{
  "index": "1964-B-5",
  "type": "NT",
  "tag": [
    "NT",
    "COMB",
    "ALG"
  ],
  "difficulty": "",
  "question": "5. Let \\( u_{n}(n=1,2,3, \\ldots) \\) denote the least common multiple of the first \\( n \\) terms of a strictly increasing sequence of positive integers (for example, the sequence \\( 1,2,3,4,5,6,10,12, \\ldots) \\). Prove that the series\n\\[\n\\sum_{n=1}^{\\infty} 1 / u_{n}\n\\]\nis convergent.",
  "solution": "First Solution. Let \\( u \\) be a positive integer. For each divisor \\( d \\) of \\( u \\) exceeding \\( \\sqrt{u} \\) there is another \\( u / d \\) that is less than \\( \\sqrt{u} \\). Hence at least half of the positive divisors of \\( u \\) are less than \\( \\sqrt{u} \\), so the number of positive divisors is at most \\( 2 \\sqrt{u} \\).\n\nNow \\( u_{n} \\), being the least common multiple of \\( n \\) distinct positive integers, has at least \\( n \\) positive divisors, so \\( 2 \\sqrt{u_{n}} \\geq n \\). Therefore, \\( \\Sigma 1 / u_{n} \\) is dominated by the convergent series \\( \\Sigma 4 / n^{2} \\) and is itself convergent.\n\nSecond Solution. The following proof, contributed by R. L. Graham, shows that the upper bound for such sums is 2 .\n\nLet \\( a_{1}, a_{2}, \\ldots \\) be a strictly increasing sequence of positive integers. Then\n\\[\n\\begin{aligned}\n\\frac{1}{a_{1}} & =\\sum_{n=2}^{\\infty}\\left(\\frac{1}{a_{n-1}}-\\frac{1}{a_{n}}\\right)=\\sum_{n=2}^{\\infty} \\frac{a_{n}-a_{n-1}}{a_{n-1} a_{n}} \\\\\n& \\geq \\sum_{n=2}^{\\infty} \\frac{\\text { g.c.d. }\\left\\{a_{n-1}, a_{n}\\right\\}}{a_{n-1} a_{n}}=\\sum_{n=2}^{\\infty} \\frac{1}{\\text { 1.c.m. }\\left\\{a_{n-1}, a_{n}\\right\\}} \\\\\n& \\geq \\sum_{n=2}^{\\infty} \\frac{1}{\\text { l.c.m. }\\left\\{a_{1}, a_{2}, \\ldots, a_{n}\\right\\}}=\\sum_{n=2}^{\\infty} \\frac{1}{u_{n}}\n\\end{aligned}\n\\]\n(Here g.c.d. stands for greatest common divisor and l.c.m. for least common multiple.) Since \\( u_{1}=a_{1} \\), we have\n\\[\n\\sum_{n=1}^{\\infty} \\frac{1}{u_{n}} \\leq \\frac{2}{a_{1}}\n\\]\n\nThere are many \\( a \\)-sequences for which \\( \\Sigma 1 / u_{n}=2 \\); for example, \\( 1,2,4 \\), \\( 8, \\ldots \\) or \\( 1,2,3,6,12, \\ldots, 3 \\cdot 2^{n-3}, \\ldots \\)",
  "vars": [
    "u",
    "u_n",
    "u_1",
    "d",
    "a_1",
    "a_2",
    "a_n",
    "a_n-1",
    "n"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "u": "baselcm",
        "u_n": "nthlcmval",
        "u_1": "firstlcm",
        "d": "divisorvar",
        "a_1": "firstaseq",
        "a_2": "secondaseq",
        "a_n": "nthaseq",
        "a_n-1": "prevaseq",
        "n": "indexvar"
      },
      "question": "5. Let \\( nthlcmval(indexvar=1,2,3, \\ldots) \\) denote the least common multiple of the first \\( indexvar \\) terms of a strictly increasing sequence of positive integers (for example, the sequence \\( 1,2,3,4,5,6,10,12, \\ldots) \\). Prove that the series\n\\[\n\\sum_{indexvar=1}^{\\infty} \\frac{1}{nthlcmval}\n\\]\nis convergent.",
      "solution": "First Solution. Let \\( baselcm \\) be a positive integer. For each divisor \\( divisorvar \\) of \\( baselcm \\) exceeding \\( \\sqrt{baselcm} \\) there is another \\( baselcm / divisorvar \\) that is less than \\( \\sqrt{baselcm} \\). Hence at least half of the positive divisors of \\( baselcm \\) are less than \\( \\sqrt{baselcm} \\), so the number of positive divisors is at most \\( 2 \\sqrt{baselcm} \\).\n\nNow \\( nthlcmval \\), being the least common multiple of \\( indexvar \\) distinct positive integers, has at least \\( indexvar \\) positive divisors, so \\( 2 \\sqrt{nthlcmval} \\geq indexvar \\). Therefore, \\( \\sum 1 / nthlcmval \\) is dominated by the convergent series \\( \\sum 4 / indexvar^{2} \\) and is itself convergent.\n\nSecond Solution. The following proof, contributed by R. L. Graham, shows that the upper bound for such sums is 2.\n\nLet \\( firstaseq, secondaseq, \\ldots \\) be a strictly increasing sequence of positive integers. Then\n\\[\n\\begin{aligned}\n\\frac{1}{firstaseq} &=\\sum_{indexvar=2}^{\\infty}\\left(\\frac{1}{prevaseq}-\\frac{1}{nthaseq}\\right)=\\sum_{indexvar=2}^{\\infty} \\frac{nthaseq-prevaseq}{prevaseq\\, nthaseq} \\\\\n&\\geq \\sum_{indexvar=2}^{\\infty} \\frac{\\text{g.c.d.}\\{prevaseq, nthaseq\\}}{prevaseq\\, nthaseq}=\\sum_{indexvar=2}^{\\infty} \\frac{1}{\\text{l.c.m.}\\{prevaseq, nthaseq\\}} \\\\\n&\\geq \\sum_{indexvar=2}^{\\infty} \\frac{1}{\\text{l.c.m.}\\{firstaseq, secondaseq, \\ldots, nthaseq\\}}=\\sum_{indexvar=2}^{\\infty} \\frac{1}{nthlcmval}\n\\end{aligned}\n\\]\n(Here g.c.d. stands for greatest common divisor and l.c.m. for least common multiple.) Since \\( firstlcm=firstaseq \\), we have\n\\[\n\\sum_{indexvar=1}^{\\infty} \\frac{1}{nthlcmval} \\leq \\frac{2}{firstaseq}\n\\]\n\nThere are many \\( a \\)-sequences for which \\( \\sum 1 / nthlcmval=2 \\); for example, \\( 1,2,4,8, \\ldots \\) or \\( 1,2,3,6,12, \\ldots, 3 \\cdot 2^{indexvar-3}, \\ldots \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "u": "marbleship",
        "u_n": "lanternage",
        "u_1": "sailormast",
        "d": "wristwatch",
        "a_1": "blueberry",
        "a_2": "strawberry",
        "a_n": "pineapple",
        "a_n-1": "chandelier",
        "n": "sandcastle"
      },
      "question": "5. Let \\( lanternage(sandcastle=1,2,3, \\ldots) \\) denote the least common multiple of the first \\( sandcastle \\) terms of a strictly increasing sequence of positive integers (for example, the sequence \\( 1,2,3,4,5,6,10,12, \\ldots) \\). Prove that the series\n\\[\n\\sum_{sandcastle=1}^{\\infty} 1 / lanternage\n\\]\nis convergent.",
      "solution": "First Solution. Let \\( marbleship \\) be a positive integer. For each divisor \\( wristwatch \\) of \\( marbleship \\) exceeding \\( \\sqrt{marbleship} \\) there is another \\( marbleship / wristwatch \\) that is less than \\( \\sqrt{marbleship} \\). Hence at least half of the positive divisors of \\( marbleship \\) are less than \\( \\sqrt{marbleship} \\), so the number of positive divisors is at most \\( 2 \\sqrt{marbleship} \\).\n\nNow \\( lanternage \\), being the least common multiple of \\( sandcastle \\) distinct positive integers, has at least \\( sandcastle \\) positive divisors, so \\( 2 \\sqrt{lanternage} \\geq sandcastle \\). Therefore, \\( \\Sigma 1 / lanternage \\) is dominated by the convergent series \\( \\Sigma 4 / sandcastle^{2} \\) and is itself convergent.\n\nSecond Solution. The following proof, contributed by R. L. Graham, shows that the upper bound for such sums is 2 .\n\nLet \\( blueberry, strawberry, \\ldots \\) be a strictly increasing sequence of positive integers. Then\n\\[\n\\begin{aligned}\n\\frac{1}{blueberry} & =\\sum_{sandcastle=2}^{\\infty}\\left(\\frac{1}{chandelier}-\\frac{1}{pineapple}\\right)=\\sum_{sandcastle=2}^{\\infty} \\frac{pineapple-chandelier}{chandelier\\, pineapple} \\\\\n& \\geq \\sum_{sandcastle=2}^{\\infty} \\frac{\\text { g.c.d. }\\left\\{chandelier, pineapple\\right\\}}{chandelier\\, pineapple}=\\sum_{sandcastle=2}^{\\infty} \\frac{1}{\\text { 1.c.m. }\\left\\{chandelier, pineapple\\right\\}} \\\\\n& \\geq \\sum_{sandcastle=2}^{\\infty} \\frac{1}{\\text { l.c.m. }\\left\\{blueberry, strawberry, \\ldots, pineapple\\right\\}}=\\sum_{sandcastle=2}^{\\infty} \\frac{1}{lanternage}\n\\end{aligned}\n\\]\n(Here g.c.d. stands for greatest common divisor and l.c.m. for least common multiple.) Since \\( sailormast=blueberry \\), we have\n\\[\n\\sum_{sandcastle=1}^{\\infty} \\frac{1}{lanternage} \\leq \\frac{2}{blueberry}\n\\]\nThere are many \\( a \\)-sequences for which \\( \\Sigma 1 / lanternage=2 \\); for example, \\( 1,2,4 \\), \\( 8, \\ldots \\) or \\( 1,2,3,6,12, \\ldots, 3 \\cdot 2^{sandcastle-3}, \\ldots \\)"
    },
    "descriptive_long_misleading": {
      "map": {
        "u": "negativenumber",
        "u_n": "gcdtermseq",
        "u_1": "gcdfirstterm",
        "d": "multiplevar",
        "a_1": "lasttermone",
        "a_2": "lasttermtwo",
        "a_n": "lasttermnth",
        "a_n-1": "lasttermnminusone",
        "n": "finishedvalue"
      },
      "question": "Problem:\n<<<\n5. Let \\( gcdtermseq(finishedvalue=1,2,3, \\ldots) \\) denote the least common multiple of the first \\( finishedvalue \\) terms of a strictly increasing sequence of positive integers (for example, the sequence \\( 1,2,3,4,5,6,10,12, \\ldots) \\). Prove that the series\n\\[\n\\sum_{finishedvalue=1}^{\\infty} 1 / gcdtermseq\n\\]\nis convergent.\n>>>\n",
      "solution": "First Solution. Let \\( negativenumber \\) be a positive integer. For each divisor \\( multiplevar \\) of \\( negativenumber \\) exceeding \\( \\sqrt{negativenumber} \\) there is another \\( negativenumber / multiplevar \\) that is less than \\( \\sqrt{negativenumber} \\). Hence at least half of the positive divisors of \\( negativenumber \\) are less than \\( \\sqrt{negativenumber} \\), so the number of positive divisors is at most \\( 2 \\sqrt{negativenumber} \\).\n\nNow \\( gcdtermseq \\), being the least common multiple of \\( finishedvalue \\) distinct positive integers, has at least \\( finishedvalue \\) positive divisors, so \\( 2 \\sqrt{gcdtermseq} \\geq finishedvalue \\). Therefore, \\( \\Sigma 1 / gcdtermseq \\) is dominated by the convergent series \\( \\Sigma 4 / finishedvalue^{2} \\) and is itself convergent.\n\nSecond Solution. The following proof, contributed by R. L. Graham, shows that the upper bound for such sums is 2.\n\nLet \\( lasttermone, lasttermtwo, \\ldots \\) be a strictly increasing sequence of positive integers. Then\n\\[\n\\begin{aligned}\n\\frac{1}{lasttermone} &= \\sum_{finishedvalue=2}^{\\infty} \\left( \\frac{1}{lasttermnminusone} - \\frac{1}{lasttermnth} \\right ) = \\sum_{finishedvalue=2}^{\\infty} \\frac{lasttermnth - lasttermnminusone}{lasttermnminusone\\, lasttermnth} \\\\\n&\\geq \\sum_{finishedvalue=2}^{\\infty} \\frac{\\text{ g.c.d. }\\{ lasttermnminusone, lasttermnth \\}}{lasttermnminusone\\, lasttermnth} = \\sum_{finishedvalue=2}^{\\infty} \\frac{1}{\\text{ 1.c.m. }\\{ lasttermnminusone, lasttermnth \\}} \\\\\n&\\geq \\sum_{finishedvalue=2}^{\\infty} \\frac{1}{\\text{ l.c.m. }\\{ lasttermone, lasttermtwo, \\ldots, lasttermnth \\}} = \\sum_{finishedvalue=2}^{\\infty} \\frac{1}{gcdtermseq}\n\\end{aligned}\n\\]\n(Here g.c.d. stands for greatest common divisor and l.c.m. for least common multiple.) Since \\( gcdfirstterm = lasttermone \\), we have\n\\[\n\\sum_{finishedvalue=1}^{\\infty} \\frac{1}{gcdtermseq} \\leq \\frac{2}{lasttermone}.\n\\]\n\nThere are many \\( a \\)-sequences for which \\( \\Sigma 1 / gcdtermseq = 2 \\); for example, \\( 1,2,4 \\), \\( 8, \\ldots \\) or \\( 1,2,3,6,12, \\ldots, 3 \\cdot 2^{finishedvalue-3}, \\ldots \\)\n"
    },
    "garbled_string": {
      "map": {
        "u": "qzxwvtnp",
        "u_n": "hjgrksla",
        "u_1": "nclvater",
        "d": "gupzrkea",
        "a_1": "vbarqiwe",
        "a_2": "fmsoklcz",
        "a_n": "tdyhepul",
        "a_n-1": "smgiolpa",
        "n": "xqavmzhe"
      },
      "question": "5. Let \\( hjgrksla(xqavmzhe=1,2,3, \\ldots) \\) denote the least common multiple of the first \\( xqavmzhe \\) terms of a strictly increasing sequence of positive integers (for example, the sequence \\( 1,2,3,4,5,6,10,12, \\ldots) \\). Prove that the series\n\\[\n\\sum_{xqavmzhe=1}^{\\infty} 1 / hjgrksla\n\\]\nis convergent.",
      "solution": "First Solution. Let \\( qzxwvtnp \\) be a positive integer. For each divisor \\( gupzrkea \\) of \\( qzxwvtnp \\) exceeding \\( \\sqrt{qzxwvtnp} \\) there is another \\( qzxwvtnp / gupzrkea \\) that is less than \\( \\sqrt{qzxwvtnp} \\). Hence at least half of the positive divisors of \\( qzxwvtnp \\) are less than \\( \\sqrt{qzxwvtnp} \\), so the number of positive divisors is at most \\( 2 \\sqrt{qzxwvtnp} \\).\n\nNow \\( hjgrksla \\), being the least common multiple of \\( xqavmzhe \\) distinct positive integers, has at least \\( xqavmzhe \\) positive divisors, so \\( 2 \\sqrt{hjgrksla} \\geq xqavmzhe \\). Therefore, \\( \\Sigma 1 / hjgrksla \\) is dominated by the convergent series \\( \\Sigma 4 / xqavmzhe^{2} \\) and is itself convergent.\n\nSecond Solution. The following proof, contributed by R. L. Graham, shows that the upper bound for such sums is 2 .\n\nLet \\( vbarqiwe, fmsoklcz, \\ldots \\) be a strictly increasing sequence of positive integers. Then\n\\[\n\\begin{aligned}\n\\frac{1}{vbarqiwe} & =\\sum_{xqavmzhe=2}^{\\infty}\\left(\\frac{1}{smgiolpa}-\\frac{1}{tdyhepul}\\right)=\\sum_{xqavmzhe=2}^{\\infty} \\frac{tdyhepul-smgiolpa}{smgiolpa tdyhepul} \\\\\n& \\geq \\sum_{xqavmzhe=2}^{\\infty} \\frac{\\text { g.c.d. }\\left\\{smgiolpa, tdyhepul\\right\\}}{smgiolpa tdyhepul}=\\sum_{xqavmzhe=2}^{\\infty} \\frac{1}{\\text { 1.c.m. }\\left\\{smgiolpa, tdyhepul\\right\\}} \\\\\n& \\geq \\sum_{xqavmzhe=2}^{\\infty} \\frac{1}{\\text { l.c.m. }\\left\\{vbarqiwe, fmsoklcz, \\ldots, tdyhepul\\right\\}}=\\sum_{xqavmzhe=2}^{\\infty} \\frac{1}{hjgrksla}\n\\end{aligned}\n\\]\n(Here g.c.d. stands for greatest common divisor and l.c.m. for least common multiple.) Since \\( nclvater=vbarqiwe \\), we have\n\\[\n\\sum_{xqavmzhe=1}^{\\infty} \\frac{1}{hjgrksla} \\leq \\frac{2}{vbarqiwe}\n\\]\n\nThere are many \\( a \\)-sequences for which \\( \\Sigma 1 / hjgrksla=2 \\); for example, \\( 1,2,4 \\), \\( 8, \\ldots \\) or \\( 1,2,3,6,12, \\ldots, 3 \\cdot 2^{xqavmzhe-3}, \\ldots \\)"
    },
    "kernel_variant": {
      "question": "Let $\\bigl(b_{m}\\bigr)_{m\\ge 1}$ be an arbitrary sequence of pair-wise distinct positive integers (no monotonicity or coprimality is required).  \nFix a real parameter $\\alpha$ with $0<\\alpha<1$ and, for every integer $n\\ge 1$, set  \n\\[\nk_{n}:=\\lfloor \\alpha n \\rfloor ,\\qquad\nL_{n}^{(\\alpha)}\n:=\\operatorname{lcm}\\!\\Bigl\\{\n           b_{i_{1}}\\times b_{i_{2}}\\times\\cdots\\times b_{i_{k_{n}}}\n           \\;\\Bigm|\\;\n           1\\le i_{1}<i_{2}<\\dots<i_{k_{n}}\\le n\n           \\Bigr\\}.\n\\]\n(Here and throughout we adopt the usual convention that the\n$\\operatorname{lcm}$ of an \\emph{empty} set of integers equals $1$; this can occur only for the finitely many $n$ with $k_{n}=0$.)\n\nIn words, $L_{n}^{(\\alpha)}$ is the least common multiple of all $k_{n}$-fold products that can be formed with the first $n$ elements of the sequence; the number $k_{n}$ of indices grows linearly with~$n$.\n\nProve that the series  \n\\[\n\\sum_{n=1}^{\\infty}\\frac{1}{L_{n}^{(\\alpha)}}\n\\]\nconverges and that its sum can be bounded above by a constant\n$K(\\alpha)$ depending on $\\alpha$ only (in particular, $K(\\alpha)$ is\nindependent of $n$ and of the particular sequence\n$\\bigl(b_{m}\\bigr)$).",
      "solution": "Throughout the proof we denote by  \n\\[\na_{1}^{(n)}<a_{2}^{(n)}<\\dots <a_{n}^{(n)}\\qquad(n\\ge 1)\n\\]\nthe numbers $b_{1},\\dots ,b_{n}$ rearranged in increasing order.\n\n\\textbf{1. A universal combinatorial lower bound for $L_{n}^{(\\alpha)}$.}  \nChoose the $k_{n}$ largest elements of the set\n$\\{a_{1}^{(n)},\\dots ,a_{n}^{(n)}\\}$, namely\n$a_{n-k_{n}+1}^{(n)},\\dots ,a_{n}^{(n)}$, and put  \n\\[\nP_{n}:=\\prod_{j=n-k_{n}+1}^{\\,n}a_{j}^{(n)}.\n\\]\nBecause $P_{n}$ is one of the products whose least common multiple is\ntaken,  \n\\[\nP_{n}\\mid L_{n}^{(\\alpha)},\\qquad L_{n}^{(\\alpha)}\\ge P_{n}.\n\\tag{1}\n\\]\nSince $a_{j}^{(n)}\\ge j$ for every $j$,  \n\\[\nP_{n}\\ge\\prod_{j=n-k_{n}+1}^{\\,n}j=\\frac{n!}{(n-k_{n})!}.\n\\tag{2}\n\\]\nSet $m_{n}:=n-k_{n}$; note that $m_{n}\\sim(1-\\alpha)n$ and\n$k_{n}\\sim\\alpha n$ as $n\\to\\infty$.\n\n\\textbf{2. Stirling’s expansion while keeping the $e^{-k_{n}}$ factor.}  \nStirling’s formula in the form\n\\[\nt!=\\sqrt{2\\pi t}\\,\\Bigl(\\tfrac{t}{e}\\Bigr)^{t}\\bigl(1+\\delta_{t}\\bigr),\n\\qquad\\lvert\\delta_{t}\\rvert\\le\\frac{C}{t},\n\\]\nyields, with $k_{n}=n-m_{n}$,\n\\[\n\\frac{n!}{m_{n}!}\n=\\sqrt{\\frac{n}{m_{n}}}\\,\n\\Bigl(\\frac{n}{m_{n}}\\Bigr)^{m_{n}}\n\\bigl(n/e\\bigr)^{k_{n}}\n\\frac{1+\\delta_{n}}{1+\\delta_{m_{n}}}.\n\\]\nBoth the square-root factor and\n$\\dfrac{1+\\delta_{n}}{1+\\delta_{m_{n}}}$ are bounded above and below by\npositive constants depending only on~$\\alpha$.  Hence there exists a\nconstant $C_{0}(\\alpha)>0$ such that\n\\[\nL_{n}^{(\\alpha)}\n\\;\\ge\\;\nC_{0}(\\alpha)\\,\n\\Bigl(\\frac{n}{m_{n}}\\Bigr)^{m_{n}}\n\\bigl(n/e\\bigr)^{k_{n}}\n\\qquad(n\\ge 1).\n\\tag{3}\n\\]\n\n\\textbf{3. Isolating the $n$–dependence inside an exponential base.}  \nBecause\n\\[\n\\frac{n}{m_{n}}\n=\\frac{1}{1-\\frac{k_{n}}{n}}\n=\\frac{1}{1-\\alpha}+O\\!\\Bigl(\\frac{1}{n}\\Bigr),\n\\]\nwe can write, for sufficiently large $n$,\n\\[\n\\Bigl(\\frac{n}{m_{n}}\\Bigr)^{m_{n}}\n\\ge\\frac12\\Bigl(\\frac{1}{1-\\alpha}\\Bigr)^{m_{n}}.\n\\]\nSince $m_{n}=\\bigl(\\tfrac{1-\\alpha}{\\alpha}+o(1)\\bigr)k_{n}$, define\n\\[\nc_{\\alpha}:=(1-\\alpha)^{-(1-\\alpha)/\\alpha}>1.\n\\]\nThen, for all large $n$,\n\\[\n\\Bigl(\\frac{n}{m_{n}}\\Bigr)^{m_{n}}\n\\ge\\frac12\\,c_{\\alpha}^{\\,k_{n}}.\n\\]\nAbsorbing the factor $\\tfrac12$ into the constant in~(3) we infer\n\\[\nL_{n}^{(\\alpha)}\n\\;\\ge\\;\nC_{1}(\\alpha)\\,\\bigl(c_{\\alpha}n/e\\bigr)^{k_{n}}\n\\qquad(n\\ge N_{0}(\\alpha)),\n\\tag{4}\n\\]\nfor some constants $C_{1}(\\alpha)>0$ and $N_{0}(\\alpha)\\in\\mathbb N$.\n\n\\textbf{4. Super-exponential decay of the reciprocals.}  \nFrom~(4) we obtain, for $n\\ge N_{0}(\\alpha)$,\n\\[\n\\frac{1}{L_{n}^{(\\alpha)}}\n\\le\n\\frac{1}{C_{1}(\\alpha)}\\,\n\\Bigl(\\frac{e}{c_{\\alpha}n}\\Bigr)^{k_{n}}\n=\nC_{2}(\\alpha)\\,\n\\exp\\!\\Bigl(-k_{n}\\bigl(\\ln n+\\ln c_{\\alpha}-1\\bigr)\\Bigr).\n\\tag{5}\n\\]\nBecause $c_{\\alpha}>1$, there exists $N_{1}(\\alpha)\\ge N_{0}(\\alpha)$\nsuch that $\\ln n+\\ln c_{\\alpha}-1\\ge\\tfrac12\\ln n$ for all\n$n\\ge N_{1}(\\alpha)$.  Since $k_{n}\\ge\\alpha n-1$, (5) gives, for\n$n\\ge N_{1}(\\alpha)$,\n\\[\n\\frac{1}{L_{n}^{(\\alpha)}}\n\\le\nC_{3}(\\alpha)\\,\n\\exp\\!\\bigl(-\\tfrac{\\alpha}{3}\\,n\\ln n\\bigr).\n\\tag{6}\n\\]\nThe right-hand side decays faster than any geometric progression, so\nthe tail $\\sum_{n\\ge N_{1}(\\alpha)}1/L_{n}^{(\\alpha)}$ converges.\n\n\\textbf{5. Completion of the argument.}  \nFor the finitely many indices $1\\le n<N_{1}(\\alpha)$ (including those\nwith $k_{n}=0$) we have $L_{n}^{(\\alpha)}\\ge 1$.  Consequently\n\\[\n\\sum_{n=1}^{\\infty}\\frac{1}{L_{n}^{(\\alpha)}}\n=\n\\sum_{n=1}^{N_{1}(\\alpha)-1}\\frac{1}{L_{n}^{(\\alpha)}}\n+\\sum_{n=N_{1}(\\alpha)}^{\\infty}\\frac{1}{L_{n}^{(\\alpha)}}\n\\le\nK(\\alpha),\n\\]\nwhere the finite constant\n\\[\nK(\\alpha)\n:=\n\\bigl(N_{1}(\\alpha)-1\\bigr)\n+\\;C_{3}(\\alpha)\\sum_{n=N_{1}(\\alpha)}^{\\infty}\n      \\exp\\!\\bigl(-\\tfrac{\\alpha}{3}\\,n\\ln n\\bigr)\n\\]\ndepends solely on~$\\alpha$.  This proves the desired convergence of the\nseries and completes the solution. \\qed",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.550212",
        "was_fixed": false,
        "difficulty_analysis": "• Higher-dimensional combinatorics  – instead of single terms we must control the lcm of {\\em all} $k$–fold products coming from the first $n$ elements; the number of objects under the lcm now grows like $\\binom{n}{k}$ rather than merely $n$.\n\n• Non-trivial divisor-counting  – the proof has to relate the rapid growth of the combinatorial quantity $\\binom{n}{k}$ to structural properties of the divisor function, requiring a careful synthesis of combinatorics and multiplicative number theory.\n\n• Parameter dependence  – the argument must work {\\em uniformly} in the free parameter $k$, forcing asymptotically sharp bounds (Stirling, $d(N)\\le 2\\sqrt N$, etc.) and a quantitative comparison between two rapidly growing sequences.\n\n• Extended reasoning chain  – establishing convergence now needs five logically distinct steps (counting products, bounding divisors, converting to a lower bound on $L_{n}^{(k)}$, estimating reciprocals, finishing with a p-series test), each of which would be unnecessary in the original problem.\n\nThese additional layers lift the variant well above the original in both technical depth and conceptual complexity."
      }
    },
    "original_kernel_variant": {
      "question": "Let $\\bigl(b_{m}\\bigr)_{m\\ge 1}$ be an arbitrary sequence of pair-wise distinct positive integers (no monotonicity or coprimality is required).  \nFix a real parameter $\\alpha$ with $0<\\alpha<1$ and, for every integer $n\\ge 1$, set  \n\\[\nk_{n}:=\\lfloor \\alpha n \\rfloor ,\\qquad\nL_{n}^{(\\alpha)}\n:=\\operatorname{lcm}\\!\\Bigl\\{\n           b_{i_{1}}\\times b_{i_{2}}\\times\\cdots\\times b_{i_{k_{n}}}\n           \\;\\Bigm|\\;\n           1\\le i_{1}<i_{2}<\\dots<i_{k_{n}}\\le n\n           \\Bigr\\}.\n\\]\n(Here and throughout we adopt the usual convention that the\n$\\operatorname{lcm}$ of an \\emph{empty} set of integers equals $1$; this can occur only for the finitely many $n$ with $k_{n}=0$.)\n\nIn words, $L_{n}^{(\\alpha)}$ is the least common multiple of all $k_{n}$-fold products that can be formed with the first $n$ elements of the sequence; the number $k_{n}$ of indices grows linearly with~$n$.\n\nProve that the series  \n\\[\n\\sum_{n=1}^{\\infty}\\frac{1}{L_{n}^{(\\alpha)}}\n\\]\nconverges and that its sum can be bounded above by a constant\n$K(\\alpha)$ depending on $\\alpha$ only (in particular, $K(\\alpha)$ is\nindependent of $n$ and of the particular sequence\n$\\bigl(b_{m}\\bigr)$).",
      "solution": "Throughout the proof we denote by  \n\\[\na_{1}^{(n)}<a_{2}^{(n)}<\\dots <a_{n}^{(n)}\\qquad(n\\ge 1)\n\\]\nthe numbers $b_{1},\\dots ,b_{n}$ rearranged in increasing order.\n\n\\textbf{1. A universal combinatorial lower bound for $L_{n}^{(\\alpha)}$.}  \nChoose the $k_{n}$ largest elements of the set\n$\\{a_{1}^{(n)},\\dots ,a_{n}^{(n)}\\}$, namely\n$a_{n-k_{n}+1}^{(n)},\\dots ,a_{n}^{(n)}$, and put  \n\\[\nP_{n}:=\\prod_{j=n-k_{n}+1}^{\\,n}a_{j}^{(n)}.\n\\]\nBecause $P_{n}$ is one of the products whose least common multiple is\ntaken,  \n\\[\nP_{n}\\mid L_{n}^{(\\alpha)},\\qquad L_{n}^{(\\alpha)}\\ge P_{n}.\n\\tag{1}\n\\]\nSince $a_{j}^{(n)}\\ge j$ for every $j$,  \n\\[\nP_{n}\\ge\\prod_{j=n-k_{n}+1}^{\\,n}j=\\frac{n!}{(n-k_{n})!}.\n\\tag{2}\n\\]\nSet $m_{n}:=n-k_{n}$; note that $m_{n}\\sim(1-\\alpha)n$ and\n$k_{n}\\sim\\alpha n$ as $n\\to\\infty$.\n\n\\textbf{2. Stirling’s expansion while keeping the $e^{-k_{n}}$ factor.}  \nStirling’s formula in the form\n\\[\nt!=\\sqrt{2\\pi t}\\,\\Bigl(\\tfrac{t}{e}\\Bigr)^{t}\\bigl(1+\\delta_{t}\\bigr),\n\\qquad\\lvert\\delta_{t}\\rvert\\le\\frac{C}{t},\n\\]\nyields, with $k_{n}=n-m_{n}$,\n\\[\n\\frac{n!}{m_{n}!}\n=\\sqrt{\\frac{n}{m_{n}}}\\,\n\\Bigl(\\frac{n}{m_{n}}\\Bigr)^{m_{n}}\n\\bigl(n/e\\bigr)^{k_{n}}\n\\frac{1+\\delta_{n}}{1+\\delta_{m_{n}}}.\n\\]\nBoth the square-root factor and\n$\\dfrac{1+\\delta_{n}}{1+\\delta_{m_{n}}}$ are bounded above and below by\npositive constants depending only on~$\\alpha$.  Hence there exists a\nconstant $C_{0}(\\alpha)>0$ such that\n\\[\nL_{n}^{(\\alpha)}\n\\;\\ge\\;\nC_{0}(\\alpha)\\,\n\\Bigl(\\frac{n}{m_{n}}\\Bigr)^{m_{n}}\n\\bigl(n/e\\bigr)^{k_{n}}\n\\qquad(n\\ge 1).\n\\tag{3}\n\\]\n\n\\textbf{3. Isolating the $n$–dependence inside an exponential base.}  \nBecause\n\\[\n\\frac{n}{m_{n}}\n=\\frac{1}{1-\\frac{k_{n}}{n}}\n=\\frac{1}{1-\\alpha}+O\\!\\Bigl(\\frac{1}{n}\\Bigr),\n\\]\nwe can write, for sufficiently large $n$,\n\\[\n\\Bigl(\\frac{n}{m_{n}}\\Bigr)^{m_{n}}\n\\ge\\frac12\\Bigl(\\frac{1}{1-\\alpha}\\Bigr)^{m_{n}}.\n\\]\nSince $m_{n}=\\bigl(\\tfrac{1-\\alpha}{\\alpha}+o(1)\\bigr)k_{n}$, define\n\\[\nc_{\\alpha}:=(1-\\alpha)^{-(1-\\alpha)/\\alpha}>1.\n\\]\nThen, for all large $n$,\n\\[\n\\Bigl(\\frac{n}{m_{n}}\\Bigr)^{m_{n}}\n\\ge\\frac12\\,c_{\\alpha}^{\\,k_{n}}.\n\\]\nAbsorbing the factor $\\tfrac12$ into the constant in~(3) we infer\n\\[\nL_{n}^{(\\alpha)}\n\\;\\ge\\;\nC_{1}(\\alpha)\\,\\bigl(c_{\\alpha}n/e\\bigr)^{k_{n}}\n\\qquad(n\\ge N_{0}(\\alpha)),\n\\tag{4}\n\\]\nfor some constants $C_{1}(\\alpha)>0$ and $N_{0}(\\alpha)\\in\\mathbb N$.\n\n\\textbf{4. Super-exponential decay of the reciprocals.}  \nFrom~(4) we obtain, for $n\\ge N_{0}(\\alpha)$,\n\\[\n\\frac{1}{L_{n}^{(\\alpha)}}\n\\le\n\\frac{1}{C_{1}(\\alpha)}\\,\n\\Bigl(\\frac{e}{c_{\\alpha}n}\\Bigr)^{k_{n}}\n=\nC_{2}(\\alpha)\\,\n\\exp\\!\\Bigl(-k_{n}\\bigl(\\ln n+\\ln c_{\\alpha}-1\\bigr)\\Bigr).\n\\tag{5}\n\\]\nBecause $c_{\\alpha}>1$, there exists $N_{1}(\\alpha)\\ge N_{0}(\\alpha)$\nsuch that $\\ln n+\\ln c_{\\alpha}-1\\ge\\tfrac12\\ln n$ for all\n$n\\ge N_{1}(\\alpha)$.  Since $k_{n}\\ge\\alpha n-1$, (5) gives, for\n$n\\ge N_{1}(\\alpha)$,\n\\[\n\\frac{1}{L_{n}^{(\\alpha)}}\n\\le\nC_{3}(\\alpha)\\,\n\\exp\\!\\bigl(-\\tfrac{\\alpha}{3}\\,n\\ln n\\bigr).\n\\tag{6}\n\\]\nThe right-hand side decays faster than any geometric progression, so\nthe tail $\\sum_{n\\ge N_{1}(\\alpha)}1/L_{n}^{(\\alpha)}$ converges.\n\n\\textbf{5. Completion of the argument.}  \nFor the finitely many indices $1\\le n<N_{1}(\\alpha)$ (including those\nwith $k_{n}=0$) we have $L_{n}^{(\\alpha)}\\ge 1$.  Consequently\n\\[\n\\sum_{n=1}^{\\infty}\\frac{1}{L_{n}^{(\\alpha)}}\n=\n\\sum_{n=1}^{N_{1}(\\alpha)-1}\\frac{1}{L_{n}^{(\\alpha)}}\n+\\sum_{n=N_{1}(\\alpha)}^{\\infty}\\frac{1}{L_{n}^{(\\alpha)}}\n\\le\nK(\\alpha),\n\\]\nwhere the finite constant\n\\[\nK(\\alpha)\n:=\n\\bigl(N_{1}(\\alpha)-1\\bigr)\n+\\;C_{3}(\\alpha)\\sum_{n=N_{1}(\\alpha)}^{\\infty}\n      \\exp\\!\\bigl(-\\tfrac{\\alpha}{3}\\,n\\ln n\\bigr)\n\\]\ndepends solely on~$\\alpha$.  This proves the desired convergence of the\nseries and completes the solution. \\qed",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.455480",
        "was_fixed": false,
        "difficulty_analysis": "• Higher-dimensional combinatorics  – instead of single terms we must control the lcm of {\\em all} $k$–fold products coming from the first $n$ elements; the number of objects under the lcm now grows like $\\binom{n}{k}$ rather than merely $n$.\n\n• Non-trivial divisor-counting  – the proof has to relate the rapid growth of the combinatorial quantity $\\binom{n}{k}$ to structural properties of the divisor function, requiring a careful synthesis of combinatorics and multiplicative number theory.\n\n• Parameter dependence  – the argument must work {\\em uniformly} in the free parameter $k$, forcing asymptotically sharp bounds (Stirling, $d(N)\\le 2\\sqrt N$, etc.) and a quantitative comparison between two rapidly growing sequences.\n\n• Extended reasoning chain  – establishing convergence now needs five logically distinct steps (counting products, bounding divisors, converting to a lower bound on $L_{n}^{(k)}$, estimating reciprocals, finishing with a p-series test), each of which would be unnecessary in the original problem.\n\nThese additional layers lift the variant well above the original in both technical depth and conceptual complexity."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}