{
  "index": "1963-B-5",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "5. Let \\( \\left\\{a_{n}\\right\\} \\) be a sequence of real numbers satisfying the inequalities\n\\[\n0 \\leq a_{k} \\leq 100 a_{n} \\text { for } n \\leq k \\leq 2 n \\text { and } n=1,2, \\ldots,\n\\]\nand such that the series\n\\[\n\\sum_{n=0}^{\\infty} a_{n}\n\\]\nconverges. Prove that\n\\[\n\\lim _{n \\rightarrow \\infty} n a_{n}=0\n\\]",
  "solution": "Solution. For each positive integer \\( k \\). let\n\\[\nS_{k}=\\sum_{n \\geq k / 2}^{k} a_{n} .\n\\]\n\nSince \\( \\Sigma a_{n} \\) converges, we have \\( S_{k} \\rightarrow 0 \\) as \\( k \\rightarrow \\infty \\).\nRewriting the given condition slightly we have\n\\[\n0 \\leq a_{k} \\leq 100 a_{n} \\text { for } \\frac{1}{2} k \\leq n \\leq k .\n\\]\n\nFor each \\( k \\) there are at least \\( k / 2 \\) integers \\( n \\) satisfying this double inequality. Adding these inequalities we have, therefore,\n\\[\n\\frac{1}{2} k a_{k} \\leq 100 S_{k} .\n\\]\n\nHence\n\\[\n\\lim \\sup k a_{k} \\leq 200 \\lim S_{k}=0 .\n\\]\n\nSince \\( a_{k} \\geq 0 \\), we conclude lim \\( k a_{k}=0 \\).",
  "vars": [
    "a_n",
    "a_k",
    "a_0",
    "n",
    "k",
    "S_k"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "a_n": "termgen",
        "a_k": "termindx",
        "a_0": "termzero",
        "n": "indexn",
        "k": "indexk",
        "S_k": "partsumk"
      },
      "question": "5. Let \\( \\left\\{termgen\\right\\} \\) be a sequence of real numbers satisfying the inequalities\n\\[\n0 \\leq termindx \\leq 100\\, termgen \\text { for } indexn \\leq indexk \\leq 2\\, indexn \\text { and } indexn=1,2, \\ldots,\n\\]\nand such that the series\n\\[\n\\sum_{indexn=0}^{\\infty} termgen\n\\]\nconverges. Prove that\n\\[\n\\lim _{indexn \\rightarrow \\infty} indexn\\, termgen = 0\n\\]",
      "solution": "Solution. For each positive integer \\( indexk \\), let\n\\[\npartsumk=\\sum_{indexn \\geq indexk / 2}^{indexk} termgen .\n\\]\n\nSince \\( \\Sigma\\, termgen \\) converges, we have \\( partsumk \\rightarrow 0 \\) as \\( indexk \\rightarrow \\infty \\).\nRewriting the given condition slightly we have\n\\[\n0 \\leq termindx \\leq 100\\, termgen \\text { for } \\frac{1}{2}\\, indexk \\leq indexn \\leq indexk .\n\\]\n\nFor each \\( indexk \\) there are at least \\( indexk / 2 \\) integers \\( indexn \\) satisfying this double inequality. Adding these inequalities we have, therefore,\n\\[\n\\frac{1}{2}\\, indexk\\, termindx \\leq 100\\, partsumk .\n\\]\n\nHence\n\\[\n\\lim \\sup indexk\\, termindx \\leq 200 \\lim partsumk = 0 .\n\\]\n\nSince \\( termindx \\geq 0 \\), we conclude \\( \\lim indexk\\, termindx = 0 \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "a_n": "mistletoe",
        "a_k": "watermelon",
        "a_0": "peppermint",
        "n": "basketball",
        "k": "playground",
        "S_k": "blacksmith"
      },
      "question": "5. Let \\( \\left\\{mistletoe\\right\\} \\) be a sequence of real numbers satisfying the inequalities\n\\[\n0 \\leq watermelon \\leq 100 mistletoe \\text { for } basketball \\leq playground \\leq 2 basketball \\text { and } basketball=1,2, \\ldots,\n\\]\nand such that the series\n\\[\n\\sum_{basketball=0}^{\\infty} mistletoe\n\\]\nconverges. Prove that\n\\[\n\\lim _{basketball \\rightarrow \\infty} basketball mistletoe=0\n\\]",
      "solution": "Solution. For each positive integer \\( playground \\). let\n\\[\nblacksmith=\\sum_{basketball \\geq playground / 2}^{playground} mistletoe .\n\\]\n\nSince \\( \\Sigma mistletoe \\) converges, we have \\( blacksmith \\rightarrow 0 \\) as \\( playground \\rightarrow \\infty \\).\nRewriting the given condition slightly we have\n\\[\n0 \\leq watermelon \\leq 100 mistletoe \\text { for } \\frac{1}{2} playground \\leq basketball \\leq playground .\n\\]\n\nFor each \\( playground \\) there are at least \\( playground / 2 \\) integers \\( basketball \\) satisfying this double inequality. Adding these inequalities we have, therefore,\n\\[\n\\frac{1}{2} playground watermelon \\leq 100 blacksmith .\n\\]\n\nHence\n\\[\n\\lim \\sup playground watermelon \\leq 200 \\lim blacksmith=0 .\n\\]\n\nSince \\( watermelon \\geq 0 \\), we conclude lim \\( playground watermelon=0 \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "a_n": "staticvalue",
        "a_k": "staticpivot",
        "a_0": "staticorigin",
        "n": "fractional",
        "k": "continuous",
        "S_k": "discrepancy"
      },
      "question": "5. Let \\( \\left\\{staticvalue\\right\\} \\) be a sequence of real numbers satisfying the inequalities\n\\[\n0 \\leq staticpivot \\leq 100 staticvalue \\text { for } fractional \\leq continuous \\leq 2 fractional \\text { and } fractional=1,2, \\ldots,\n\\]\nand such that the series\n\\[\n\\sum_{fractional=0}^{\\infty} staticvalue\n\\]\nconverges. Prove that\n\\[\n\\lim _{fractional \\rightarrow \\infty} fractional staticvalue=0\n\\]",
      "solution": "Solution. For each positive integer \\( continuous \\). let\n\\[\ndiscrepancy=\\sum_{fractional \\geq continuous / 2}^{continuous} staticvalue .\n\\]\n\nSince \\( \\Sigma staticvalue \\) converges, we have \\( discrepancy \\rightarrow 0 \\) as \\( continuous \\rightarrow \\infty \\).\nRewriting the given condition slightly we have\n\\[\n0 \\leq staticpivot \\leq 100 staticvalue \\text { for } \\frac{1}{2} continuous \\leq fractional \\leq continuous .\n\\]\n\nFor each \\( continuous \\) there are at least \\( continuous / 2 \\) integers \\( fractional \\) satisfying this double inequality. Adding these inequalities we have, therefore,\n\\[\n\\frac{1}{2} continuous staticpivot \\leq 100 discrepancy .\n\\]\n\nHence\n\\[\n\\lim \\sup continuous staticpivot \\leq 200 \\lim discrepancy=0 .\n\\]\n\nSince \\( staticpivot \\geq 0 \\), we conclude lim \\( continuous staticpivot=0 \\)."
    },
    "garbled_string": {
      "map": {
        "a_n": "qzxwvtnp",
        "a_k": "hjgrksla",
        "a_0": "nmdyqhra",
        "n": "fusdhjla",
        "k": "pqowirvn",
        "S_k": "vchlnrma"
      },
      "question": "5. Let \\( \\left\\{qzxwvtnp\\right\\} \\) be a sequence of real numbers satisfying the inequalities\n\\[\n0 \\leq hjgrksla \\leq 100 qzxwvtnp \\text { for } fusdhjla \\leq pqowirvn \\leq 2 fusdhjla \\text { and } fusdhjla=1,2, \\ldots,\n\\]\nand such that the series\n\\[\n\\sum_{fusdhjla=0}^{\\infty} qzxwvtnp\n\\]\nconverges. Prove that\n\\[\n\\lim _{fusdhjla \\rightarrow \\infty} fusdhjla qzxwvtnp=0\n\\]",
      "solution": "Solution. For each positive integer \\( pqowirvn \\). let\n\\[\nvchlnrma=\\sum_{fusdhjla \\geq pqowirvn / 2}^{pqowirvn} qzxwvtnp .\n\\]\n\nSince \\( \\Sigma qzxwvtnp \\) converges, we have \\( vchlnrma \\rightarrow 0 \\) as \\( pqowirvn \\rightarrow \\infty \\).\nRewriting the given condition slightly we have\n\\[\n0 \\leq hjgrksla \\leq 100 qzxwvtnp \\text { for } \\frac{1}{2} pqowirvn \\leq fusdhjla \\leq pqowirvn .\n\\]\n\nFor each \\( pqowirvn \\) there are at least \\( pqowirvn / 2 \\) integers \\( fusdhjla \\) satisfying this double inequality. Adding these inequalities we have, therefore,\n\\[\n\\frac{1}{2} pqowirvn hjgrksla \\leq 100 vchlnrma .\n\\]\n\nHence\n\\[\n\\lim \\sup pqowirvn hjgrksla \\leq 200 \\lim vchlnrma=0 .\n\\]\n\nSince \\( hjgrksla \\geq 0 \\), we conclude lim \\( pqowirvn hjgrksla=0 \\)."
    },
    "kernel_variant": {
      "question": "Let $d\\ge 2$ be a fixed integer and let $M>1$ be a constant.  \nFor $n=(n_{1},\\dots ,n_{d})\\in\\mathbb N^{d}$ define  \n\\[\nQ(n)=\\Bigl\\{\\,k=(k_{1},\\dots ,k_{d})\\in\\mathbb N^{d}\\;:\\;\nn_{i}\\le k_{i}\\le 2n_{i}\\quad(1\\le i\\le d)\\Bigr\\}.\n\\]\n\nAssume that a non–negative array $(a_{n})_{n\\in\\mathbb N^{d}}$ satisfies  \n\\[\n\\sum_{n_{1}=1}^{\\infty}\\;\\dots\\;\\sum_{n_{d}=1}^{\\infty} a_{n}<\\infty,\n\\qquad \n\\sum_{k\\in Q(n)} a_{k}\\le M\\,a_{n}\\quad(n\\in\\mathbb N^{d}).\n\\tag{$\\star$}\n\\]\n\n(a)  Prove that\n\\[\n\\lim_{\\min_{1\\le i\\le d}n_{i}\\to\\infty}\n\\,(n_{1}n_{2}\\dots n_{d})\\,a_{n}=0 .\n\\tag{A}\n\\]\n\n(b)  Show that the weighted series\n\\[\n\\sum_{n\\in\\mathbb N^{d}}(n_{1}n_{2}\\dots n_{d})\\,a_{n}\n\\]\nis convergent.\n\n(The second assertion is strictly stronger than (A); its proof\nrequires a refined use of the counting argument employed in part (a).)",
      "solution": "Throughout the proof the symbols $C,C_{1},C_{2},\\dots$ denote\npositive constants depending only on $d$ and on the Carleson\nconstant $M$.\n\n\\medskip\n1.\\;A counting lemma.  \nFor $k=(k_{1},\\dots ,k_{d})\\in\\mathbb N^{d}$ put\n\\[\n\\lVert k\\rVert_{\\infty}:=\\max_{1\\le i\\le d}k_{i},\\qquad\n\\mathcal N(k):=\\bigl\\{\\,n\\in\\mathbb N^{d}\\;:\\;k\\in Q(n)\\bigr\\}.\n\\]\nIf $k_{i}\\ge 4$ for every $i$ then\n\\[\n\\#\\mathcal N(k)=\\prod_{i=1}^{d}\n\\bigl(k_{i}-\\lceil k_{i}/2\\rceil+1\\bigr)\n\\ge 2^{-d}\\,k_{1}\\dots k_{d}.\n\\tag{1.1}\n\\]\n\n\\medskip\n2.\\;A double–counting inequality.  \nFix $N\\in\\mathbb N$ and sum the inequalities in $(\\star)$ over\nall $n$ with $\\lVert n\\rVert_{\\infty}\\le N$:\n\\[\n\\sum_{\\lVert n\\rVert_{\\infty}\\le N}\\,\n\\sum_{k\\in Q(n)}a_{k}\\le\nM\\sum_{\\lVert n\\rVert_{\\infty}\\le N}a_{n}.\n\\]\nInterchanging the order of summation and using (1.1) yields\n\\[\n\\sum_{\\lVert k\\rVert_{\\infty}\\le N}\n(k_{1}\\dots k_{d})\\,a_{k}\n\\;\\le\\; C_{2}\\sum_{\\lVert n\\rVert_{\\infty}\\le N} a_{n}.\n\\tag{2.1}\n\\]\n\n\\medskip\n3.\\;Proof of part (a).  \nLet $S=\\sum_{n}a_{n}<\\infty$ and set\n\\[\nT_{N}:=\\sum_{\\lVert k\\rVert_{\\infty}=N}\n(k_{1}\\dots k_{d})\\,a_{k}\\qquad(N\\ge 1).\n\\]\nSumming (2.1) over $m=1,\\dots ,N$ gives\n\\[\n\\sum_{m=1}^{N}T_{m}\\le C_{2}S\\quad(N\\ge 1),\n\\]\nhence $(T_{N})_{N\\ge 1}$ is a summable, and therefore null, sequence:\n$T_{N}\\to 0$ as $N\\to\\infty$.\n\nFor an arbitrary index $k=(k_{1},\\dots ,k_{d})$ choose\n$N=\\lVert k\\rVert_{\\infty}$; then\n\\[\n(k_{1}\\dots k_{d})\\,a_{k}\\le T_{N}\\longrightarrow 0.\n\\]\nConsequently (A) is proved.\n\n\\bigskip\n4.\\;Proof of part (b).  \nBy letting $N\\to\\infty$ in (2.1) we obtain\n\\[\n\\sum_{k\\in\\mathbb N^{d}}\n(k_{1}\\dots k_{d})\\,a_{k}\\le C_{2}\n\\sum_{n\\in\\mathbb N^{d}}a_{n}=C_{2}S<\\infty,\n\\]\nwhich shows the desired convergence.\n\n\\medskip\nThis completes the solution of both parts of the problem.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.545075",
        "was_fixed": false,
        "difficulty_analysis": "Compared with the original one–dimensional, point-wise bounded problem, the enhanced variant is appreciably harder for several reasons:\n\n1. Higher dimension:  The indices live in ℕᵈ, forcing multi–parameter summations and combinatorial counting of lattice points inside high–dimensional boxes.\n\n2. Weaker hypothesis:  The bound (★) controls only the *aggregate* of many terms, not the individual terms themselves.  Recovering information about a single a_k from such coarse data demands a non–trivial double-counting argument.\n\n3. Necessity of sharp counting:  One must estimate from below how many n satisfy k∈Q(n); this calls for a careful combinatorial analysis (Step 1) that has no analogue in the original exercise.\n\n4. Tail management:  Because (★) couples far–apart indices, the proof must exploit delicate tail estimates (Step 3) rather than the straightforward “local” comparison used originally.\n\n5. Optimality construction:  Part (b) asks for a concrete counter-example showing that any smaller exponent fails.  Designing such an array and verifying both convergence and condition (★) requires additional creative insight.\n\nTaken together these features oblige the solver to combine combinatorial geometry, rearrangement of multiple sums, delicate limiting arguments and constructive examples, greatly elevating both the technical and conceptual load compared with the original problem."
      }
    },
    "original_kernel_variant": {
      "question": "Let $d\\ge 2$ be a fixed integer and let $M>1$ be a constant.  \nFor $n=(n_{1},\\dots ,n_{d})\\in\\mathbb N^{d}$ define  \n\\[\nQ(n)=\\Bigl\\{\\,k=(k_{1},\\dots ,k_{d})\\in\\mathbb N^{d}\\;:\\;\nn_{i}\\le k_{i}\\le 2n_{i}\\quad(1\\le i\\le d)\\Bigr\\}.\n\\]\n\nAssume that a non–negative array $(a_{n})_{n\\in\\mathbb N^{d}}$ satisfies  \n\\[\n\\sum_{n_{1}=1}^{\\infty}\\;\\dots\\;\\sum_{n_{d}=1}^{\\infty} a_{n}<\\infty,\n\\qquad \n\\sum_{k\\in Q(n)} a_{k}\\le M\\,a_{n}\\quad(n\\in\\mathbb N^{d}).\n\\tag{$\\star$}\n\\]\n\n(a)  Prove that\n\\[\n\\lim_{\\min_{1\\le i\\le d}n_{i}\\to\\infty}\n\\,(n_{1}n_{2}\\dots n_{d})\\,a_{n}=0 .\n\\tag{A}\n\\]\n\n(b)  Show that the weighted series\n\\[\n\\sum_{n\\in\\mathbb N^{d}}(n_{1}n_{2}\\dots n_{d})\\,a_{n}\n\\]\nis convergent.\n\n(The second assertion is strictly stronger than (A); its proof\nrequires a refined use of the counting argument employed in part (a).)",
      "solution": "Throughout the proof the symbols $C,C_{1},C_{2},\\dots$ denote\npositive constants depending only on $d$ and on the Carleson\nconstant $M$.\n\n\\medskip\n1.\\;A counting lemma.  \nFor $k=(k_{1},\\dots ,k_{d})\\in\\mathbb N^{d}$ put\n\\[\n\\lVert k\\rVert_{\\infty}:=\\max_{1\\le i\\le d}k_{i},\\qquad\n\\mathcal N(k):=\\bigl\\{\\,n\\in\\mathbb N^{d}\\;:\\;k\\in Q(n)\\bigr\\}.\n\\]\nIf $k_{i}\\ge 4$ for every $i$ then\n\\[\n\\#\\mathcal N(k)=\\prod_{i=1}^{d}\n\\bigl(k_{i}-\\lceil k_{i}/2\\rceil+1\\bigr)\n\\ge 2^{-d}\\,k_{1}\\dots k_{d}.\n\\tag{1.1}\n\\]\n\n\\medskip\n2.\\;A double–counting inequality.  \nFix $N\\in\\mathbb N$ and sum the inequalities in $(\\star)$ over\nall $n$ with $\\lVert n\\rVert_{\\infty}\\le N$:\n\\[\n\\sum_{\\lVert n\\rVert_{\\infty}\\le N}\\,\n\\sum_{k\\in Q(n)}a_{k}\\le\nM\\sum_{\\lVert n\\rVert_{\\infty}\\le N}a_{n}.\n\\]\nInterchanging the order of summation and using (1.1) yields\n\\[\n\\sum_{\\lVert k\\rVert_{\\infty}\\le N}\n(k_{1}\\dots k_{d})\\,a_{k}\n\\;\\le\\; C_{2}\\sum_{\\lVert n\\rVert_{\\infty}\\le N} a_{n}.\n\\tag{2.1}\n\\]\n\n\\medskip\n3.\\;Proof of part (a).  \nLet $S=\\sum_{n}a_{n}<\\infty$ and set\n\\[\nT_{N}:=\\sum_{\\lVert k\\rVert_{\\infty}=N}\n(k_{1}\\dots k_{d})\\,a_{k}\\qquad(N\\ge 1).\n\\]\nSumming (2.1) over $m=1,\\dots ,N$ gives\n\\[\n\\sum_{m=1}^{N}T_{m}\\le C_{2}S\\quad(N\\ge 1),\n\\]\nhence $(T_{N})_{N\\ge 1}$ is a summable, and therefore null, sequence:\n$T_{N}\\to 0$ as $N\\to\\infty$.\n\nFor an arbitrary index $k=(k_{1},\\dots ,k_{d})$ choose\n$N=\\lVert k\\rVert_{\\infty}$; then\n\\[\n(k_{1}\\dots k_{d})\\,a_{k}\\le T_{N}\\longrightarrow 0.\n\\]\nConsequently (A) is proved.\n\n\\bigskip\n4.\\;Proof of part (b).  \nBy letting $N\\to\\infty$ in (2.1) we obtain\n\\[\n\\sum_{k\\in\\mathbb N^{d}}\n(k_{1}\\dots k_{d})\\,a_{k}\\le C_{2}\n\\sum_{n\\in\\mathbb N^{d}}a_{n}=C_{2}S<\\infty,\n\\]\nwhich shows the desired convergence.\n\n\\medskip\nThis completes the solution of both parts of the problem.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.451686",
        "was_fixed": false,
        "difficulty_analysis": "Compared with the original one–dimensional, point-wise bounded problem, the enhanced variant is appreciably harder for several reasons:\n\n1. Higher dimension:  The indices live in ℕᵈ, forcing multi–parameter summations and combinatorial counting of lattice points inside high–dimensional boxes.\n\n2. Weaker hypothesis:  The bound (★) controls only the *aggregate* of many terms, not the individual terms themselves.  Recovering information about a single a_k from such coarse data demands a non–trivial double-counting argument.\n\n3. Necessity of sharp counting:  One must estimate from below how many n satisfy k∈Q(n); this calls for a careful combinatorial analysis (Step 1) that has no analogue in the original exercise.\n\n4. Tail management:  Because (★) couples far–apart indices, the proof must exploit delicate tail estimates (Step 3) rather than the straightforward “local” comparison used originally.\n\n5. Optimality construction:  Part (b) asks for a concrete counter-example showing that any smaller exponent fails.  Designing such an array and verifying both convergence and condition (★) requires additional creative insight.\n\nTaken together these features oblige the solver to combine combinatorial geometry, rearrangement of multiple sums, delicate limiting arguments and constructive examples, greatly elevating both the technical and conceptual load compared with the original problem."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}