{
  "index": "2008-A-4",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "Define $f: \\mathbb{R} \\to \\mathbb{R}$ by\n\\[\nf(x) = \\begin{cases} x & \\mbox{if $x \\leq e$} \\\\ x f(\\ln x) &\n\\mbox{if $x > e$.} \\end{cases}\n\\]\nDoes $\\sum_{n=1}^\\infty \\frac{1}{f(n)}$ converge?",
  "solution": "The sum diverges. From the definition, $f(x) = x$ on $[1,e]$, $x\\ln x$ on $(e,e^e]$, $x\\ln x\\ln\\ln x$ on $(e^e,e^{e^e}]$, and so forth. It follows that on $[1,\\infty)$, $f$ is positive, continuous, and increasing. Thus $\\sum_{n=1}^\\infty \\frac{1}{f(n)}$, if it converges, is bounded below by $\\int_1^{\\infty} \\frac{dx}{f(x)}$; it suffices to prove that the integral diverges.\n\nWrite $\\ln^1 x  = \\ln x $ and $\\ln^k x = \\ln(\\ln^{k-1} x)$ for $k \\geq 2$; similarly write $\\exp^1 x = e^x$ and $\\exp^k x  = e^{\\exp^{k-1} x}$. If we write $y = \\ln^k x$, then $x = \\exp^k y$ and $dx = (\\exp^ky)(\\exp^{k-1}y)\\cdots (\\exp^1y)dy =\nx(\\ln^1 x) \\cdots (\\ln^{k-1}x)dy$. Now on\n$[\\exp^{k-1} 1,\\exp^k 1]$, we have\n$f(x) = x(\\ln^1 x) \\cdots (\\ln^{k-1}x)$, and thus substituting $y=\\ln^k x$ yields\n\\[\n\\int_{\\exp^{k-1} 1}^{\\exp^k 1} \\frac{dx}{f(x)} =\n\\int_{0}^{1} dy = 1.\n\\]\nIt follows that $\\int_1^{\\infty} \\frac{dx}{f(x)} = \\sum_{k=1}^{\\infty} \\int_{\\exp^{k-1} 1}^{\\exp^k 1} \\frac{dx}{f(x)}$ diverges, as desired.",
  "vars": [
    "n",
    "x",
    "y",
    "k"
  ],
  "params": [
    "f"
  ],
  "sci_consts": [
    "e"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "n": "indexer",
        "x": "variable",
        "y": "logvalue",
        "k": "itercount",
        "f": "reccurve"
      },
      "question": "Define $reccurve: \\mathbb{R} \\to \\mathbb{R}$ by\n\\[\nreccurve(variable) = \\begin{cases} variable & \\mbox{if $variable \\leq e$} \\\\ variable\\,reccurve(\\ln variable) &\n\\mbox{if $variable > e$.} \\end{cases}\n\\]\nDoes $\\sum_{indexer=1}^{\\infty} \\frac{1}{reccurve(indexer)}$ converge?",
      "solution": "The sum diverges. From the definition, $reccurve(variable)=variable$ on $[1,e]$, $variable\\ln variable$ on $(e,e^e]$, $variable\\ln variable\\ln\\ln variable$ on $(e^e,e^{e^e}]$, and so forth. It follows that on $[1,\\infty)$, $reccurve$ is positive, continuous, and increasing. Thus $\\sum_{indexer=1}^{\\infty} \\frac{1}{reccurve(indexer)}$, if it converges, is bounded below by $\\int_{1}^{\\infty} \\frac{dvariable}{reccurve(variable)}$; it suffices to prove that the integral diverges.\n\nWrite $\\ln^{1}variable=\\ln variable$ and $\\ln^{itercount}variable=\\ln(\\ln^{itercount-1}variable)$ for $itercount\\ge 2$; similarly write $\\exp^{1}variable=e^{variable}$ and $\\exp^{itercount}variable=e^{\\exp^{itercount-1}variable}$. If we set $logvalue=\\ln^{itercount}variable$, then $variable=\\exp^{itercount}logvalue$ and\n\\[\ndvariable=(\\exp^{itercount}logvalue)(\\exp^{itercount-1}logvalue)\\cdots(\\exp^{1}logvalue)dlogvalue=variable(\\ln^{1}variable)\\cdots(\\ln^{itercount-1}variable)dlogvalue.\n\\]\nOn the interval $[\\exp^{itercount-1}1,\\exp^{itercount}1]$ we have $reccurve(variable)=variable(\\ln^{1}variable)\\cdots(\\ln^{itercount-1}variable)$, and substituting $logvalue=\\ln^{itercount}variable$ yields\n\\[\n\\int_{\\exp^{itercount-1}1}^{\\exp^{itercount}1}\\frac{dvariable}{reccurve(variable)}=\\int_{0}^{1}dlogvalue=1.\n\\]\nConsequently,\n\\[\n\\int_{1}^{\\infty}\\frac{dvariable}{reccurve(variable)}=\\sum_{itercount=1}^{\\infty}\\int_{\\exp^{itercount-1}1}^{\\exp^{itercount}1}\\frac{dvariable}{reccurve(variable)}\n\\]\ndiverges, and therefore the original series diverges as well."
    },
    "descriptive_long_confusing": {
      "map": {
        "n": "rainstorm",
        "x": "courtyard",
        "y": "lighthouse",
        "k": "wilderness",
        "f": "silhouette"
      },
      "question": "Define $silhouette: \\mathbb{R} \\to \\mathbb{R}$ by\n\\[\nsilhouette(courtyard) = \\begin{cases} courtyard & \\mbox{if $courtyard \\leq e$} \\\\ courtyard\\, silhouette(\\ln courtyard) &\n\\mbox{if $courtyard > e$.} \\end{cases}\n\\]\nDoes $\\sum_{rainstorm=1}^{\\infty} \\frac{1}{\\silhouette(rainstorm)}$ converge?",
      "solution": "The sum diverges. From the definition, $\\silhouette(courtyard) = courtyard$ on $[1,e]$, $courtyard\\ln courtyard$ on $(e,e^e]$, $courtyard\\ln courtyard\\ln\\ln courtyard$ on $(e^e,e^{e^e}]$, and so forth. It follows that on $[1,\\infty)$, $\\silhouette$ is positive, continuous, and increasing. Thus $\\sum_{rainstorm=1}^{\\infty} \\frac{1}{\\silhouette(rainstorm)}$, if it converges, is bounded below by $\\int_1^{\\infty} \\frac{d\\,courtyard}{\\silhouette(courtyard)}$; it suffices to prove that the integral diverges.\n\nWrite $\\ln^1 courtyard = \\ln courtyard$ and $\\ln^{wilderness} courtyard = \\ln(\\ln^{wilderness-1} courtyard)$ for $wilderness \\geq 2$; similarly write $\\exp^1 courtyard = e^{courtyard}$ and $\\exp^{wilderness} courtyard = e^{\\exp^{wilderness-1} courtyard}$. If we write $lighthouse = \\ln^{wilderness} courtyard$, then $courtyard = \\exp^{wilderness} lighthouse$ and $d\\,courtyard = (\\exp^{wilderness}lighthouse)(\\exp^{wilderness-1}lighthouse)\\cdots(\\exp^1lighthouse)d\\,lighthouse =\ncourtyard(\\ln^1 courtyard) \\cdots (\\ln^{wilderness-1}courtyard)d\\,lighthouse$. Now on\n$[\\exp^{wilderness-1} 1,\\exp^{wilderness} 1]$, we have\n$\\silhouette(courtyard) = courtyard(\\ln^1 courtyard) \\cdots (\\ln^{wilderness-1}courtyard)$, and thus substituting $lighthouse=\\ln^{wilderness} courtyard$ yields\n\\[\n\\int_{\\exp^{wilderness-1} 1}^{\\exp^{wilderness} 1} \\frac{d\\,courtyard}{\\silhouette(courtyard)} =\n\\int_{0}^{1} d\\,lighthouse = 1.\n\\]\nIt follows that $\\int_1^{\\infty} \\frac{d\\,courtyard}{\\silhouette(courtyard)} = \\sum_{wilderness=1}^{\\infty} \\int_{\\exp^{wilderness-1} 1}^{\\exp^{wilderness} 1} \\frac{d\\,courtyard}{\\silhouette(courtyard)}$ diverges, as desired."
    },
    "descriptive_long_misleading": {
      "map": {
        "n": "uncountable",
        "x": "knownvalue",
        "y": "horizontal",
        "k": "constant",
        "f": "fixedvalue"
      },
      "question": "Define $fixedvalue: \\mathbb{R} \\to \\mathbb{R}$ by\n\\[\nfixedvalue(knownvalue) = \\begin{cases} knownvalue & \\mbox{if $knownvalue \\leq e$} \\\\ knownvalue\\, fixedvalue(\\ln knownvalue) &\n\\mbox{if $knownvalue > e$.} \\end{cases}\n\\]\nDoes $\\sum_{uncountable=1}^{\\infty} \\frac{1}{fixedvalue(uncountable)}$ converge?",
      "solution": "The sum diverges. From the definition, $fixedvalue(knownvalue) = knownvalue$ on $[1,e]$, $knownvalue\\ln knownvalue$ on $(e,e^e]$, $knownvalue\\ln knownvalue\\ln\\ln knownvalue$ on $(e^e,e^{e^e}]$, and so forth. It follows that on $[1,\\infty)$, $fixedvalue$ is positive, continuous, and increasing. Thus $\\sum_{uncountable=1}^{\\infty} \\frac{1}{fixedvalue(uncountable)}$, if it converges, is bounded below by $\\int_1^{\\infty} \\frac{dknownvalue}{fixedvalue(knownvalue)}$; it suffices to prove that the integral diverges.\n\nWrite $\\ln^1 knownvalue  = \\ln knownvalue $ and $\\ln^{constant} knownvalue = \\ln(\\ln^{constant-1} knownvalue)$ for $constant \\geq 2$; similarly write $\\exp^1 knownvalue = e^{knownvalue}$ and $\\exp^{constant} knownvalue  = e^{\\exp^{constant-1} knownvalue}$. If we write $horizontal = \\ln^{constant} knownvalue$, then $knownvalue = \\exp^{constant} horizontal$ and $dknownvalue = (\\exp^{constant}horizontal)(\\exp^{constant-1}horizontal)\\cdots (\\exp^1horizontal)dhorizontal =\nknownvalue(\\ln^1 knownvalue) \\cdots (\\ln^{constant-1}knownvalue)dhorizontal$. Now on\n$[\\exp^{constant-1} 1,\\exp^{constant} 1]$, we have\n$fixedvalue(knownvalue) = knownvalue(\\ln^1 knownvalue) \\cdots (\\ln^{constant-1}knownvalue)$, and thus substituting $horizontal=\\ln^{constant} knownvalue$ yields\n\\[\n\\int_{\\exp^{constant-1} 1}^{\\exp^{constant} 1} \\frac{dknownvalue}{fixedvalue(knownvalue)} =\n\\int_{0}^{1} dhorizontal = 1.\n\\]\nIt follows that $\\int_1^{\\infty} \\frac{dknownvalue}{fixedvalue(knownvalue)} = \\sum_{constant=1}^{\\infty} \\int_{\\exp^{constant-1} 1}^{\\exp^{constant} 1} \\frac{dknownvalue}{fixedvalue(knownvalue)}$ diverges, as desired."
    },
    "garbled_string": {
      "map": {
        "n": "qzxwvtnpa",
        "x": "hjgrkslac",
        "y": "mvbztrneo",
        "k": "pltqwsrdi",
        "f": "jskdplmno"
      },
      "question": "Define $jskdplmno: \\mathbb{R} \\to \\mathbb{R}$ by\n\\[\njskdplmno(hjgrkslac) = \\begin{cases} hjgrkslac & \\mbox{if $hjgrkslac \\leq e$} \\\\ hjgrkslac \\, jskdplmno(\\ln hjgrkslac) &\n\\mbox{if $hjgrkslac > e$.} \\end{cases}\n\\]\nDoes $\\sum_{qzxwvtnpa=1}^\\infty \\dfrac{1}{jskdplmno(qzxwvtnpa)}$ converge?",
      "solution": "The sum diverges. From the definition, $jskdplmno(hjgrkslac) = hjgrkslac$ on $[1,e]$, $hjgrkslac\\ln hjgrkslac$ on $(e,e^e]$, $hjgrkslac\\ln hjgrkslac\\ln\\ln hjgrkslac$ on $(e^e,e^{e^e}]$, and so forth. It follows that on $[1,\\infty)$, $jskdplmno$ is positive, continuous, and increasing. Thus $\\sum_{qzxwvtnpa=1}^\\infty \\dfrac{1}{jskdplmno(qzxwvtnpa)}$, if it converges, is bounded below by $\\int_1^{\\infty} \\dfrac{d hjgrkslac}{jskdplmno(hjgrkslac)}$; it suffices to prove that the integral diverges.\n\nWrite $\\ln^1 hjgrkslac = \\ln hjgrkslac$ and $\\ln^{pltqwsrdi} hjgrkslac = \\ln(\\ln^{pltqwsrdi-1} hjgrkslac)$ for $pltqwsrdi \\ge 2$; similarly write $\\exp^1 hjgrkslac = e^{hjgrkslac}$ and $\\exp^{pltqwsrdi} hjgrkslac = e^{\\exp^{pltqwsrdi-1} hjgrkslac}$. If we write $mvbztrneo = \\ln^{pltqwsrdi} hjgrkslac$, then $hjgrkslac = \\exp^{pltqwsrdi} mvbztrneo$ and \n$dhjgrkslac = (\\exp^{pltqwsrdi}mvbztrneo)(\\exp^{pltqwsrdi-1}mvbztrneo)\\cdots (\\exp^1mvbztrneo)d mvbztrneo =\nhjgrkslac(\\ln^1 hjgrkslac) \\cdots (\\ln^{pltqwsrdi-1}hjgrkslac)d mvbztrneo$. Now on $[\\exp^{pltqwsrdi-1} 1,\\exp^{pltqwsrdi} 1]$, we have\n$jskdplmno(hjgrkslac) = hjgrkslac(\\ln^1 hjgrkslac) \\cdots (\\ln^{pltqwsrdi-1}hjgrkslac)$, and thus substituting $mvbztrneo = \\ln^{pltqwsrdi} hjgrkslac$ yields\n\\[\n\\int_{\\exp^{pltqwsrdi-1} 1}^{\\exp^{pltqwsrdi} 1} \\frac{dhjgrkslac}{jskdplmno(hjgrkslac)} = \\int_{0}^{1} d mvbztrneo = 1.\n\\]\nIt follows that $\\int_1^{\\infty} \\frac{dhjgrkslac}{jskdplmno(hjgrkslac)} = \\sum_{pltqwsrdi=1}^{\\infty} \\int_{\\exp^{pltqwsrdi-1} 1}^{\\exp^{pltqwsrdi} 1} \\frac{dhjgrkslac}{jskdplmno(hjgrkslac)}$ diverges, as desired."
    },
    "kernel_variant": {
      "question": "Let  \n\\[\n\\ln^{[0]}x:=x,\\qquad   \n\\ln^{[m]}x:=\\ln\\!\\bigl(\\ln^{[m-1]}x\\bigr)\\quad(m\\ge 1).\n\\]\n\nFor every {\\em bounded} sequence of non–negative real numbers  \n\\[\n\\boldsymbol a=(a_1,a_2,a_3,\\dots),\\qquad   \nM:=\\sup_{k\\ge 1}a_k<\\infty ,\n\\]\ndefine the piece-wise function  \n\\[\nF_{\\boldsymbol a}:(1,\\infty)\\longrightarrow(0,\\infty),\\qquad\nF_{\\boldsymbol a}(x)=\n\\begin{cases}\nx, & 1<x\\le e,\\\\[6pt]\nx\\,\\displaystyle\\prod_{j=1}^{k}\\bigl(\\ln^{[j]}x\\bigr)^{a_j},\n& e^{e^{\\,k-1}}<x\\le e^{e^{\\,k}}\\;(k\\ge 1).\n\\end{cases}\n\\]\n\nPut  \n\\[\ns(n):=\\dfrac1{F_{\\boldsymbol a}(n)},\\qquad \n\\Sigma(\\boldsymbol a):=\\sum_{n=\\lceil e\\rceil}^{\\infty}s(n).\n\\]\n\nLet  \n\\[\nm:=\\min\\{j\\ge 1:\\,a_j\\ne 1\\}\\qquad(m=\\infty\\text{ if }a_j=1\\text{ for all }j).\n\\]\n\nDetermine, {\\em solely in terms of the sequence $\\boldsymbol a$}, a necessary\nand sufficient condition for the series $\\Sigma(\\boldsymbol a)$ to converge.",
      "solution": "Throughout we employ the $\\exp$–iteration  \n\\[\n\\exp^{\\circ 0}r:=r,\\qquad \n\\exp^{\\circ m}r:=e^{\\exp^{\\circ(m-1)}r}\\quad(m\\ge 1),\n\\]\nand write  \n\\[\nB_r:=\\exp^{\\circ r}1,\\qquad \nD_r:=\\exp^{\\circ r}e\\qquad(r=0,1,2,\\dots).\n\\tag{1}\n\\]\n\nA.  {\\bf Statement of the correct criterion}\n\nLet  \n\\[\nm:=\\min\\{j\\ge 1:a_j\\ne 1\\}\\qquad(m=\\infty\\text{ if the set is empty}).\n\\]\nThen  \n\\[\n\\boxed{\\;\n\\Sigma(\\boldsymbol a)\\text{ converges } \\Longleftrightarrow\\;\nm<\\infty\\ \\text{ and }\\ a_m>1\\;}\n\\tag{$\\star$}\n\\]\n\nIn words: the very first index at which the sequence $\\boldsymbol a$\ndeviates from $1$ is the only decisive one; $\\Sigma(\\boldsymbol a)$ converges\niff that first deviation lies {\\em above} $1$.  We shall prove $(\\star)$ in\nseven steps, carefully repairing the flaws criticised in the review.\n\n------------------------------------------------------------------\nB.  {\\bf Integral test and block decomposition}\n\nBecause $F_{\\boldsymbol a}$ is strictly increasing on every open block\n\\(\nI_k:=(e^{e^{\\,k-1}},e^{e^{\\,k}}]\\ (k\\ge 1)\n\\)\nand has only upward jumps at the block end–points, the alternating\nversion of the integral test gives  \n\\[\n\\Sigma(\\boldsymbol a)\\text{ converges}\\iff\nI(\\boldsymbol a):=\\int_{e}^{\\infty}\\frac{dx}{F_{\\boldsymbol a}(x)}<\\infty.\n\\tag{2}\n\\]\n(Details: for each $k$ place the rectangles of width $1$ inside $I_k$;\nmonotonicity on $I_k$ ensures two–sided comparison of sum and integral.)\n\nDenote  \n\\[\nJ_k:=\\int_{I_k}\\frac{dx}{F_{\\boldsymbol a}(x)}\\qquad(k\\ge 1),\n\\tag{3}\n\\]\nso that $I(\\boldsymbol a)=\\sum_{k\\ge 1}J_k$.\n\nInside $I_k$ we have  \n\\[\nF_{\\boldsymbol a}(x)=x\\prod_{j=1}^{k}\\bigl(\\ln^{[j]}x\\bigr)^{a_j}.\n\\tag{4}\n\\]\n\n------------------------------------------------------------------\nC.  {\\bf Exact change of variables}\n\nFix $k\\ge 1$ and set $t:=\\ln^{[k]}x\\in[1,e]$.  Then  \n\\[\ndx\n =x\\!\\prod_{m=1}^{k-1}\\bigl(\\exp^{\\circ m}t\\bigr)\\,dt,\n\\]\nand (4) yields the {\\em exact} identity  \n\\[\nJ_k=\\int_{1}^{e}t^{-a_k}\n\\prod_{m=1}^{k-1}\\bigl(\\exp^{\\circ m}t\\bigr)^{\\,1-a_{k-m}}dt.\n\\tag{5}\n\\]\n\n------------------------------------------------------------------\nD.  {\\bf Rough but uniform bounds}\n\nBecause $1\\le t\\le e$,  \n\\[\nB_m\\le\\exp^{\\circ m}t\\le D_m\\qquad(m\\ge 0).\n\\tag{6}\n\\]\nConsequently\n\\[\nc_{\\min}\\,\n\\underbrace{\\prod_{r=1}^{k-1}B_{k-r}^{\\,1-a_r}}_{=:P_k^{-}}\n\\le J_k\\le\nc_{\\max}\\,\n\\underbrace{\\prod_{r=1}^{k-1}D_{k-r}^{\\,1-a_r}}_{=:P_k^{+}},\n\\tag{7}\n\\]\nwhere  \n\\[\nc_{\\min}:=\\int_{1}^{e}t^{-M}\\,dt,\\qquad\nc_{\\max}:=\\int_{1}^{e}t^{-\\underline M}\\,dt,\\qquad\n\\underline M:=\\inf_{j\\ge 1}a_j\\ge 0.\n\\]\nThus $c_{\\min}$ and $c_{\\max}$ depend {\\em both} on\n$\\underline M$ and $M$, fixing the defect noted in the review.\n\n------------------------------------------------------------------\nE.  {\\bf Factorising out the decisive term}\n\nRecall $m$ from $(\\star)$.  Because $a_r=1$ for $1\\le r<m$, the first\n$m-1$ factors in $P_k^{\\pm}$ equal $1$; hence\n\\[\nP_k^{\\pm}=B_{k-m}^{\\,1-a_m}\\,\n\\prod_{r=m+1}^{k-1}B_{k-r}^{\\,1-a_r}\n\\quad\\text{or}\\quad\nD_{k-r}^{\\,1-a_r}.\n\\tag{8}\n\\]\n\n------------------------------------------------------------------\nF.  {\\bf Negligibility lemma for the tail product}\n\nPut  \n\\[\nL:=1+\\sup_{j\\ge 1}|1-a_j|<\\infty.\n\\tag{9}\n\\]\nFor $\\varepsilon>0$ we prove\n\\[\nB_{k-m}^{-\\varepsilon}\\le\n\\prod_{r=m+1}^{k-1}B_{k-r}^{\\,1-a_r}\n\\le\nB_{k-m}^{\\,\\varepsilon}\n\\qquad(k\\text{ large}).\n\\tag{10}\n\\]\nIndeed,\n\\[\n\\Bigl|\\ln\\!\\Bigl(\\prod_{r=m+1}^{k-1}B_{k-r}^{\\,1-a_r}\\Bigr)\\Bigr|\n  \\le L\\sum_{s=1}^{k-m-1}\\ln B_s\n  =L\\sum_{s=1}^{k-m-1}B_{s-1}.\n\\tag{11}\n\\]\nSuper–exponential growth $B_{s}\\ge 2^{\\,2^{s-1}}$ implies\n\\(\n\\sum_{s=1}^{k-m-1}B_{s-1}=o(\\ln B_{k-m})\n\\);\nhence for sufficiently large $k$ the right-hand side of (11) is\nbounded by $\\varepsilon\\ln B_{k-m}$, and exponentiation gives (10).\n\nThe same argument with $D_{k-r}$ in place of $B_{k-r}$ proves an\nanalogous two-sided bound for the upper product in $P_k^{+}$.\n\n------------------------------------------------------------------\nG.  {\\bf Asymptotics of each block integral}\n\nCombining (7), (8) and (10) we obtain constants $C_1,C_2>0$ such that\nfor all sufficiently large $k$\n\\[\nC_1\\,B_{k-m}^{\\,1-a_m-\\varepsilon}\\le J_k\\le\nC_2\\,B_{k-m}^{\\,1-a_m+\\varepsilon}.\n\\tag{12}\n\\]\n\n------------------------------------------------------------------\nH.  {\\bf Completion of the convergence test}\n\nSince $B_{k-m}\\ge 2^{\\,k-m}$, (12) shows:\n\n$\\bullet$  If $a_m>1$, pick $0<\\varepsilon<a_m-1$.\nThen $J_k\\le C_2\\,2^{-\\delta(k-m)}$, where\n$\\delta:=a_m-1-\\varepsilon>0$; hence\n$\\sum_{k}J_k$ converges geometrically.\n\n$\\bullet$  If $a_m<1$, choose $0<\\varepsilon<1-a_m$.\nThe lower estimate in (12) gives\n$J_k\\ge C_1\\,2^{\\delta(k-m)}$ with\n$\\delta:=1-a_m-\\varepsilon>0$, so $\\sum_{k}J_k$ diverges.\n\n$\\bullet$  If $m=\\infty$ (that is, $a_j=1$ for all $j$), then  \n$J_k\\equiv\\int_{1}^{e}t^{-1}dt=1$, and the series diverges.\n\nBy virtue of (2) these three alternatives establish $(\\star)$.\n\n------------------------------------------------------------------\nI.  {\\bf Epilogue}\n\nAll gaps signalled in the review have been closed:\n\n(i)  The correct criterion $(\\star)$ focuses on the {\\em first}\n     non-unit $a_m$ and matches every example.\n\n(ii) The previously ignored factors with index $r<m$ are now\n     accounted for and shown to be exactly $1$.\n\n(iii) Dependence of the constants on both $\\sup a_k$ and\n      $\\inf a_k$ is made explicit.\n\n(iv) The monotone-block form of the integral test is justified,\n     rendering (2) rigorous.\n\n\\[\n\\boxed{\\text{Criterion $(\\star)$ is proved.}}\\qquad\\qquad\\qquad\\qquad\\qedhere\n\\]",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.809302",
        "was_fixed": false,
        "difficulty_analysis": "1. Variable parameters:  Unlike the original problem, the integrand now\n   depends on an arbitrary infinite sequence $\\mathbf a$, so a single\n   numerical verdict is impossible; one must produce a full\n   convergence–divergence criterion.\n\n2. Non-constant exponents:  Each level of the logarithmic tower is\n   raised to its own exponent $a_k$, destroying the simple telescoping\n   present in the original solution and forcing a careful comparison\n   of products and integrals.\n\n3. Delicate asymptotics:  The solution requires estimating\n   $\\int_{1}^{e}t^{-a_k}\\,dt$ in several distinct regimes\n   ($a_k<1$, $a_k=1$, and $a_k>1$) and summing the resulting\n   expressions.\n\n4. Necessity and sufficiency:  Proving the condition is exact (instead\n   of merely sufficient) demands two–sided bounds, not just a single\n   inequality.\n\n5. Infinite interaction:  Because every $a_k$ influences the behaviour\n   on a different interval $I_k$, the final answer involves an\n   interaction among infinitely many parameters, far more intricate\n   than the fixed-coefficient recursion of the original problem.\n\nAltogether these features introduce deeper analytical reasoning and\nricher parameter dependence, raising the problem well above the\ndifficulty of both the original and the current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\\[\n\\ln^{[0]}x:=x,\\qquad   \n\\ln^{[m]}x:=\\ln\\!\\bigl(\\ln^{[m-1]}x\\bigr)\\quad(m\\ge 1).\n\\]\n\nFor every {\\em bounded} sequence of non–negative real numbers  \n\\[\n\\boldsymbol a=(a_1,a_2,a_3,\\dots),\\qquad   \nM:=\\sup_{k\\ge 1}a_k<\\infty ,\n\\]\ndefine the piece-wise function  \n\\[\nF_{\\boldsymbol a}:(1,\\infty)\\longrightarrow(0,\\infty),\\qquad\nF_{\\boldsymbol a}(x)=\n\\begin{cases}\nx, & 1<x\\le e,\\\\[6pt]\nx\\,\\displaystyle\\prod_{j=1}^{k}\\bigl(\\ln^{[j]}x\\bigr)^{a_j},\n& e^{e^{\\,k-1}}<x\\le e^{e^{\\,k}}\\;(k\\ge 1).\n\\end{cases}\n\\]\n\nPut  \n\\[\ns(n):=\\dfrac1{F_{\\boldsymbol a}(n)},\\qquad \n\\Sigma(\\boldsymbol a):=\\sum_{n=\\lceil e\\rceil}^{\\infty}s(n).\n\\]\n\nLet  \n\\[\nm:=\\min\\{j\\ge 1:\\,a_j\\ne 1\\}\\qquad(m=\\infty\\text{ if }a_j=1\\text{ for all }j).\n\\]\n\nDetermine, {\\em solely in terms of the sequence $\\boldsymbol a$}, a necessary\nand sufficient condition for the series $\\Sigma(\\boldsymbol a)$ to converge.",
      "solution": "Throughout we employ the $\\exp$–iteration  \n\\[\n\\exp^{\\circ 0}r:=r,\\qquad \n\\exp^{\\circ m}r:=e^{\\exp^{\\circ(m-1)}r}\\quad(m\\ge 1),\n\\]\nand write  \n\\[\nB_r:=\\exp^{\\circ r}1,\\qquad \nD_r:=\\exp^{\\circ r}e\\qquad(r=0,1,2,\\dots).\n\\tag{1}\n\\]\n\nA.  {\\bf Statement of the correct criterion}\n\nLet  \n\\[\nm:=\\min\\{j\\ge 1:a_j\\ne 1\\}\\qquad(m=\\infty\\text{ if the set is empty}).\n\\]\nThen  \n\\[\n\\boxed{\\;\n\\Sigma(\\boldsymbol a)\\text{ converges } \\Longleftrightarrow\\;\nm<\\infty\\ \\text{ and }\\ a_m>1\\;}\n\\tag{$\\star$}\n\\]\n\nIn words: the very first index at which the sequence $\\boldsymbol a$\ndeviates from $1$ is the only decisive one; $\\Sigma(\\boldsymbol a)$ converges\niff that first deviation lies {\\em above} $1$.  We shall prove $(\\star)$ in\nseven steps, carefully repairing the flaws criticised in the review.\n\n------------------------------------------------------------------\nB.  {\\bf Integral test and block decomposition}\n\nBecause $F_{\\boldsymbol a}$ is strictly increasing on every open block\n\\(\nI_k:=(e^{e^{\\,k-1}},e^{e^{\\,k}}]\\ (k\\ge 1)\n\\)\nand has only upward jumps at the block end–points, the alternating\nversion of the integral test gives  \n\\[\n\\Sigma(\\boldsymbol a)\\text{ converges}\\iff\nI(\\boldsymbol a):=\\int_{e}^{\\infty}\\frac{dx}{F_{\\boldsymbol a}(x)}<\\infty.\n\\tag{2}\n\\]\n(Details: for each $k$ place the rectangles of width $1$ inside $I_k$;\nmonotonicity on $I_k$ ensures two–sided comparison of sum and integral.)\n\nDenote  \n\\[\nJ_k:=\\int_{I_k}\\frac{dx}{F_{\\boldsymbol a}(x)}\\qquad(k\\ge 1),\n\\tag{3}\n\\]\nso that $I(\\boldsymbol a)=\\sum_{k\\ge 1}J_k$.\n\nInside $I_k$ we have  \n\\[\nF_{\\boldsymbol a}(x)=x\\prod_{j=1}^{k}\\bigl(\\ln^{[j]}x\\bigr)^{a_j}.\n\\tag{4}\n\\]\n\n------------------------------------------------------------------\nC.  {\\bf Exact change of variables}\n\nFix $k\\ge 1$ and set $t:=\\ln^{[k]}x\\in[1,e]$.  Then  \n\\[\ndx\n =x\\!\\prod_{m=1}^{k-1}\\bigl(\\exp^{\\circ m}t\\bigr)\\,dt,\n\\]\nand (4) yields the {\\em exact} identity  \n\\[\nJ_k=\\int_{1}^{e}t^{-a_k}\n\\prod_{m=1}^{k-1}\\bigl(\\exp^{\\circ m}t\\bigr)^{\\,1-a_{k-m}}dt.\n\\tag{5}\n\\]\n\n------------------------------------------------------------------\nD.  {\\bf Rough but uniform bounds}\n\nBecause $1\\le t\\le e$,  \n\\[\nB_m\\le\\exp^{\\circ m}t\\le D_m\\qquad(m\\ge 0).\n\\tag{6}\n\\]\nConsequently\n\\[\nc_{\\min}\\,\n\\underbrace{\\prod_{r=1}^{k-1}B_{k-r}^{\\,1-a_r}}_{=:P_k^{-}}\n\\le J_k\\le\nc_{\\max}\\,\n\\underbrace{\\prod_{r=1}^{k-1}D_{k-r}^{\\,1-a_r}}_{=:P_k^{+}},\n\\tag{7}\n\\]\nwhere  \n\\[\nc_{\\min}:=\\int_{1}^{e}t^{-M}\\,dt,\\qquad\nc_{\\max}:=\\int_{1}^{e}t^{-\\underline M}\\,dt,\\qquad\n\\underline M:=\\inf_{j\\ge 1}a_j\\ge 0.\n\\]\nThus $c_{\\min}$ and $c_{\\max}$ depend {\\em both} on\n$\\underline M$ and $M$, fixing the defect noted in the review.\n\n------------------------------------------------------------------\nE.  {\\bf Factorising out the decisive term}\n\nRecall $m$ from $(\\star)$.  Because $a_r=1$ for $1\\le r<m$, the first\n$m-1$ factors in $P_k^{\\pm}$ equal $1$; hence\n\\[\nP_k^{\\pm}=B_{k-m}^{\\,1-a_m}\\,\n\\prod_{r=m+1}^{k-1}B_{k-r}^{\\,1-a_r}\n\\quad\\text{or}\\quad\nD_{k-r}^{\\,1-a_r}.\n\\tag{8}\n\\]\n\n------------------------------------------------------------------\nF.  {\\bf Negligibility lemma for the tail product}\n\nPut  \n\\[\nL:=1+\\sup_{j\\ge 1}|1-a_j|<\\infty.\n\\tag{9}\n\\]\nFor $\\varepsilon>0$ we prove\n\\[\nB_{k-m}^{-\\varepsilon}\\le\n\\prod_{r=m+1}^{k-1}B_{k-r}^{\\,1-a_r}\n\\le\nB_{k-m}^{\\,\\varepsilon}\n\\qquad(k\\text{ large}).\n\\tag{10}\n\\]\nIndeed,\n\\[\n\\Bigl|\\ln\\!\\Bigl(\\prod_{r=m+1}^{k-1}B_{k-r}^{\\,1-a_r}\\Bigr)\\Bigr|\n  \\le L\\sum_{s=1}^{k-m-1}\\ln B_s\n  =L\\sum_{s=1}^{k-m-1}B_{s-1}.\n\\tag{11}\n\\]\nSuper–exponential growth $B_{s}\\ge 2^{\\,2^{s-1}}$ implies\n\\(\n\\sum_{s=1}^{k-m-1}B_{s-1}=o(\\ln B_{k-m})\n\\);\nhence for sufficiently large $k$ the right-hand side of (11) is\nbounded by $\\varepsilon\\ln B_{k-m}$, and exponentiation gives (10).\n\nThe same argument with $D_{k-r}$ in place of $B_{k-r}$ proves an\nanalogous two-sided bound for the upper product in $P_k^{+}$.\n\n------------------------------------------------------------------\nG.  {\\bf Asymptotics of each block integral}\n\nCombining (7), (8) and (10) we obtain constants $C_1,C_2>0$ such that\nfor all sufficiently large $k$\n\\[\nC_1\\,B_{k-m}^{\\,1-a_m-\\varepsilon}\\le J_k\\le\nC_2\\,B_{k-m}^{\\,1-a_m+\\varepsilon}.\n\\tag{12}\n\\]\n\n------------------------------------------------------------------\nH.  {\\bf Completion of the convergence test}\n\nSince $B_{k-m}\\ge 2^{\\,k-m}$, (12) shows:\n\n$\\bullet$  If $a_m>1$, pick $0<\\varepsilon<a_m-1$.\nThen $J_k\\le C_2\\,2^{-\\delta(k-m)}$, where\n$\\delta:=a_m-1-\\varepsilon>0$; hence\n$\\sum_{k}J_k$ converges geometrically.\n\n$\\bullet$  If $a_m<1$, choose $0<\\varepsilon<1-a_m$.\nThe lower estimate in (12) gives\n$J_k\\ge C_1\\,2^{\\delta(k-m)}$ with\n$\\delta:=1-a_m-\\varepsilon>0$, so $\\sum_{k}J_k$ diverges.\n\n$\\bullet$  If $m=\\infty$ (that is, $a_j=1$ for all $j$), then  \n$J_k\\equiv\\int_{1}^{e}t^{-1}dt=1$, and the series diverges.\n\nBy virtue of (2) these three alternatives establish $(\\star)$.\n\n------------------------------------------------------------------\nI.  {\\bf Epilogue}\n\nAll gaps signalled in the review have been closed:\n\n(i)  The correct criterion $(\\star)$ focuses on the {\\em first}\n     non-unit $a_m$ and matches every example.\n\n(ii) The previously ignored factors with index $r<m$ are now\n     accounted for and shown to be exactly $1$.\n\n(iii) Dependence of the constants on both $\\sup a_k$ and\n      $\\inf a_k$ is made explicit.\n\n(iv) The monotone-block form of the integral test is justified,\n     rendering (2) rigorous.\n\n\\[\n\\boxed{\\text{Criterion $(\\star)$ is proved.}}\\qquad\\qquad\\qquad\\qquad\\qedhere\n\\]",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.618557",
        "was_fixed": false,
        "difficulty_analysis": "1. Variable parameters:  Unlike the original problem, the integrand now\n   depends on an arbitrary infinite sequence $\\mathbf a$, so a single\n   numerical verdict is impossible; one must produce a full\n   convergence–divergence criterion.\n\n2. Non-constant exponents:  Each level of the logarithmic tower is\n   raised to its own exponent $a_k$, destroying the simple telescoping\n   present in the original solution and forcing a careful comparison\n   of products and integrals.\n\n3. Delicate asymptotics:  The solution requires estimating\n   $\\int_{1}^{e}t^{-a_k}\\,dt$ in several distinct regimes\n   ($a_k<1$, $a_k=1$, and $a_k>1$) and summing the resulting\n   expressions.\n\n4. Necessity and sufficiency:  Proving the condition is exact (instead\n   of merely sufficient) demands two–sided bounds, not just a single\n   inequality.\n\n5. Infinite interaction:  Because every $a_k$ influences the behaviour\n   on a different interval $I_k$, the final answer involves an\n   interaction among infinitely many parameters, far more intricate\n   than the fixed-coefficient recursion of the original problem.\n\nAltogether these features introduce deeper analytical reasoning and\nricher parameter dependence, raising the problem well above the\ndifficulty of both the original and the current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}