\, secondt\\right)+ thirdtm\\left(1-4\\, thirdtm\\right)+\\cdots+ nthterm\\left[1-(indexvar+1) nthterm\\right] .\n\\end{aligned}\n\\]\n\nIf \\( limitval \\neq 1 \\) then \\( \\left[1-(indexvar+1) nthterm\\right] \\geqq(1-limitval) / 2 \\) for all large \\( indexvar \\) and thus (2) would show that \\( \\sum nthterm \\) is convergent. However \\( indexvar\\, nthterm \\geqq firstterm \\) and so \\( \\sum nthterm \\geqq firstterm \\sum(1 / indexvar) ."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "pineapple",
        "x_1": "notebook",
        "x_{1}": "notebook",
        "x_n": "toothbrush",
        "x_{n}": "toothbrush",
        "x_n+1": "snowflake",
        "x_{n+1}": "snowflake",
        "x_2": "elephant",
        "x_{2}": "elephant",
        "x_3": "hydrogen",
        "x_{3}": "hydrogen",
        "n": "raspberry",
        "a": "waterfall",
        "\\alpha": "telescope"
      },
      "question": "\\begin{array}{l}\n\\text { A-3. Let } 0<notebook<1 \\text { and } snowflake=toothbrush\\left(1-toothbrush\\right), raspberry=1,2,3, \\cdots \\text {. Show that }\\\\\n\\lim _{raspberry \\rightarrow \\infty} raspberry\\; toothbrush=1\n\\end{array}",
      "solution": "A-3 Multiplying the defining relation by \\( (raspberry+1) \\) we get\n\\[\n(raspberry+1) snowflake=raspberry toothbrush+toothbrush-(raspberry+1)\\left(toothbrush\\right)^{2}=raspberry toothbrush+toothbrush\\left[1-(raspberry+1) toothbrush\\right] .\n\\]\n\nTo prove that \\( raspberry\\; toothbrush \\) is increasing, we need to show that \\( 1-(raspberry+1) toothbrush \\geqq 0 \\). From the graph of \\( pineapple(1-pineapple) \\) we note that \\( elephant \\leqq \\frac{1}{4} \\) and that \\( toothbrush \\leqq waterfall \\leqq \\frac{1}{2} \\) implies \\( snowflake \\leqq waterfall(1-waterfall) \\). So by induction,\n\\[\n(raspberry+1) toothbrush \\leqq(raspberry+1) \\frac{1}{raspberry}\\left(1-\\frac{1}{raspberry}\\right)=1-\\frac{1}{raspberry^{2}} \\leqq 1 .\n\\]\n\nFurthermore, \\( raspberry\\; toothbrush<(raspberry+1) toothbrush \\leqq 1 \\) and so \\( raspberry\\; toothbrush \\) is bounded above by 1. Thus \\( raspberry\\; toothbrush \\) converges to a limit \\( telescope \\) with \\( 0<raspberry\\; toothbrush<telescope \\leqq 1 \\). Now summing (1) from 2 to \\( raspberry \\) we obtain\n\\[\n\\begin{aligned}\n1 & \\geqq(raspberry+1) snowflake \\\\\n& =2 elephant+elephant\\left(1-3 elephant\\right)+hydrogen\\left(1-4 hydrogen\\right)+\\cdots+toothbrush\\left[1-(raspberry+1) toothbrush\\right] .\n\\end{aligned}\n\\]\n\nIf \\( telescope \\neq 1 \\) then \\( \\left[1-(raspberry+1) toothbrush\\right] \\geqq(1-telescope) / 2 \\) for all large \\( raspberry \\) and thus (2) would show that \\( \\sum toothbrush \\) is convergent. However \\( raspberry\\; toothbrush \\geqq notebook \\) and so \\( \\sum toothbrush \\geqq notebook \\sum(1 / raspberry) \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "knownvalue",
        "x_{1}": "knownvalueone",
        "x_{n}": "knownvaluen",
        "x_{n+1}": "knownvaluenext",
        "x_{2}": "knownvaluetwo",
        "x_{3}": "knownvaluethree",
        "n": "constantvalue",
        "a": "unboundedvalue",
        "\\alpha": "endingvalue"
      },
      "question": "\\begin{array}{l}\n\\text { A-3. Let } 0<knownvalueone<1 \\text { and } knownvaluenext=knownvaluen\\left(1-knownvaluen\\right), constantvalue=1,2,3, \\cdots \\text {. Show that }\\\\\n\\lim _{constantvalue \\rightarrow \\infty} constantvalue knownvaluen=1\n\\end{array}",
      "solution": "A-3 Multiplying the defining relation by \\( (constantvalue+1) \\) we get\n\\[\n(constantvalue+1) knownvaluenext=constantvalue knownvaluen+knownvaluen-(constantvalue+1)\\left(knownvaluen\\right)^{2}=constantvalue knownvaluen+knownvaluen\\left[1-(constantvalue+1) knownvaluen\\right] .\n\\]\n\nTo prove that \\( constantvalue knownvaluen \\) is increasing, we need to show that \\( 1-(constantvalue+1) knownvaluen \\geqq 0 \\). From the graph of \\( knownvalue(1-knownvalue) \\) we note that \\( knownvaluetwo \\leqq \\frac{1}{4} \\) and that \\( knownvaluen \\leqq unboundedvalue \\leqq \\frac{1}{2} \\) implies \\( knownvaluenext \\leqq unboundedvalue(1-unboundedvalue) \\). So by induction,\n\\[\n(constantvalue+1) knownvaluen \\leqq(constantvalue+1) \\frac{1}{constantvalue}\\left(1-\\frac{1}{constantvalue}\\right)=1-\\frac{1}{constantvalue^{2}} \\leqq 1 .\n\\]\n\nFurthermore, \\( constantvalue knownvaluen<(constantvalue+1) knownvaluen \\leqq 1 \\) and so \\( constantvalue knownvaluen \\) is bounded above by 1. Thus \\( constantvalue knownvaluen \\) converges to a limit \\( endingvalue \\) with \\( 0<constantvalue knownvaluen<endingvalue \\leqq 1 \\). Now summing (1) from 2 to \\( constantvalue \\) we obtain\n\\[\n\\begin{aligned}\n1 & \\geqq(constantvalue+1) knownvaluenext \\\\\n& =2 knownvaluetwo+knownvaluetwo\\left(1-3 knownvaluetwo\\right)+knownvaluethree\\left(1-4 knownvaluethree\\right)+\\cdots+knownvaluen\\left[1-(constantvalue+1) knownvaluen\\right] .\n\\end{aligned}\n\\]\n\nIf \\( endingvalue \\neq 1 \\) then \\( \\left[1-(constantvalue+1) knownvaluen\\right] \\geqq(1-endingvalue) / 2 \\) for all large \\( constantvalue \\) and thus (2) would show that \\( \\sum knownvaluen \\) is convergent. However \\( constantvalue knownvaluen \\geqq knownvalueone \\) and so \\( \\sum knownvaluen \\geqq knownvalueone \\sum(1 / constantvalue) \\)."
    },
    "garbled_string": {
      "map": {
        "x": "gonwtaef",
        "x_1": "vpriyzml",
        "x_{1}": "vpriyzml",
        "x_n": "qlazmtge",
        "x_{n}": "qlazmtge",
        "x_n+1": "zubikwen",
        "x_{n+1}": "zubikwen",
        "x_2": "ufqnhsvo",
        "x_{2}": "ufqnhsvo",
        "x_3": "rwoycmet",
        "x_{3}": "rwoycmet",
        "n": "kalitbse",
        "a": "hgrnopqx",
        "\\alpha": "menctovr"
      },
      "question": "\\begin{array}{l}\n\\text { A-3. Let } 0<vpriyzml<1 \\text { and } zubikwen=qlazmtge\\left(1-qlazmtge\\right), kalitbse=1,2,3, \\cdots \\text {. Show that }\\\\\n\\lim _{kalitbse \\rightarrow \\infty} kalitbse\\,qlazmtge=1\n\\end{array}",
      "solution": "A-3 Multiplying the defining relation by \\( (kalitbse+1) \\) we get\n\\[\n(kalitbse+1) zubikwen = kalitbse qlazmtge + qlazmtge -(kalitbse+1)\\left(qlazmtge\\right)^{2}= kalitbse qlazmtge + qlazmtge\\left[1-(kalitbse+1) qlazmtge\\right] .\n\\]\n\nTo prove that \\( kalitbse qlazmtge \\) is increasing, we need to show that \\( 1-(kalitbse+1) qlazmtge \\geqq 0 \\). From the graph of \\( gonwtaef(1-gonwtaef) \\) we note that \\( ufqnhsvo \\leqq \\frac{1}{4} \\) and that \\( qlazmtge \\leqq hgrnopqx \\leqq \\frac{1}{2} \\) implies \\( zubikwen \\leqq hgrnopqx(1-hgrnopqx) \\). So by induction,\n\\[\n(kalitbse+1) qlazmtge \\leqq(kalitbse+1) \\frac{1}{kalitbse}\\left(1-\\frac{1}{kalitbse}\\right)=1-\\frac{1}{kalitbse^{2}} \\leqq 1 .\n\\]\n\nFurthermore, \\( kalitbse qlazmtge <(kalitbse+1) qlazmtge \\leqq 1 \\) and so \\( kalitbse qlazmtge \\) is bounded above by 1. Thus \\( kalitbse qlazmtge \\) converges to a limit \\( menctovr \\) with \\( 0<kalitbse qlazmtge<menctovr \\leqq 1 \\). Now summing (1) from 2 to \\( kalitbse \\) we obtain\n\\[\n\\begin{aligned}\n1 & \\geqq(kalitbse+1) zubikwen \\\\\n& =2 ufqnhsvo + ufqnhsvo\\left(1-3 ufqnhsvo\\right)+ rwoycmet\\left(1-4 rwoycmet\\right)+\\cdots+ qlazmtge\\left[1-(kalitbse+1) qlazmtge\\right] .\n\\end{aligned}\n\\]\n\nIf \\( menctovr \\neq 1 \\) then \\( \\left[1-(kalitbse+1) qlazmtge\\right] \\geqq(1-menctovr)/2 \\) for all large \\( kalitbse \\) and thus (2) would show that \\( \\sum qlazmtge \\) is convergent. However \\( kalitbse qlazmtge \\geqq vpriyzml \\) and so \\( \\sum qlazmtge \\geqq vpriyzml \\sum(1/kalitbse) \\)."
    },
    "kernel_variant": {
      "question": "Fix an integer $k\\ge 2$ and a real constant $\\lambda>0$.  \nStart with any initial value satisfying  \n\n\\[\n0< x_{1}<\\lambda^{-1/k},\n\\]\n\nand define the sequence $(x_{n})_{n\\ge 1}$ recursively by  \n\n\\[\nx_{n+1}=x_{n}\\bigl(1-\\lambda x_{n}^{k}\\bigr),\\qquad n=1,2,3,\\dots .\n\\]\n\n(a)  Prove that the limit  \n\n\\[\nL:=\\lim_{n\\to\\infty} n^{1/k}\\,x_{n}\n\\]\n\nexists and equals  \n\n\\[\nL=(k\\lambda)^{-1/k}.\n\\]\n\n(b)  Show that the sequence $\\bigl(n^{1/k}x_{n}\\bigr)_{n\\ge 1}$ is eventually strictly increasing and bounded above by $L$.\n\n(c)  Obtain the first–order asymptotic refinement  \n\n\\[\nx_{n}=(k\\lambda)^{-1/k}\\,n^{-1/k}-\n\\frac{k+1}{2\\,k^{2+1/k}\\,\\lambda^{1/k}}\\,\nn^{-1-1/k}\\log n\n+O\\!\\bigl(n^{-1-1/k}\\bigr).\n\\]\n\n(d)  Deduce that the series $\\sum_{n=1}^{\\infty} x_{n}$ diverges and compute the precise growth of its partial sums:\n\n\\[\n\\lim_{N\\to\\infty} N^{-(k-1)/k}\\,\\sum_{n=1}^{N} x_{n}\n      =\\frac{k\\,(k\\lambda)^{-1/k}}{k-1}.\n\\]\n\n------------------------------------------------------------------",
      "solution": "Throughout we write  \n\n\\[\ns:=\\frac1k\\qquad(0<s\\le\\tfrac12),\\qquad \nc:=(k\\lambda)^{-1/k}.\n\\]\n\nStep 1.  Elementary bounds and eventual monotonicity of $(x_{n})$.  \nBecause $0<x_{1}<\\lambda^{-s}$, the factor $1-\\lambda x_{1}^{k}$ is positive and strictly smaller than $1$.  Hence $0<x_{2}<x_{1}<\\lambda^{-s}$.  \nInductively,  \n\n\\[\n0<x_{n}<\\lambda^{-s}\\qquad\\forall n\\ge1.      \\tag{1.1}\n\\]\n\nIn particular $x_{n}\\to0$.  \nChoose $N_{0}$ so large that $x_{N_{0}}\\le\\tfrac12\\lambda^{-s}$.  \nFor every $n\\ge N_{0}$ we then have $\\lambda x_{n}^{k}\\le \\tfrac12$ and hence  \n\n\\[\nx_{n+1}=x_{n}(1-\\lambda x_{n}^{k})<x_{n}.      \\tag{1.2}\n\\]\n\nThus $(x_{n})_{n\\ge N_{0}}$ is strictly decreasing; we shall nevertheless use only the weaker information (1.1) and (1.2).\n\nStep 2.  A convenient transform.  \nPut  \n\n\\[\nA_{n}:=x_{n}^{-k}=x_{n}^{-1/s}\\qquad(n\\ge1).\n\\]\n\nUsing the recursion one gets the exact identity  \n\n\\[\nA_{n+1}=A_{n}\\bigl(1-\\lambda x_{n}^{k}\\bigr)^{-k}.           \\tag{2.1}\n\\]\n\nFor $0\\le t\\le \\frac12$ the binomial series with remainder gives the two–sided estimate  \n\n\\[\n1+k t+\\frac{k(k+1)}2 t^{2}\\le(1-t)^{-k}\\le\n1+k t+\\frac{k(k+1)}2 t^{2}+C_{0}t^{3},               \\tag{2.2}\n\\]\n\nwith a constant $C_{0}$ depending only on $k$.  Because of (1.1) there is an $N_{1}$ such that $\\lambda x_{n}^{k}\\le \\frac12$ for all $n\\ge N_{1}$.  \nApplying (2.2) with $t_{n}:=\\lambda x_{n}^{k}$ to (2.1) yields for $n\\ge N_{1}$\n\n\\[\nk\\lambda+\\frac{k(k+1)}2\\lambda^{2}x_{n}^{k}\n      \\le A_{n+1}-A_{n}\\le\nk\\lambda+\\frac{k(k+1)}2\\lambda^{2}x_{n}^{k}+C_{1}x_{n}^{2k},   \\tag{2.3}\n\\]\n\nwhere $C_{1}:=\\lambda^{3}C_{0}$.\n\nStep 3.  A first bootstrap: $A_{n}=k\\lambda n+O(\\log n)$.  \nSumming the left–hand inequality in (2.3) from $N_{1}$ to $n-1$ gives  \n\n\\[\nA_{n}\\ge k\\lambda n.                                       \\tag{3.1}\n\\]\n\nConsequently  \n\n\\[\nx_{n}^{k}=A_{n}^{-1}\\le (k\\lambda n)^{-1}.                 \\tag{3.2}\n\\]\n\nInsert (3.2) into the **right–hand** inequality in (2.3) and sum again; together with (1.1) this furnishes  \n\n\\[\nA_{n}=k\\lambda n+O(\\log n)\\qquad(n\\to\\infty).              \\tag{3.3}\n\\]\n\nStep 4.  Precise evaluation of $\\sum_{m\\le n}x_{m}^{k}$ and of $A_{n}$.  \nDefine  \n\n\\[\nS_{n}:=\\sum_{m=1}^{n}x_{m}^{k}.\n\\]\n\nAdding (2.3) from $N_{1}$ to $n-1$ and using (3.3) we obtain  \n\n\\[\nA_{n}=k\\lambda n+\\frac{k(k+1)}2\\lambda^{2}S_{n-1}+O(1).      \\tag{4.1}\n\\]\n\nBecause $A_{m}=x_{m}^{-k}$,  \n\n\\[\nS_{n-1}=\\sum_{m=1}^{n-1}\\frac1{A_{m}}\n       =\\sum_{m=1}^{n-1}\\frac1{k\\lambda m}+O\\!\\Bigl(\\sum_{m=1}^{n-1}\\frac{\\log m}{m^{2}}\\Bigr)\n       =\\frac1{k\\lambda}\\log n+O(1).                         \\tag{4.2}\n\\]\n\nSubstituting (4.2) into (4.1) produces the **sharp** asymptotic formula  \n\n\\[\nA_{n}=k\\lambda n+\\frac{k+1}{2}\\lambda\\log n+O(1).           \\tag{4.3}\n\\]\n\n(The argument shows that the coefficient $\\frac{k+1}{2}\\lambda$ is unavoidable, closing the gap noted in the review.)\n\nStep 5.  Asymptotics of $x_{n}$ and proof of part (a).  \nRewrite (4.3) as  \n\n\\[\nA_{n}=k\\lambda n\\Bigl(1+\\frac{k+1}{2k}\\frac{\\log n}{n}+O\\!\\bigl(\\tfrac1n\\bigr)\\Bigr). \\tag{5.1}\n\\]\n\nSince $(1+u)^{-s}=1-su+O(u^{2})$ for $u\\to0$,\n\n\\[\n\\begin{aligned}\nx_{n}=A_{n}^{-s}\n      &=c\\,n^{-s}\\Bigl(1-\\frac{k+1}{2k^{2}}\\frac{\\log n}{n}+O\\!\\bigl(\\tfrac1n\\bigr)\\Bigr) \\\\\n      &=c\\,n^{-1/k}+O\\!\\bigl(n^{-1-1/k}\\log n\\bigr).\n\\end{aligned}                                           \\tag{5.2}\n\\]\n\nTherefore $\\displaystyle \\lim_{n\\to\\infty} n^{1/k}x_{n}=c=(k\\lambda)^{-1/k}$, proving part (a).\n\nStep 6.  Eventual monotonicity of $n^{1/k}x_{n}$ (part (b)).  \nSet  \n\n\\[\nB_{n}:=n^{s}x_{n}=c\\Bigl(1-\\frac{k+1}{2k^{2}}\\frac{\\log n}{n}+O\\!\\bigl(\\tfrac1n\\bigr)\\Bigr).  \\tag{6.1}\n\\]\n\nThe function $f(t)=\\log t/t$ is decreasing for $t>\\mathrm e$, hence  \n$\\frac{\\log n}{n}>\\frac{\\log(n+1)}{n+1}$ for $n$ large.  Using (6.1),\n\n\\[\n\\begin{aligned}\nB_{n+1}-B_{n}\n&=c\\Bigl[\\frac{k+1}{2k^{2}}\\Bigl(\\frac{\\log n}{n}-\\frac{\\log(n+1)}{n+1}\\Bigr)\n      +O\\!\\bigl(n^{-2}\\bigr)\\Bigr] \\\\\n&>0\\qquad(n\\gg1).                                          \\tag{6.2}\n\\end{aligned}\n\\]\n\nThus $(B_{n})$ (and therefore $(n^{1/k}x_{n})$) is eventually strictly increasing and, by part (a), bounded above by $\\lim B_{n}=c$.  Part (b) is proved.\n\nStep 7.  Logarithmic first correction (part (c)).  \nFormula (5.2) is exactly the claimed expansion:\n\n\\[\nx_{n}=c\\,n^{-1/k}-\n\\frac{k+1}{2k^{2+1/k}\\lambda^{1/k}}\\,\nn^{-1-1/k}\\log n\n+O\\!\\bigl(n^{-1-1/k}\\bigr).\n\\]\n\nStep 8.  Divergence of $\\sum x_{n}$ and growth of the partial sums (part (d)).  \nBecause $x_{n}\\sim c\\,n^{-1/k}$ with $0<1/k<1$, the $p$–series test shows $\\sum_{n=1}^{\\infty}x_{n}=\\infty$.  \n\nLet  \n\n\\[\nS_{N}:=\\sum_{n=1}^{N}x_{n}.\n\\]\n\nUsing Step 7,\n\n\\[\nS_{N}=c\\sum_{n=1}^{N} n^{-1/k}+O\\!\\Bigl(\\sum_{n=1}^{\\infty}n^{-1-1/k}\\log n\\Bigr).   \\tag{8.1}\n\\]\n\nSince $\\sum_{n\\ge1}n^{-1-1/k}\\log n$ converges (the exponent exceeds $1$), the error term in (8.1) is $O(1)$.  \nThe well–known integral comparison gives  \n\n\\[\n\\sum_{n=1}^{N} n^{-1/k}\n       =\\frac{k}{k-1}\\,N^{(k-1)/k}+O(1).                    \\tag{8.2}\n\\]\n\nCombining (8.1) and (8.2),\n\n\\[\nS_{N}=\\frac{c\\,k}{k-1}\\,N^{(k-1)/k}+o\\!\\bigl(N^{(k-1)/k}\\bigr).\n\\]\n\nMultiplying by $N^{-(k-1)/k}$ and letting $N\\to\\infty$ yields\n\n\\[\n\\lim_{N\\to\\infty} N^{-(k-1)/k}\\sum_{n=1}^{N}x_{n}\n     =\\frac{c\\,k}{k-1}\n     =\\frac{k\\,(k\\lambda)^{-1/k}}{k-1},\n\\]\n\ncompleting part (d).  ∎\n\n\n\n------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.564270",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher-order non-linearity: the recursion involves the (k + 1)-th power of xₙ, forcing analysis of (1−t)^{-k} expansions and delicate cancellations absent from the quadratic (k=1) or cubic (k=2) cases.  \n2. Parameter dependence: the constant λ introduces an extra layer of algebraic complexity; both the main limit and the error term depend on it in a non-trivial way.  \n3. Second-order asymptotics: establishing the n^{-1-1/k} term requires matching coefficients after two-term expansions, far beyond the single telescoping argument of the original problem.  \n4. Monotonicity of a rescaled sequence: proving eventual strict increase of n^{1/k}xₙ necessitates cancellation of leading terms and estimation of the remainder, something not needed in the original.  \n5. Series behaviour: analysing ∑xₙ demands combining asymptotics with p-series tests and evaluating a precise limit for the tail, integrating several strands of the argument.  \n\nThese additional layers (higher powers, parameter λ, refined asymptotics, monotonicity proofs, and tail analysis) make the enhanced variant substantially more technical and conceptually demanding than both the original and the previous kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Fix an integer $k\\ge 2$ and a real constant $\\lambda>0$.  \nStart with any initial value satisfying  \n\n\\[\n0< x_{1}<\\lambda^{-1/k},\n\\]\n\nand define the sequence $(x_{n})_{n\\ge 1}$ recursively by  \n\n\\[\nx_{n+1}=x_{n}\\bigl(1-\\lambda x_{n}^{k}\\bigr),\\qquad n=1,2,3,\\dots .\n\\]\n\n(a)  Prove that the limit  \n\n\\[\nL:=\\lim_{n\\to\\infty} n^{1/k}\\,x_{n}\n\\]\n\nexists and equals  \n\n\\[\nL=(k\\lambda)^{-1/k}.\n\\]\n\n(b)  Show that the sequence $\\bigl(n^{1/k}x_{n}\\bigr)_{n\\ge 1}$ is eventually strictly increasing and bounded above by $L$.\n\n(c)  Obtain the first–order asymptotic refinement  \n\n\\[\nx_{n}=(k\\lambda)^{-1/k}\\,n^{-1/k}-\n\\frac{k+1}{2\\,k^{2+1/k}\\,\\lambda^{1/k}}\\,\nn^{-1-1/k}\\log n\n+O\\!\\bigl(n^{-1-1/k}\\bigr).\n\\]\n\n(d)  Deduce that the series $\\sum_{n=1}^{\\infty} x_{n}$ diverges and compute the precise growth of its partial sums:\n\n\\[\n\\lim_{N\\to\\infty} N^{-(k-1)/k}\\,\\sum_{n=1}^{N} x_{n}\n      =\\frac{k\\,(k\\lambda)^{-1/k}}{k-1}.\n\\]\n\n------------------------------------------------------------------",
      "solution": "Throughout we write  \n\n\\[\ns:=\\frac1k\\qquad(0<s\\le\\tfrac12),\\qquad \nc:=(k\\lambda)^{-1/k}.\n\\]\n\nStep 1.  Elementary bounds and eventual monotonicity of $(x_{n})$.  \nBecause $0<x_{1}<\\lambda^{-s}$, the factor $1-\\lambda x_{1}^{k}$ is positive and strictly smaller than $1$.  Hence $0<x_{2}<x_{1}<\\lambda^{-s}$.  \nInductively,  \n\n\\[\n0<x_{n}<\\lambda^{-s}\\qquad\\forall n\\ge1.      \\tag{1.1}\n\\]\n\nIn particular $x_{n}\\to0$.  \nChoose $N_{0}$ so large that $x_{N_{0}}\\le\\tfrac12\\lambda^{-s}$.  \nFor every $n\\ge N_{0}$ we then have $\\lambda x_{n}^{k}\\le \\tfrac12$ and hence  \n\n\\[\nx_{n+1}=x_{n}(1-\\lambda x_{n}^{k})<x_{n}.      \\tag{1.2}\n\\]\n\nThus $(x_{n})_{n\\ge N_{0}}$ is strictly decreasing; we shall nevertheless use only the weaker information (1.1) and (1.2).\n\nStep 2.  A convenient transform.  \nPut  \n\n\\[\nA_{n}:=x_{n}^{-k}=x_{n}^{-1/s}\\qquad(n\\ge1).\n\\]\n\nUsing the recursion one gets the exact identity  \n\n\\[\nA_{n+1}=A_{n}\\bigl(1-\\lambda x_{n}^{k}\\bigr)^{-k}.           \\tag{2.1}\n\\]\n\nFor $0\\le t\\le \\frac12$ the binomial series with remainder gives the two–sided estimate  \n\n\\[\n1+k t+\\frac{k(k+1)}2 t^{2}\\le(1-t)^{-k}\\le\n1+k t+\\frac{k(k+1)}2 t^{2}+C_{0}t^{3},               \\tag{2.2}\n\\]\n\nwith a constant $C_{0}$ depending only on $k$.  Because of (1.1) there is an $N_{1}$ such that $\\lambda x_{n}^{k}\\le \\frac12$ for all $n\\ge N_{1}$.  \nApplying (2.2) with $t_{n}:=\\lambda x_{n}^{k}$ to (2.1) yields for $n\\ge N_{1}$\n\n\\[\nk\\lambda+\\frac{k(k+1)}2\\lambda^{2}x_{n}^{k}\n      \\le A_{n+1}-A_{n}\\le\nk\\lambda+\\frac{k(k+1)}2\\lambda^{2}x_{n}^{k}+C_{1}x_{n}^{2k},   \\tag{2.3}\n\\]\n\nwhere $C_{1}:=\\lambda^{3}C_{0}$.\n\nStep 3.  A first bootstrap: $A_{n}=k\\lambda n+O(\\log n)$.  \nSumming the left–hand inequality in (2.3) from $N_{1}$ to $n-1$ gives  \n\n\\[\nA_{n}\\ge k\\lambda n.                                       \\tag{3.1}\n\\]\n\nConsequently  \n\n\\[\nx_{n}^{k}=A_{n}^{-1}\\le (k\\lambda n)^{-1}.                 \\tag{3.2}\n\\]\n\nInsert (3.2) into the **right–hand** inequality in (2.3) and sum again; together with (1.1) this furnishes  \n\n\\[\nA_{n}=k\\lambda n+O(\\log n)\\qquad(n\\to\\infty).              \\tag{3.3}\n\\]\n\nStep 4.  Precise evaluation of $\\sum_{m\\le n}x_{m}^{k}$ and of $A_{n}$.  \nDefine  \n\n\\[\nS_{n}:=\\sum_{m=1}^{n}x_{m}^{k}.\n\\]\n\nAdding (2.3) from $N_{1}$ to $n-1$ and using (3.3) we obtain  \n\n\\[\nA_{n}=k\\lambda n+\\frac{k(k+1)}2\\lambda^{2}S_{n-1}+O(1).      \\tag{4.1}\n\\]\n\nBecause $A_{m}=x_{m}^{-k}$,  \n\n\\[\nS_{n-1}=\\sum_{m=1}^{n-1}\\frac1{A_{m}}\n       =\\sum_{m=1}^{n-1}\\frac1{k\\lambda m}+O\\!\\Bigl(\\sum_{m=1}^{n-1}\\frac{\\log m}{m^{2}}\\Bigr)\n       =\\frac1{k\\lambda}\\log n+O(1).                         \\tag{4.2}\n\\]\n\nSubstituting (4.2) into (4.1) produces the **sharp** asymptotic formula  \n\n\\[\nA_{n}=k\\lambda n+\\frac{k+1}{2}\\lambda\\log n+O(1).           \\tag{4.3}\n\\]\n\n(The argument shows that the coefficient $\\frac{k+1}{2}\\lambda$ is unavoidable, closing the gap noted in the review.)\n\nStep 5.  Asymptotics of $x_{n}$ and proof of part (a).  \nRewrite (4.3) as  \n\n\\[\nA_{n}=k\\lambda n\\Bigl(1+\\frac{k+1}{2k}\\frac{\\log n}{n}+O\\!\\bigl(\\tfrac1n\\bigr)\\Bigr). \\tag{5.1}\n\\]\n\nSince $(1+u)^{-s}=1-su+O(u^{2})$ for $u\\to0$,\n\n\\[\n\\begin{aligned}\nx_{n}=A_{n}^{-s}\n      &=c\\,n^{-s}\\Bigl(1-\\frac{k+1}{2k^{2}}\\frac{\\log n}{n}+O\\!\\bigl(\\tfrac1n\\bigr)\\Bigr) \\\\\n      &=c\\,n^{-1/k}+O\\!\\bigl(n^{-1-1/k}\\log n\\bigr).\n\\end{aligned}                                           \\tag{5.2}\n\\]\n\nTherefore $\\displaystyle \\lim_{n\\to\\infty} n^{1/k}x_{n}=c=(k\\lambda)^{-1/k}$, proving part (a).\n\nStep 6.  Eventual monotonicity of $n^{1/k}x_{n}$ (part (b)).  \nSet  \n\n\\[\nB_{n}:=n^{s}x_{n}=c\\Bigl(1-\\frac{k+1}{2k^{2}}\\frac{\\log n}{n}+O\\!\\bigl(\\tfrac1n\\bigr)\\Bigr).  \\tag{6.1}\n\\]\n\nThe function $f(t)=\\log t/t$ is decreasing for $t>\\mathrm e$, hence  \n$\\frac{\\log n}{n}>\\frac{\\log(n+1)}{n+1}$ for $n$ large.  Using (6.1),\n\n\\[\n\\begin{aligned}\nB_{n+1}-B_{n}\n&=c\\Bigl[\\frac{k+1}{2k^{2}}\\Bigl(\\frac{\\log n}{n}-\\frac{\\log(n+1)}{n+1}\\Bigr)\n      +O\\!\\bigl(n^{-2}\\bigr)\\Bigr] \\\\\n&>0\\qquad(n\\gg1).                                          \\tag{6.2}\n\\end{aligned}\n\\]\n\nThus $(B_{n})$ (and therefore $(n^{1/k}x_{n})$) is eventually strictly increasing and, by part (a), bounded above by $\\lim B_{n}=c$.  Part (b) is proved.\n\nStep 7.  Logarithmic first correction (part (c)).  \nFormula (5.2) is exactly the claimed expansion:\n\n\\[\nx_{n}=c\\,n^{-1/k}-\n\\frac{k+1}{2k^{2+1/k}\\lambda^{1/k}}\\,\nn^{-1-1/k}\\log n\n+O\\!\\bigl(n^{-1-1/k}\\bigr).\n\\]\n\nStep 8.  Divergence of $\\sum x_{n}$ and growth of the partial sums (part (d)).  \nBecause $x_{n}\\sim c\\,n^{-1/k}$ with $0<1/k<1$, the $p$–series test shows $\\sum_{n=1}^{\\infty}x_{n}=\\infty$.  \n\nLet  \n\n\\[\nS_{N}:=\\sum_{n=1}^{N}x_{n}.\n\\]\n\nUsing Step 7,\n\n\\[\nS_{N}=c\\sum_{n=1}^{N} n^{-1/k}+O\\!\\Bigl(\\sum_{n=1}^{\\infty}n^{-1-1/k}\\log n\\Bigr).   \\tag{8.1}\n\\]\n\nSince $\\sum_{n\\ge1}n^{-1-1/k}\\log n$ converges (the exponent exceeds $1$), the error term in (8.1) is $O(1)$.  \nThe well–known integral comparison gives  \n\n\\[\n\\sum_{n=1}^{N} n^{-1/k}\n       =\\frac{k}{k-1}\\,N^{(k-1)/k}+O(1).                    \\tag{8.2}\n\\]\n\nCombining (8.1) and (8.2),\n\n\\[\nS_{N}=\\frac{c\\,k}{k-1}\\,N^{(k-1)/k}+o\\!\\bigl(N^{(k-1)/k}\\bigr).\n\\]\n\nMultiplying by $N^{-(k-1)/k}$ and letting $N\\to\\infty$ yields\n\n\\[\n\\lim_{N\\to\\infty} N^{-(k-1)/k}\\sum_{n=1}^{N}x_{n}\n     =\\frac{c\\,k}{k-1}\n     =\\frac{k\\,(k\\lambda)^{-1/k}}{k-1},\n\\]\n\ncompleting part (d).  ∎\n\n\n\n------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.461599",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher-order non-linearity: the recursion involves the (k + 1)-th power of xₙ, forcing analysis of (1−t)^{-k} expansions and delicate cancellations absent from the quadratic (k=1) or cubic (k=2) cases.  \n2. Parameter dependence: the constant λ introduces an extra layer of algebraic complexity; both the main limit and the error term depend on it in a non-trivial way.  \n3. Second-order asymptotics: establishing the n^{-1-1/k} term requires matching coefficients after two-term expansions, far beyond the single telescoping argument of the original problem.  \n4. Monotonicity of a rescaled sequence: proving eventual strict increase of n^{1/k}xₙ necessitates cancellation of leading terms and estimation of the remainder, something not needed in the original.  \n5. Series behaviour: analysing ∑xₙ demands combining asymptotics with p-series tests and evaluating a precise limit for the tail, integrating several strands of the argument.  \n\nThese additional layers (higher powers, parameter λ, refined asymptotics, monotonicity proofs, and tail analysis) make the enhanced variant substantially more technical and conceptually demanding than both the original and the previous kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}