{
  "index": "1969-B-5",
  "type": "ANA",
  "tag": [
    "ANA",
    "NT"
  ],
  "difficulty": "",
  "question": "B-5. Let \\( a_{1}<a_{2}<a_{3}<\\cdots \\) be an increasing sequence of positive integers. Let the series\n\\[\n\\sum_{n=1}^{\\infty} 1 / a_{n}\n\\]\nbe convergent. For any number \\( x \\), let \\( k(x) \\) be the number of the \\( a_{n} \\) 's which do not exceed \\( x \\). Show that \\( \\lim _{x \\rightarrow \\infty} k(x) / x=0 \\).",
  "solution": "B-5 The following proof shows it is not necessary to stipulate that the \\( a_{n} \\) be integers. Suppose for some \\( \\epsilon>0 \\) there are \\( x_{j} \\rightarrow \\infty \\) with \\( k\\left(x_{j}\\right) / x_{j} \\geqq \\epsilon \\). Note that if \\( 1 \\leqq n \\leqq k\\left(x_{j}\\right) \\), then (because the \\( a_{n} \\) increase) \\( a_{n} \\leqq a_{k\\left(x_{j}\\right)} \\leqq x_{j} \\) and \\( 1 / a_{n} \\geqq 1 / x_{j} \\). Now for any positive integer \\( N \\),\n\\[\n\\sum_{n=N}^{\\infty} 1 / a_{n} \\geqq \\sup _{j} \\sum_{n=N}^{k\\left(x_{j}\\right)} 1 / a_{n} \\geqq \\sup _{j} \\frac{k\\left(x_{j}\\right)-N}{x_{j}} \\geqq \\sup _{j}\\left(\\epsilon-N / x_{j}\\right)=\\epsilon .\n\\]\n\nBut this contradicts the convergence of \\( \\sum_{n=1}^{\\infty} 1 / a_{n} \\), which implies\n\\[\n\\lim _{N \\rightarrow \\infty} \\sum_{n=N}^{\\infty} 1 / a_{n}=0\n\\]",
  "vars": [
    "n",
    "x",
    "j",
    "N"
  ],
  "params": [
    "a_n",
    "k",
    "\\\\epsilon",
    "x_j"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "n": "indexvar",
        "x": "variablex",
        "j": "indexj",
        "N": "cutoffn",
        "a_n": "serieselem",
        "k": "countfunc",
        "\\\\epsilon": "smallpos",
        "x_j": "samplept"
      },
      "question": "B-5. Let \\( serieselem_{1}<serieselem_{2}<serieselem_{3}<\\cdots \\) be an increasing sequence of positive integers. Let the series\n\\[\n\\sum_{indexvar=1}^{\\infty} 1 / serieselem_{indexvar}\n\\]\nbe convergent. For any number \\( variablex \\), let \\( countfunc(variablex) \\) be the number of the \\( serieselem_{indexvar} \\) 's which do not exceed \\( variablex \\). Show that \\( \\lim _{variablex \\rightarrow \\infty} countfunc(variablex) / variablex=0 \\).",
      "solution": "B-5 The following proof shows it is not necessary to stipulate that the \\( serieselem_{indexvar} \\) be integers. Suppose for some \\( smallpos>0 \\) there are \\( samplept \\rightarrow \\infty \\) with \\( countfunc\\left(samplept\\right) / samplept \\geqq smallpos \\). Note that if \\( 1 \\leqq indexvar \\leqq countfunc\\left(samplept\\right) \\), then (because the \\( serieselem_{indexvar} \\) increase) \\( serieselem_{indexvar} \\leqq serieselem_{countfunc\\left(samplept\\right)} \\leqq samplept \\) and \\( 1 / serieselem_{indexvar} \\geqq 1 / samplept \\). Now for any positive integer \\( cutoffn \\),\n\\[\n\\sum_{indexvar=cutoffn}^{\\infty} 1 / serieselem_{indexvar} \\geqq \\sup _{indexj} \\sum_{indexvar=cutoffn}^{countfunc\\left(samplept\\right)} 1 / serieselem_{indexvar} \\geqq \\sup _{indexj} \\frac{countfunc\\left(samplept\\right)-cutoffn}{samplept} \\geqq \\sup _{indexj}\\left(smallpos-cutoffn / samplept\\right)=smallpos .\n\\]\n\nBut this contradicts the convergence of \\( \\sum_{indexvar=1}^{\\infty} 1 / serieselem_{indexvar} \\), which implies\n\\[\n\\lim _{cutoffn \\rightarrow \\infty} \\sum_{indexvar=cutoffn}^{\\infty} 1 / serieselem_{indexvar}=0\n\\]\n"
    },
    "descriptive_long_confusing": {
      "map": {
        "n": "caterpillar",
        "x": "mulberry",
        "j": "watermelon",
        "N": "dragonfly",
        "a_n": "pumpkinseed",
        "k": "peppermint",
        "\\epsilon": "butterscotch",
        "x_j": "cheesecake"
      },
      "question": "B-5. Let \\( pumpkinseed_{1}<pumpkinseed_{2}<pumpkinseed_{3}<\\cdots \\) be an increasing sequence of positive integers. Let the series\n\\[\n\\sum_{caterpillar=1}^{\\infty} 1 / pumpkinseed_{caterpillar}\n\\]\nbe convergent. For any number \\( mulberry \\), let \\( peppermint(mulberry) \\) be the number of the \\( pumpkinseed_{caterpillar} \\)'s which do not exceed \\( mulberry \\). Show that \\( \\lim _{mulberry \\rightarrow \\infty} peppermint(mulberry) / mulberry=0 \\).",
      "solution": "B-5 The following proof shows it is not necessary to stipulate that the \\( pumpkinseed_{caterpillar} \\) be integers. Suppose for some \\( butterscotch>0 \\) there are \\( cheesecake \\rightarrow \\infty \\) with \\( peppermint\\left(cheesecake\\right) / cheesecake \\geqq butterscotch \\). Note that if \\( 1 \\leqq caterpillar \\leqq peppermint\\left(cheesecake\\right) \\), then (because the \\( pumpkinseed_{caterpillar} \\) increase) \\( pumpkinseed_{caterpillar} \\leqq pumpkinseed_{peppermint\\left(cheesecake\\right)} \\leqq cheesecake \\) and \\( 1 / pumpkinseed_{caterpillar} \\geqq 1 / cheesecake \\). Now for any positive integer \\( dragonfly \\),\n\\[\n\\sum_{caterpillar=dragonfly}^{\\infty} 1 / pumpkinseed_{caterpillar} \\geqq \\sup _{watermelon} \\sum_{caterpillar=dragonfly}^{peppermint\\left(cheesecake\\right)} 1 / pumpkinseed_{caterpillar} \\geqq \\sup _{watermelon} \\frac{peppermint\\left(cheesecake\\right)-dragonfly}{cheesecake} \\geqq \\sup _{watermelon}\\left(butterscotch-dragonfly / cheesecake\\right)=butterscotch .\n\\]\n\nBut this contradicts the convergence of \\( \\sum_{caterpillar=1}^{\\infty} 1 / pumpkinseed_{caterpillar} \\), which implies\n\\[\n\\lim _{dragonfly \\rightarrow \\infty} \\sum_{caterpillar=dragonfly}^{\\infty} 1 / pumpkinseed_{caterpillar}=0\n\\]"
    },
    "descriptive_long_misleading": {
      "map": {
        "n": "terminator",
        "x": "constant",
        "j": "principal",
        "N": "startpoint",
        "a_n": "continuous",
        "k": "neglector",
        "\\\\epsilon": "enormous",
        "x_j": "stationary"
      },
      "question": "B-5. Let \\( continuous_{1}<continuous_{2}<continuous_{3}<\\cdots \\) be an increasing sequence of positive integers. Let the series\n\\[\n\\sum_{terminator=1}^{\\infty} 1 / continuous_{terminator}\n\\]\nbe convergent. For any number \\( constant \\), let \\( neglector(constant) \\) be the number of the \\( continuous_{terminator} \\)'s which do not exceed \\( constant \\). Show that \\( \\lim _{constant \\rightarrow \\infty} neglector(constant) / constant=0 \\).",
      "solution": "B-5 The following proof shows it is not necessary to stipulate that the \\( continuous_{terminator} \\) be integers. Suppose for some \\( enormous>0 \\) there are \\( stationary \\rightarrow \\infty \\) with \\( neglector\\left(stationary\\right) / stationary \\geqq enormous \\). Note that if \\( 1 \\leqq terminator \\leqq neglector\\left(stationary\\right) \\), then (because the \\( continuous_{terminator} \\) increase) \\( continuous_{terminator} \\leqq continuous_{neglector\\left(stationary\\right)} \\leqq stationary \\) and \\( 1 / continuous_{terminator} \\geqq 1 / stationary \\). Now for any positive integer \\( startpoint \\),\n\\[\n\\sum_{terminator=startpoint}^{\\infty} 1 / continuous_{terminator} \\geqq \\sup _{principal} \\sum_{terminator=startpoint}^{neglector\\left(stationary\\right)} 1 / continuous_{terminator} \\geqq \\sup _{principal} \\frac{neglector\\left(stationary\\right)-startpoint}{stationary} \\geqq \\sup _{principal}\\left(enormous-startpoint / stationary\\right)=enormous .\n\\]\n\nBut this contradicts the convergence of \\( \\sum_{terminator=1}^{\\infty} 1 / continuous_{terminator} \\), which implies\n\\[\n\\lim _{startpoint \\rightarrow \\infty} \\sum_{terminator=startpoint}^{\\infty} 1 / continuous_{terminator}=0\n\\]"
    },
    "garbled_string": {
      "map": {
        "n": "zykmentha",
        "x": "plorzmuni",
        "j": "frebatquo",
        "N": "sibvarkol",
        "a_n": "qzxwvtnp",
        "k": "pajksldf",
        "\\epsilon": "vqomczrj",
        "x_j": "pouydmkl"
      },
      "question": "B-5. Let \\( qzxwvtnp_{1}<qzxwvtnp_{2}<qzxwvtnp_{3}<\\cdots \\) be an increasing sequence of positive integers. Let the series\n\\[\n\\sum_{zykmentha=1}^{\\infty} 1 / qzxwvtnp_{zykmentha}\n\\]\nbe convergent. For any number \\( plorzmuni \\), let \\( pajksldf(plorzmuni) \\) be the number of the \\( qzxwvtnp_{zykmentha} \\)'s which do not exceed \\( plorzmuni \\). Show that \\( \\lim _{plorzmuni \\rightarrow \\infty} pajksldf(plorzmuni) / plorzmuni=0 \\).",
      "solution": "B-5 The following proof shows it is not necessary to stipulate that the \\( qzxwvtnp_{zykmentha} \\) be integers. Suppose for some \\( vqomczrj>0 \\) there are \\( pouydmkl_{frebatquo} \\rightarrow \\infty \\) with \\( pajksldf\\left(pouydmkl_{frebatquo}\\right) / pouydmkl_{frebatquo} \\geqq vqomczrj \\). Note that if \\( 1 \\leqq zykmentha \\leqq pajksldf\\left(pouydmkl_{frebatquo}\\right) \\), then (because the \\( qzxwvtnp_{zykmentha} \\) increase) \\( qzxwvtnp_{zykmentha} \\leqq qzxwvtnp_{pajksldf\\left(pouydmkl_{frebatquo}\\right)} \\leqq pouydmkl_{frebatquo} \\) and \\( 1 / qzxwvtnp_{zykmentha} \\geqq 1 / pouydmkl_{frebatquo} \\). Now for any positive integer \\( sibvarkol \\),\n\\[\n\\sum_{zykmentha=sibvarkol}^{\\infty} 1 / qzxwvtnp_{zykmentha} \\geqq \\sup _{frebatquo} \\sum_{zykmentha=sibvarkol}^{pajksldf\\left(pouydmkl_{frebatquo}\\right)} 1 / qzxwvtnp_{zykmentha} \\geqq \\sup _{frebatquo} \\frac{pajksldf\\left(pouydmkl_{frebatquo}\\right)-sibvarkol}{pouydmkl_{frebatquo}} \\geqq \\sup _{frebatquo}\\left(vqomczrj-sibvarkol / pouydmkl_{frebatquo}\\right)=vqomczrj .\n\\]\n\nBut this contradicts the convergence of \\( \\sum_{zykmentha=1}^{\\infty} 1 / qzxwvtnp_{zykmentha} \\), which implies\n\\[\n\\lim _{sibvarkol \\rightarrow \\infty} \\sum_{zykmentha=sibvarkol}^{\\infty} 1 / qzxwvtnp_{zykmentha}=0\n\\]"
    },
    "kernel_variant": {
      "question": "Let d∈ℕ be fixed and let β>0.  \nConsider a countably-infinite set  \n\n S = {x₁,x₂,x₃,…} ⊂ ℝᵈ  \n\nwhose elements are listed in non-decreasing order of their Euclidean norms\n\n 0<r₁≤r₂≤r₃≤…, rₙ:=‖xₙ‖₂.\n\nAssume that the logarithmically weighted d-dimensional energy of S is finite  \n\n (★)  ∑_{n=1}^{∞} 1 / ( rₙ^{d}·(log(e+rₙ))^{1+β} ) < ∞.\n\n(Throughout log denotes the natural logarithm.)\n\nFor R≥e define the radial counting function  \n\n N(R):= #{ n : rₙ ≤ R }.\n\n(a)  Prove that  \n  lim_{R→∞} N(R)·(log R)^{1+β} / R^{d} = 0.\n\n(b)  Show that for every ε>0 there exists R₀(ε) such that  \n  N(R) ≤ ε·R^{d}/(log R)^{1+β} for all R≥R₀(ε).\n\n(c)  Prove optimality of the exponent 1+β:  \n  For every 0<γ<1+β construct a set S_γ ⊂ ℝᵈ that still satisfies (★) but for which  \n\n   limsup_{R→∞} N_γ(R)·(log R)^{γ}/R^{d}=∞,  \n\n  where N_γ denotes the counting function of S_γ.  \n  (Consequently, in (a) and (b) the power 1+β of log R cannot be replaced by any smaller exponent.)\n\n\n───────────────────────────────────────────────────",
      "solution": "Step 0 Notation  \nWrite log for the natural logarithm and restrict all radii to R≥e so that log R>0.  The symbols C, C′,… denote positive constants depending only on d, β (their value may change from line to line).\n\n────────────────────────  \nStep 1 Elementary comparison  \n\nBecause (rₙ) is non-decreasing, rₙ≤R holds for all n≤N(R).  Hence for these n\n\n 1 /( rₙ^{d}(log(e+rₙ))^{1+β} )\n  ≥ C /( R^{d}(log R)^{1+β} )         (1)\n\nwith C:=2^{-(1+β)}, since log(e+R)≤log 2+log R.\n\n────────────────────────  \nStep 2 Proof of (a)\n\nSuppose the conclusion fails.  Then there exist ε₀>0 and radii R_k→∞ such that  \n\n N(R_k) ≥ ε₀·R_k^{d}/(log R_k)^{1+β}.                                      (2)\n\nFix any M∈ℕ.  Using (1) and (2),\n\n ∑_{n=M}^{N(R_k)} 1 /( rₙ^{d}(log(e+rₙ))^{1+β} )\n  ≥ C·(N(R_k)−M)/(R_k^{d}(log R_k)^{1+β})\n  ≥ (Cε₀)/2                                  for all large k.\n\nPassing k→∞ contradicts the convergence of the series (★), because every tail\n∑_{n=M}^{∞} … would be bounded below by (Cε₀)/2>0.  Therefore the limit in (a) is indeed 0.\n\n────────────────────────  \nStep 3 Proof of (b)\n\nDefine F(R):=N(R)(log R)^{1+β}/R^{d}.  Step 2 shows lim_{R→∞}F(R)=0, hence for any ε>0 there is R₀(ε) such that F(R)<ε whenever R≥R₀(ε).  Rearranging yields\n\n N(R) ≤ ε·R^{d}/(log R)^{1+β}, R≥R₀(ε).\n\n────────────────────────  \nStep 4 Construction of S_γ for part (c)\n\nFix γ with 0<γ<1+β and put σ:=1+β−γ>0.  Choose shell radii  \n\n R_k := exp(exp k), k=1,2,….\n\nThen log R_k = e^{k} and the ratio R_{k+1}/R_k grows super-exponentially.\n\nPopulate the sphere of radius R_k with  \n\n m_k := ⌊ k·R_k^{d}/(log R_k)^{γ} ⌋                                    (3)\n\ndistinct, well-separated points (e.g. equally spaced on the sphere and then inflated to radius R_k).  Let  \n\n S_γ := ⋃_{k=1}^{∞} {those m_k points}.\n\nIndex the elements of S_γ in any order compatible with their radii; call the corresponding counting function N_γ.\n\n────────────────────────  \nStep 5 Energy condition (★) for S_γ\n\nAll points created in the k-th shell have norm R_k, so their total contribution to (★) equals\n\n m_k /( R_k^{d}(log(e+R_k))^{1+β} )\n ≤ C′· m_k /( R_k^{d}(log R_k)^{1+β} ).                               (4)\n\nUsing (3),\n\n m_k /( R_k^{d}(log R_k)^{1+β} )\n ≤ k/(log R_k)^{1+β+γ}.                                              (5)\n\nBecause log R_k=e^{k}, relation (5) becomes\n\n k·e^{-k(1+β+γ)}.\n\nSince 1+β+γ>1, the series ∑_{k≥1} k·e^{-k(1+β+γ)} converges geometrically, so the total energy is finite.  Thus S_γ satisfies (★).\n\n────────────────────────  \nStep 6 Violation of the logarithmic bound\n\nEvaluate the counting ratio solely at the special radii R_k.  For such R=R_k we have N_γ(R_k)≥m_k, hence by (3)\n\n N_γ(R_k)(log R_k)^{γ}/R_k^{d}\n  ≥ m_k (log R_k)^{γ}/R_k^{d}\n  ≥ k → ∞.                                                 (6)\n\nTherefore  \n\n limsup_{R→∞} N_γ(R)(log R)^{γ}/R^{d} = ∞,\n\nand part (c) is proved.  Because γ can be chosen arbitrarily close to 1+β from below, the exponent 1+β in parts (a) and (b) is optimal.\n\n───────────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.587629",
        "was_fixed": false,
        "difficulty_analysis": "•  Higher dimension:  Instead of a one-dimensional sequence, the problem is set in ℝᵈ with arbitrary fixed d, introducing geometric considerations absent from the original task.\n\n•  Logarithmic weights:  The summability condition includes a nonlinear logarithmic factor (log(e+‖x‖))^{1+β}, adding subtle growth-rate interplay that does not appear in either the original problem or the simple kernel variant.\n\n•  Multiple assertions:  The enhanced problem demands three separate results:\n   – an asymptotic density estimate [(a)];\n   – a uniform quantitative bound [(b)];\n   – and an optimality construction [(c)] showing that the exponent cannot be improved.\n  Each part requires a different technique (contradiction via tails of a series, limit-definition manipulation, and explicit sequence construction).\n\n•  Deeper theoretical tools:  The solution employs shell decomposition, tail estimates, and series comparison, and it constructs sparse sets by balancing point counts against weighted energy—concepts drawn from geometric measure theory and harmonic analysis rather than elementary series tests alone.\n\n•  Non-trivial optimality:  Part (c) forces the solver not only to prove upper bounds but also to recognize their sharpness and exhibit extremal examples, a level of sophistication far beyond the demands of the original Olympiad problem.\n\nTaken together, these extensions raise the conceptual, technical, and creative workload substantially, satisfying the requirement that the new kernel variant be “SIGNIFICANTLY harder” than both previous versions."
      }
    },
    "original_kernel_variant": {
      "question": "Let d∈ℕ be fixed and let β>0.  \nConsider a countably-infinite set  \n\n S = {x₁,x₂,x₃,…} ⊂ ℝᵈ  \n\nwhose elements are listed in non-decreasing order of their Euclidean norms\n\n 0<r₁≤r₂≤r₃≤…, rₙ:=‖xₙ‖₂.\n\nAssume that the logarithmically weighted d-dimensional energy of S is finite  \n\n (★)  ∑_{n=1}^{∞} 1 / ( rₙ^{d}·(log(e+rₙ))^{1+β} ) < ∞.\n\n(Throughout log denotes the natural logarithm.)\n\nFor R≥e define the radial counting function  \n\n N(R):= #{ n : rₙ ≤ R }.\n\n(a)  Prove that  \n  lim_{R→∞} N(R)·(log R)^{1+β} / R^{d} = 0.\n\n(b)  Show that for every ε>0 there exists R₀(ε) such that  \n  N(R) ≤ ε·R^{d}/(log R)^{1+β} for all R≥R₀(ε).\n\n(c)  Prove optimality of the exponent 1+β:  \n  For every 0<γ<1+β construct a set S_γ ⊂ ℝᵈ that still satisfies (★) but for which  \n\n   limsup_{R→∞} N_γ(R)·(log R)^{γ}/R^{d}=∞,  \n\n  where N_γ denotes the counting function of S_γ.  \n  (Consequently, in (a) and (b) the power 1+β of log R cannot be replaced by any smaller exponent.)\n\n\n───────────────────────────────────────────────────",
      "solution": "Step 0 Notation  \nWrite log for the natural logarithm and restrict all radii to R≥e so that log R>0.  The symbols C, C′,… denote positive constants depending only on d, β (their value may change from line to line).\n\n────────────────────────  \nStep 1 Elementary comparison  \n\nBecause (rₙ) is non-decreasing, rₙ≤R holds for all n≤N(R).  Hence for these n\n\n 1 /( rₙ^{d}(log(e+rₙ))^{1+β} )\n  ≥ C /( R^{d}(log R)^{1+β} )         (1)\n\nwith C:=2^{-(1+β)}, since log(e+R)≤log 2+log R.\n\n────────────────────────  \nStep 2 Proof of (a)\n\nSuppose the conclusion fails.  Then there exist ε₀>0 and radii R_k→∞ such that  \n\n N(R_k) ≥ ε₀·R_k^{d}/(log R_k)^{1+β}.                                      (2)\n\nFix any M∈ℕ.  Using (1) and (2),\n\n ∑_{n=M}^{N(R_k)} 1 /( rₙ^{d}(log(e+rₙ))^{1+β} )\n  ≥ C·(N(R_k)−M)/(R_k^{d}(log R_k)^{1+β})\n  ≥ (Cε₀)/2                                  for all large k.\n\nPassing k→∞ contradicts the convergence of the series (★), because every tail\n∑_{n=M}^{∞} … would be bounded below by (Cε₀)/2>0.  Therefore the limit in (a) is indeed 0.\n\n────────────────────────  \nStep 3 Proof of (b)\n\nDefine F(R):=N(R)(log R)^{1+β}/R^{d}.  Step 2 shows lim_{R→∞}F(R)=0, hence for any ε>0 there is R₀(ε) such that F(R)<ε whenever R≥R₀(ε).  Rearranging yields\n\n N(R) ≤ ε·R^{d}/(log R)^{1+β}, R≥R₀(ε).\n\n────────────────────────  \nStep 4 Construction of S_γ for part (c)\n\nFix γ with 0<γ<1+β and put σ:=1+β−γ>0.  Choose shell radii  \n\n R_k := exp(exp k), k=1,2,….\n\nThen log R_k = e^{k} and the ratio R_{k+1}/R_k grows super-exponentially.\n\nPopulate the sphere of radius R_k with  \n\n m_k := ⌊ k·R_k^{d}/(log R_k)^{γ} ⌋                                    (3)\n\ndistinct, well-separated points (e.g. equally spaced on the sphere and then inflated to radius R_k).  Let  \n\n S_γ := ⋃_{k=1}^{∞} {those m_k points}.\n\nIndex the elements of S_γ in any order compatible with their radii; call the corresponding counting function N_γ.\n\n────────────────────────  \nStep 5 Energy condition (★) for S_γ\n\nAll points created in the k-th shell have norm R_k, so their total contribution to (★) equals\n\n m_k /( R_k^{d}(log(e+R_k))^{1+β} )\n ≤ C′· m_k /( R_k^{d}(log R_k)^{1+β} ).                               (4)\n\nUsing (3),\n\n m_k /( R_k^{d}(log R_k)^{1+β} )\n ≤ k/(log R_k)^{1+β+γ}.                                              (5)\n\nBecause log R_k=e^{k}, relation (5) becomes\n\n k·e^{-k(1+β+γ)}.\n\nSince 1+β+γ>1, the series ∑_{k≥1} k·e^{-k(1+β+γ)} converges geometrically, so the total energy is finite.  Thus S_γ satisfies (★).\n\n────────────────────────  \nStep 6 Violation of the logarithmic bound\n\nEvaluate the counting ratio solely at the special radii R_k.  For such R=R_k we have N_γ(R_k)≥m_k, hence by (3)\n\n N_γ(R_k)(log R_k)^{γ}/R_k^{d}\n  ≥ m_k (log R_k)^{γ}/R_k^{d}\n  ≥ k → ∞.                                                 (6)\n\nTherefore  \n\n limsup_{R→∞} N_γ(R)(log R)^{γ}/R^{d} = ∞,\n\nand part (c) is proved.  Because γ can be chosen arbitrarily close to 1+β from below, the exponent 1+β in parts (a) and (b) is optimal.\n\n───────────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.472042",
        "was_fixed": false,
        "difficulty_analysis": "•  Higher dimension:  Instead of a one-dimensional sequence, the problem is set in ℝᵈ with arbitrary fixed d, introducing geometric considerations absent from the original task.\n\n•  Logarithmic weights:  The summability condition includes a nonlinear logarithmic factor (log(e+‖x‖))^{1+β}, adding subtle growth-rate interplay that does not appear in either the original problem or the simple kernel variant.\n\n•  Multiple assertions:  The enhanced problem demands three separate results:\n   – an asymptotic density estimate [(a)];\n   – a uniform quantitative bound [(b)];\n   – and an optimality construction [(c)] showing that the exponent cannot be improved.\n  Each part requires a different technique (contradiction via tails of a series, limit-definition manipulation, and explicit sequence construction).\n\n•  Deeper theoretical tools:  The solution employs shell decomposition, tail estimates, and series comparison, and it constructs sparse sets by balancing point counts against weighted energy—concepts drawn from geometric measure theory and harmonic analysis rather than elementary series tests alone.\n\n•  Non-trivial optimality:  Part (c) forces the solver not only to prove upper bounds but also to recognize their sharpness and exhibit extremal examples, a level of sophistication far beyond the demands of the original Olympiad problem.\n\nTaken together, these extensions raise the conceptual, technical, and creative workload substantially, satisfying the requirement that the new kernel variant be “SIGNIFICANTLY harder” than both previous versions."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}