{
  "index": "1961-A-3",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "3. Evaluate\n\\[\n\\lim _{n \\rightarrow \\infty} \\sum_{j=1}^{n^{2}} \\frac{n}{n^{2}+j^{2}} .\n\\]",
  "solution": "Solution. We write the sum in the form\n\\[\nS_{n}=\\frac{1}{n} \\sum_{i=1}^{n^{2}} \\frac{1}{1+\\left(\\frac{i}{n}\\right)^{2}} .\n\\]\n\nSince\n\\[\n\\int_{i / n}^{(i+1) / n} \\frac{d x}{1+x^{2}}<\\frac{1}{n} \\frac{1}{1+\\left(\\frac{i}{n}\\right)^{2}}<\\int_{(i-1) / n}^{1 / n} \\frac{d x}{1+x^{2}} .\n\\]\nwe get\n\\[\n\\int_{1 / n}^{\\left(n^{2}+1\\right) / n} \\frac{d x}{1+x^{2}}<S_{n}<\\int_{0}^{n} \\frac{d x}{1+x^{2}} .\n\\]\n\nNow\n\\[\n\\begin{array}{l}\n\\int_{0}^{n} \\frac{d x}{1+x^{2}}=\\arctan n . \\quad \\lim _{n-\\infty} \\arctan n=\\frac{\\pi}{2} \\\\\n\\int_{1 \\cdot n}^{n+\\left(11^{\\prime} n\\right)} \\frac{d x}{1+x^{2}}=\\arctan \\left(n+\\frac{1}{n}\\right)-\\arctan \\left(\\frac{1}{n}\\right) .\n\\end{array}\n\\]\nand\n\\[\n\\lim _{n \\rightarrow \\infty}\\left(\\arctan \\left(n+\\frac{1}{n}\\right)-\\arctan \\left(\\frac{1}{n}\\right)\\right)=\\frac{\\pi}{2} .\n\\]\n\nThus both the left and right members of (1) have the limit \\( \\pi / 2 \\), so\n\\[\n\\lim _{n \\rightarrow \\infty} S_{n}=\\frac{\\pi}{2}\n\\]",
  "vars": [
    "n",
    "j",
    "S_n",
    "i",
    "x"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "n": "biginteger",
        "j": "indexvar",
        "S_n": "sumvalue",
        "i": "iterator",
        "x": "realaxis"
      },
      "question": "3. Evaluate\n\\[\n\\lim _{biginteger \\rightarrow \\infty} \\sum_{indexvar=1}^{biginteger^{2}} \\frac{biginteger}{biginteger^{2}+indexvar^{2}} .\n\\]",
      "solution": "Solution. We write the sum in the form\n\\[\nsumvalue=\\frac{1}{biginteger} \\sum_{iterator=1}^{biginteger^{2}} \\frac{1}{1+\\left(\\frac{iterator}{biginteger}\\right)^{2}} .\n\\]\n\nSince\n\\[\n\\int_{iterator / biginteger}^{(iterator+1) / biginteger} \\frac{d realaxis}{1+realaxis^{2}}<\\frac{1}{biginteger} \\frac{1}{1+\\left(\\frac{iterator}{biginteger}\\right)^{2}}<\\int_{(iterator-1) / biginteger}^{1 / biginteger} \\frac{d realaxis}{1+realaxis^{2}} .\n\\]\nwe get\n\\[\n\\int_{1 / biginteger}^{\\left(biginteger^{2}+1\\right) / biginteger} \\frac{d realaxis}{1+realaxis^{2}}<sumvalue<\\int_{0}^{biginteger} \\frac{d realaxis}{1+realaxis^{2}} .\n\\]\n\nNow\n\\[\n\\begin{array}{l}\n\\int_{0}^{biginteger} \\frac{d realaxis}{1+realaxis^{2}}=\\arctan biginteger . \\quad \\lim _{biginteger-\\infty} \\arctan biginteger=\\frac{\\pi}{2} \\\\\n\\int_{1 \\cdot biginteger}^{biginteger+\\left(11^{\\prime} biginteger\\right)} \\frac{d realaxis}{1+realaxis^{2}}=\\arctan \\left(biginteger+\\frac{1}{biginteger}\\right)-\\arctan \\left(\\frac{1}{biginteger}\\right) .\n\\end{array}\n\\]\nand\n\\[\n\\lim _{biginteger \\rightarrow \\infty}\\left(\\arctan \\left(biginteger+\\frac{1}{biginteger}\\right)-\\arctan \\left(\\frac{1}{biginteger}\\right)\\right)=\\frac{\\pi}{2} .\n\\]\n\nThus both the left and right members of (1) have the limit \\( \\pi / 2 \\), so\n\\[\n\\lim _{biginteger \\rightarrow \\infty} sumvalue=\\frac{\\pi}{2}\n\\]"
    },
    "descriptive_long_confusing": {
      "map": {
        "n": "elephant",
        "j": "harmonica",
        "S_n": "watermelon",
        "x": "zeppelin"
      },
      "question": "\\[\n\\lim _{elephant \\rightarrow \\infty} \\sum_{harmonica=1}^{elephant^{2}} \\frac{elephant}{elephant^{2}+harmonica^{2}} .\n\\]",
      "solution": "Solution. We write the sum in the form\n\\[\nwatermelon=\\frac{1}{elephant} \\sum_{i=1}^{elephant^{2}} \\frac{1}{1+\\left(\\frac{i}{elephant}\\right)^{2}} .\n\\]\n\nSince\n\\[\n\\int_{i / elephant}^{(i+1) / elephant} \\frac{d \\zeppelin}{1+\\zeppelin^{2}}<\\frac{1}{elephant} \\frac{1}{1+\\left(\\frac{i}{elephant}\\right)^{2}}<\\int_{(i-1) / elephant}^{1 / elephant} \\frac{d \\zeppelin}{1+\\zeppelin^{2}} .\n\\]\nwe get\n\\[\n\\int_{1 / elephant}^{\\left(elephant^{2}+1\\right) / elephant} \\frac{d \\zeppelin}{1+\\zeppelin^{2}}<watermelon<\\int_{0}^{elephant} \\frac{d \\zeppelin}{1+\\zeppelin^{2}} .\n\\]\n\nNow\n\\[\n\\begin{array}{l}\n\\int_{0}^{elephant} \\frac{d \\zeppelin}{1+\\zeppelin^{2}}=\\arctan elephant . \\quad \\lim _{elephant-\\infty} \\arctan elephant=\\frac{\\pi}{2} \\\\\n\\int_{1 \\cdot elephant}^{elephant+\\left(11^{\\prime} elephant\\right)} \\frac{d \\zeppelin}{1+\\zeppelin^{2}}=\\arctan \\left(elephant+\\frac{1}{elephant}\\right)-\\arctan \\left(\\frac{1}{elephant}\\right) .\n\\end{array}\n\\]\nand\n\\[\n\\lim _{elephant \\rightarrow \\infty}\\left(\\arctan \\left(elephant+\\frac{1}{elephant}\\right)-\\arctan \\left(\\frac{1}{elephant}\\right)\\right)=\\frac{\\pi}{2} .\n\\]\n\nThus both the left and right members of (1) have the limit \\( \\pi / 2 \\), so\n\\[\n\\lim _{elephant \\rightarrow \\infty} watermelon=\\frac{\\pi}{2}\n\\]"
    },
    "descriptive_long_misleading": {
      "map": {
        "n": "infinitesimal",
        "j": "colossal",
        "S_n": "difference",
        "i": "enormous",
        "x": "dependent"
      },
      "question": "3. Evaluate\n\\[\n\\lim _{infinitesimal \\rightarrow \\infty} \\sum_{colossal=1}^{infinitesimal^{2}} \\frac{infinitesimal}{infinitesimal^{2}+colossal^{2}} .\n\\]\n",
      "solution": "Solution. We write the sum in the form\n\\[\ndifference=\\frac{1}{infinitesimal} \\sum_{enormous=1}^{infinitesimal^{2}} \\frac{1}{1+\\left(\\frac{enormous}{infinitesimal}\\right)^{2}} .\n\\]\n\nSince\n\\[\n\\int_{enormous / infinitesimal}^{(enormous+1) / infinitesimal} \\frac{d dependent}{1+dependent^{2}}<\\frac{1}{infinitesimal} \\frac{1}{1+\\left(\\frac{enormous}{infinitesimal}\\right)^{2}}<\\int_{(enormous-1) / infinitesimal}^{1 / infinitesimal} \\frac{d dependent}{1+dependent^{2}} .\n\\]\nwe get\n\\[\n\\int_{1 / infinitesimal}^{\\left(infinitesimal^{2}+1\\right) / infinitesimal} \\frac{d dependent}{1+dependent^{2}}<difference<\\int_{0}^{infinitesimal} \\frac{d dependent}{1+dependent^{2}} .\n\\]\n\nNow\n\\[\n\\begin{array}{l}\n\\int_{0}^{infinitesimal} \\frac{d dependent}{1+dependent^{2}}=\\arctan infinitesimal . \\quad \\lim _{infinitesimal-\\infty} \\arctan infinitesimal=\\frac{\\pi}{2} \\\\\n\\int_{1 \\cdot infinitesimal}^{infinitesimal+\\left(11^{\\prime} infinitesimal\\right)} \\frac{d dependent}{1+dependent^{2}}=\\arctan \\left(infinitesimal+\\frac{1}{infinitesimal}\\right)-\\arctan \\left(\\frac{1}{infinitesimal}\\right) .\n\\end{array}\n\\]\nand\n\\[\n\\lim _{infinitesimal \\rightarrow \\infty}\\left(\\arctan \\left(infinitesimal+\\frac{1}{infinitesimal}\\right)-\\arctan \\left(\\frac{1}{infinitesimal}\\right)\\right)=\\frac{\\pi}{2} .\n\\]\n\nThus both the left and right members of (1) have the limit \\( \\pi / 2 \\), so\n\\[\n\\lim _{infinitesimal \\rightarrow \\infty} difference=\\frac{\\pi}{2}\n\\]\n"
    },
    "garbled_string": {
      "map": {
        "n": "qzxwvtnp",
        "j": "hjgrksla",
        "S_n": "vbxkqplm",
        "i": "cnsdyhrg",
        "x": "twpqlkjs"
      },
      "question": "\\[\n\\lim _{qzxwvtnp \\rightarrow \\infty} \\sum_{hjgrksla=1}^{qzxwvtnp^{2}} \\frac{qzxwvtnp}{qzxwvtnp^{2}+hjgrksla^{2}} .\n\\]",
      "solution": "Solution. We write the sum in the form\n\\[\nvbxkqplm=\\frac{1}{qzxwvtnp} \\sum_{cnsdyhrg=1}^{qzxwvtnp^{2}} \\frac{1}{1+\\left(\\frac{cnsdyhrg}{qzxwvtnp}\\right)^{2}} .\n\\]\n\nSince\n\\[\n\\int_{cnsdyhrg / qzxwvtnp}^{(cnsdyhrg+1) / qzxwvtnp} \\frac{d twpqlkjs}{1+twpqlkjs^{2}}<\\frac{1}{qzxwvtnp} \\frac{1}{1+\\left(\\frac{cnsdyhrg}{qzxwvtnp}\\right)^{2}}<\\int_{(cnsdyhrg-1) / qzxwvtnp}^{1 / qzxwvtnp} \\frac{d twpqlkjs}{1+twpqlkjs^{2}} .\n\\]\nwe get\n\\[\n\\int_{1 / qzxwvtnp}^{\\left(qzxwvtnp^{2}+1\\right) / qzxwvtnp} \\frac{d twpqlkjs}{1+twpqlkjs^{2}}<vbxkqplm<\\int_{0}^{qzxwvtnp} \\frac{d twpqlkjs}{1+twpqlkjs^{2}} .\n\\]\n\nNow\n\\[\n\\begin{array}{l}\n\\int_{0}^{qzxwvtnp} \\frac{d twpqlkjs}{1+twpqlkjs^{2}}=\\arctan qzxwvtnp . \\quad \\lim _{qzxwvtnp-\\infty} \\arctan qzxwvtnp=\\frac{\\pi}{2} \\\\\n\\int_{1 \\cdot qzxwvtnp}^{qzxwvtnp+\\left(11^{\\prime} qzxwvtnp\\right)} \\frac{d twpqlkjs}{1+twpqlkjs^{2}}=\\arctan \\left(qzxwvtnp+\\frac{1}{qzxwvtnp}\\right)-\\arctan \\left(\\frac{1}{qzxwvtnp}\\right) .\n\\end{array}\n\\]\nand\n\\[\n\\lim _{qzxwvtnp \\rightarrow \\infty}\\left(\\arctan \\left(qzxwvtnp+\\frac{1}{qzxwvtnp}\\right)-\\arctan \\left(\\frac{1}{qzxwvtnp}\\right)\\right)=\\frac{\\pi}{2} .\n\\]\n\nThus both the left and right members of (1) have the limit \\( \\pi / 2 \\), so\n\\[\n\\lim _{qzxwvtnp \\rightarrow \\infty} vbxkqplm=\\frac{\\pi}{2}\n\\]"
    },
    "kernel_variant": {
      "question": "Let  \n\\[\nf(x)=\\frac{1}{1+x^{2}},\\qquad x\\ge 0 .\n\\]\n\nFor every positive integer \\(n\\) put  \n\\[\nS_{n}\\;=\\;\\frac{1}{n^{2}}\n\\sum_{\\substack{1\\le i,j\\le n^{2}\\\\\\gcd(i,j)=1}}\nf\\!\\Bigl(\\tfrac{i}{n}\\Bigr)\\,\nf\\!\\Bigl(\\tfrac{j}{n}\\Bigr).\n\\]\n\nEvaluate the limit  \n\\[\nL \\;=\\;\\lim_{n\\to\\infty} S_{n}.\n\\]\n\n(The Möbius function is denoted by \\(\\mu\\), the Riemann zeta–function by \\(\\zeta\\).)\n\n--------------------------------------------------------------------",
      "solution": "Throughout \\(C,C_{1},\\dots\\) denote positive absolute constants that may vary from line to line, and  \n\\[\n\\int f:=\\int_{0}^{\\infty}\\!f(x)\\,dx=\\frac{\\pi}{2}.\n\\]\n\nStep 1. Möbius inversion.  \nUsing the identity \\(\\mathbf{1}_{(\\gcd(i,j)=1)}=\\sum_{d\\mid i,\\,d\\mid j}\\mu(d)\\), we obtain  \n\\[\nS_{n}\n=\\frac{1}{n^{2}}\n\\sum_{d=1}^{n^{2}}\\mu(d)\n\\sum_{k=1}^{\\lfloor n^{2}/d\\rfloor}\n\\sum_{\\ell=1}^{\\lfloor n^{2}/d\\rfloor}\nf\\!\\Bigl(\\tfrac{dk}{n}\\Bigr)\\,\nf\\!\\Bigl(\\tfrac{d\\ell}{n}\\Bigr). \\tag{1}\n\\]\n\nPut \\(m_{d}(n):=\\lfloor n^{2}/d\\rfloor\\).  \nThe inner double sum factors:\n\\[\n\\sum_{k,\\ell\\le m_{d}(n)}f\\!\\Bigl(\\tfrac{dk}{n}\\Bigr)\nf\\!\\Bigl(\\tfrac{d\\ell}{n}\\Bigr)\n=\\Bigl[\\;\\sum_{k=1}^{m_{d}(n)}f\\!\\Bigl(\\tfrac{dk}{n}\\Bigr)\\Bigr]^{2}. \\tag{2}\n\\]\n\nStep 2. A uniform Riemann–sum estimate.  \nDefine  \n\\[\nA_{d,n}:=\\frac{d}{n}\\sum_{k=1}^{m_{d}(n)}f\\!\\Bigl(\\tfrac{dk}{n}\\Bigr).\n\\]\n\nBecause \\(f\\) is positive, decreasing and continuous, for every \\(k\\ge 1\\) we have  \n\\[\n\\int_{(k-1)d/n}^{kd/n}\\!f(x)\\,dx\n\\;\\ge\\;\\frac{d}{n}f\\!\\Bigl(\\tfrac{kd}{n}\\Bigr)\n\\;\\ge\\;\n\\int_{kd/n}^{(k+1)d/n}\\!f(x)\\,dx .\n\\]\n\nSumming \\(k=1,\\dots,m_{d}(n)\\) yields  \n\\[\n\\int_{d/n}^{(m_{d}(n)+1)d/n}\\!f\n\\;\\le\\;A_{d,n}\n\\;\\le\\;\n\\int_{0}^{m_{d}(n)d/n}\\!f. \\tag{3}\n\\]\n\nHence  \n\\[\n\\Bigl|A_{d,n}-\\!\\!\\int_{0}^{m_{d}(n)d/n}\\!f\\Bigr|\n\\le 2\\frac{d}{n}. \\tag{4}\n\\]\n\nWe now show that\n\\[\nm_{d}(n)\\,\\frac{d}{n}\\;\\ge\\;\\frac{n}{2}\\qquad(n\\ge 2,\\;1\\le d\\le n^{2}). \\tag{5}\n\\]\n\nIndeed, since \\(m_{d}(n)=\\lfloor n^{2}/d\\rfloor\\ge n^{2}/d-1\\),  \n\\[\nm_{d}(n)\\frac{d}{n}\\;\\ge\\;\\Bigl(\\frac{n^{2}}{d}-1\\Bigr)\\frac{d}{n}\n=\\;n-\\frac{d}{n}.\n\\]\nIf \\(d\\le n^{2}/2\\) the right–hand side is at least \\(n/2\\).  \nIf \\(d> n^{2}/2\\) then \\(m_{d}(n)=1\\) and \\(m_{d}(n)\\tfrac{d}{n}=d/n\\ge n^{2}/(2n)=n/2\\).  \nThus (5) holds in all cases.\n\nBecause \\(f(x)\\le x^{-2}\\) for \\(x\\ge 1\\) and \\(n/2\\ge 1\\) when \\(n\\ge 2\\), (5) gives  \n\\[\n\\int_{m_{d}(n)d/n}^{\\infty}f(x)\\,dx\n\\;\\le\\;\\int_{n/2}^{\\infty}x^{-2}\\,dx\n=\\frac{2}{n}. \\tag{6}\n\\]\n\nCombining (4) and (6) we obtain  \n\\[\n\\bigl|A_{d,n}-\\tfrac{\\pi}{2}\\bigr|\n\\le 2\\frac{d}{n}+\\frac{2}{n}\\le 4\\frac{d}{n}. \\tag{7}\n\\]\n\nA second fact needed later is the trivial bound  \n\\[\n0\\le A_{d,n}\\le\\int_{0}^{\\infty}f=\\frac{\\pi}{2}. \\tag{8}\n\\]\n\nStep 3. Separating the main term.  \nRewrite the inner sum as  \n\\[\n\\sum_{k=1}^{m_{d}(n)}f\\!\\Bigl(\\tfrac{dk}{n}\\Bigr)\n=\\frac{n}{d}\\Bigl(\\frac{\\pi}{2}+\\varepsilon_{d,n}\\Bigr),\\qquad\n|\\varepsilon_{d,n}|\\le 4\\frac{d}{n}, \\tag{9}\n\\]\nby (7).  \nInsert (9) into (1)–(2):\n\\[\nS_{n}\n=\\sum_{d=1}^{n^{2}}\\frac{\\mu(d)}{d^{2}}\n\\Bigl(\\frac{\\pi}{2}+\\varepsilon_{d,n}\\Bigr)^{2}. \\tag{10}\n\\]\n\nStep 4. Passage to the limit.\n\n(i) Main term.  \n\\[\n\\sum_{d\\le n^{2}}\\frac{\\mu(d)}{d^{2}}\n\\;\\xrightarrow[n\\to\\infty]{}\\;\n\\sum_{d=1}^{\\infty}\\frac{\\mu(d)}{d^{2}}\n=\\frac{1}{\\zeta(2)}\n=\\frac{6}{\\pi^{2}}. \\tag{11}\n\\]\n\n(ii) Linear error term.  \nUsing \\(|\\varepsilon_{d,n}|\\le 4d/n\\),\n\\[\n\\Bigl|\\sum_{d\\le n^{2}}\\frac{\\mu(d)}{d^{2}}\\varepsilon_{d,n}\\Bigr|\n\\le\\frac{4}{n}\\sum_{d\\le n^{2}}\\frac{1}{d}\n=O\\!\\Bigl(\\tfrac{\\log n}{n}\\Bigr)\\xrightarrow[n\\to\\infty]{}0. \\tag{12}\n\\]\n\n(iii) Quadratic error term.  \nDefine  \n\\[\nQ_{n}:=\\sum_{d\\le n^{2}}\\frac{\\mu(d)}{d^{2}}\\varepsilon_{d,n}^{2}. \\tag{13}\n\\]\n\nAbsolute convergence of \\(\\sum_{d\\ge 1}|\\,\\mu(d)|/d^{2}\\) is crucial.  \nFrom (8) we have \\(|\\varepsilon_{d,n}|\\le M:=\\pi/2\\).  \nFix \\(\\eta>0\\); choose \\(D=D(\\eta)\\) such that  \n\\[\n\\sum_{d>D}\\frac{|\\,\\mu(d)|}{d^{2}}<\\frac{\\eta}{M^{2}}. \\tag{14}\n\\]\n\nSplit (13) at \\(D\\):\n\\[\nQ_{n}=Q_{n}^{(1)}+Q_{n}^{(2)},\\quad\nQ_{n}^{(1)}:=\\sum_{d\\le D}\\frac{\\mu(d)}{d^{2}}\\varepsilon_{d,n}^{2},\\;\nQ_{n}^{(2)}:=\\sum_{D<d\\le n^{2}}\\frac{\\mu(d)}{d^{2}}\\varepsilon_{d,n}^{2}.\n\\]\n\nFor each fixed \\(d\\) we have \\(\\varepsilon_{d,n}\\to 0\\) by (9); hence \\(Q_{n}^{(1)}\\to 0\\).  \nFor the tail use (14):\n\\[\n|Q_{n}^{(2)}|\\le M^{2}\\sum_{d>D}\\frac{|\\,\\mu(d)|}{d^{2}}<\\eta. \\tag{15}\n\\]\n\nBecause \\(\\eta\\) is arbitrary, (15) forces \\(Q_{n}^{(2)}\\to 0\\), thus \\(Q_{n}\\to 0\\).\n\nStep 5. Conclusion.  \nCombining (10)–(12) and the above limit for \\(Q_{n}\\) we get  \n\\[\n\\lim_{n\\to\\infty}S_{n}\n=\\Bigl(\\frac{\\pi}{2}\\Bigr)^{2}\\cdot\\frac{1}{\\zeta(2)}\n=\\frac{\\pi^{2}}{4}\\cdot\\frac{6}{\\pi^{2}}\n=\\frac{3}{2}.\n\\]\n\\[\n\\boxed{L=\\dfrac{3}{2}}\n\\qquad\\qquad\\qquad\\qquad\\qquad\\qquad\\qquad\\qquad\\qquad\\qquad\\;\\;\\;\\;\\square\n\\]\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.527593",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension: the one–dimensional Riemann sum is replaced by a two–dimensional sum, drastically enlarging the combinatorial range.  \n• Additional constraint: the requirement \\(\\gcd(i,j)=1\\) introduces multiplicative number-theoretic structure, forcing use of Möbius inversion and the Riemann zeta function.  \n• Interaction of concepts: analysis (Riemann sums, dominated convergence) must be blended with analytic number theory (Dirichlet series, density of coprime pairs).  \n• Deeper theory: evaluating the limit hinges on understanding \\(\\sum\\mu(d)/d^{2}=1/\\zeta(2)\\), a non-elementary constant.  \n• More steps: the solution demands reformulation, inversion, asymptotic analysis of weighted sums, uniform error control, and identification of an Euler product—substantially beyond the single-integral estimate in the original problem."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\\[\nf(x)=\\frac{1}{1+x^{2}},\\qquad x\\ge 0 .\n\\]\n\nFor every positive integer \\(n\\) put  \n\\[\nS_{n}\\;=\\;\\frac{1}{n^{2}}\n\\sum_{\\substack{1\\le i,j\\le n^{2}\\\\\\gcd(i,j)=1}}\nf\\!\\Bigl(\\tfrac{i}{n}\\Bigr)\\,\nf\\!\\Bigl(\\tfrac{j}{n}\\Bigr).\n\\]\n\nEvaluate the limit  \n\\[\nL \\;=\\;\\lim_{n\\to\\infty} S_{n}.\n\\]\n\n(The Möbius function is denoted by \\(\\mu\\), the Riemann zeta–function by \\(\\zeta\\).)\n\n--------------------------------------------------------------------",
      "solution": "Throughout \\(C,C_{1},\\dots\\) denote positive absolute constants that may vary from line to line, and  \n\\[\n\\int f:=\\int_{0}^{\\infty}\\!f(x)\\,dx=\\frac{\\pi}{2}.\n\\]\n\nStep 1. Möbius inversion.  \nUsing the identity \\(\\mathbf{1}_{(\\gcd(i,j)=1)}=\\sum_{d\\mid i,\\,d\\mid j}\\mu(d)\\), we obtain  \n\\[\nS_{n}\n=\\frac{1}{n^{2}}\n\\sum_{d=1}^{n^{2}}\\mu(d)\n\\sum_{k=1}^{\\lfloor n^{2}/d\\rfloor}\n\\sum_{\\ell=1}^{\\lfloor n^{2}/d\\rfloor}\nf\\!\\Bigl(\\tfrac{dk}{n}\\Bigr)\\,\nf\\!\\Bigl(\\tfrac{d\\ell}{n}\\Bigr). \\tag{1}\n\\]\n\nPut \\(m_{d}(n):=\\lfloor n^{2}/d\\rfloor\\).  \nThe inner double sum factors:\n\\[\n\\sum_{k,\\ell\\le m_{d}(n)}f\\!\\Bigl(\\tfrac{dk}{n}\\Bigr)\nf\\!\\Bigl(\\tfrac{d\\ell}{n}\\Bigr)\n=\\Bigl[\\;\\sum_{k=1}^{m_{d}(n)}f\\!\\Bigl(\\tfrac{dk}{n}\\Bigr)\\Bigr]^{2}. \\tag{2}\n\\]\n\nStep 2. A uniform Riemann–sum estimate.  \nDefine  \n\\[\nA_{d,n}:=\\frac{d}{n}\\sum_{k=1}^{m_{d}(n)}f\\!\\Bigl(\\tfrac{dk}{n}\\Bigr).\n\\]\n\nBecause \\(f\\) is positive, decreasing and continuous, for every \\(k\\ge 1\\) we have  \n\\[\n\\int_{(k-1)d/n}^{kd/n}\\!f(x)\\,dx\n\\;\\ge\\;\\frac{d}{n}f\\!\\Bigl(\\tfrac{kd}{n}\\Bigr)\n\\;\\ge\\;\n\\int_{kd/n}^{(k+1)d/n}\\!f(x)\\,dx .\n\\]\n\nSumming \\(k=1,\\dots,m_{d}(n)\\) yields  \n\\[\n\\int_{d/n}^{(m_{d}(n)+1)d/n}\\!f\n\\;\\le\\;A_{d,n}\n\\;\\le\\;\n\\int_{0}^{m_{d}(n)d/n}\\!f. \\tag{3}\n\\]\n\nHence  \n\\[\n\\Bigl|A_{d,n}-\\!\\!\\int_{0}^{m_{d}(n)d/n}\\!f\\Bigr|\n\\le 2\\frac{d}{n}. \\tag{4}\n\\]\n\nWe now show that\n\\[\nm_{d}(n)\\,\\frac{d}{n}\\;\\ge\\;\\frac{n}{2}\\qquad(n\\ge 2,\\;1\\le d\\le n^{2}). \\tag{5}\n\\]\n\nIndeed, since \\(m_{d}(n)=\\lfloor n^{2}/d\\rfloor\\ge n^{2}/d-1\\),  \n\\[\nm_{d}(n)\\frac{d}{n}\\;\\ge\\;\\Bigl(\\frac{n^{2}}{d}-1\\Bigr)\\frac{d}{n}\n=\\;n-\\frac{d}{n}.\n\\]\nIf \\(d\\le n^{2}/2\\) the right–hand side is at least \\(n/2\\).  \nIf \\(d> n^{2}/2\\) then \\(m_{d}(n)=1\\) and \\(m_{d}(n)\\tfrac{d}{n}=d/n\\ge n^{2}/(2n)=n/2\\).  \nThus (5) holds in all cases.\n\nBecause \\(f(x)\\le x^{-2}\\) for \\(x\\ge 1\\) and \\(n/2\\ge 1\\) when \\(n\\ge 2\\), (5) gives  \n\\[\n\\int_{m_{d}(n)d/n}^{\\infty}f(x)\\,dx\n\\;\\le\\;\\int_{n/2}^{\\infty}x^{-2}\\,dx\n=\\frac{2}{n}. \\tag{6}\n\\]\n\nCombining (4) and (6) we obtain  \n\\[\n\\bigl|A_{d,n}-\\tfrac{\\pi}{2}\\bigr|\n\\le 2\\frac{d}{n}+\\frac{2}{n}\\le 4\\frac{d}{n}. \\tag{7}\n\\]\n\nA second fact needed later is the trivial bound  \n\\[\n0\\le A_{d,n}\\le\\int_{0}^{\\infty}f=\\frac{\\pi}{2}. \\tag{8}\n\\]\n\nStep 3. Separating the main term.  \nRewrite the inner sum as  \n\\[\n\\sum_{k=1}^{m_{d}(n)}f\\!\\Bigl(\\tfrac{dk}{n}\\Bigr)\n=\\frac{n}{d}\\Bigl(\\frac{\\pi}{2}+\\varepsilon_{d,n}\\Bigr),\\qquad\n|\\varepsilon_{d,n}|\\le 4\\frac{d}{n}, \\tag{9}\n\\]\nby (7).  \nInsert (9) into (1)–(2):\n\\[\nS_{n}\n=\\sum_{d=1}^{n^{2}}\\frac{\\mu(d)}{d^{2}}\n\\Bigl(\\frac{\\pi}{2}+\\varepsilon_{d,n}\\Bigr)^{2}. \\tag{10}\n\\]\n\nStep 4. Passage to the limit.\n\n(i) Main term.  \n\\[\n\\sum_{d\\le n^{2}}\\frac{\\mu(d)}{d^{2}}\n\\;\\xrightarrow[n\\to\\infty]{}\\;\n\\sum_{d=1}^{\\infty}\\frac{\\mu(d)}{d^{2}}\n=\\frac{1}{\\zeta(2)}\n=\\frac{6}{\\pi^{2}}. \\tag{11}\n\\]\n\n(ii) Linear error term.  \nUsing \\(|\\varepsilon_{d,n}|\\le 4d/n\\),\n\\[\n\\Bigl|\\sum_{d\\le n^{2}}\\frac{\\mu(d)}{d^{2}}\\varepsilon_{d,n}\\Bigr|\n\\le\\frac{4}{n}\\sum_{d\\le n^{2}}\\frac{1}{d}\n=O\\!\\Bigl(\\tfrac{\\log n}{n}\\Bigr)\\xrightarrow[n\\to\\infty]{}0. \\tag{12}\n\\]\n\n(iii) Quadratic error term.  \nDefine  \n\\[\nQ_{n}:=\\sum_{d\\le n^{2}}\\frac{\\mu(d)}{d^{2}}\\varepsilon_{d,n}^{2}. \\tag{13}\n\\]\n\nAbsolute convergence of \\(\\sum_{d\\ge 1}|\\,\\mu(d)|/d^{2}\\) is crucial.  \nFrom (8) we have \\(|\\varepsilon_{d,n}|\\le M:=\\pi/2\\).  \nFix \\(\\eta>0\\); choose \\(D=D(\\eta)\\) such that  \n\\[\n\\sum_{d>D}\\frac{|\\,\\mu(d)|}{d^{2}}<\\frac{\\eta}{M^{2}}. \\tag{14}\n\\]\n\nSplit (13) at \\(D\\):\n\\[\nQ_{n}=Q_{n}^{(1)}+Q_{n}^{(2)},\\quad\nQ_{n}^{(1)}:=\\sum_{d\\le D}\\frac{\\mu(d)}{d^{2}}\\varepsilon_{d,n}^{2},\\;\nQ_{n}^{(2)}:=\\sum_{D<d\\le n^{2}}\\frac{\\mu(d)}{d^{2}}\\varepsilon_{d,n}^{2}.\n\\]\n\nFor each fixed \\(d\\) we have \\(\\varepsilon_{d,n}\\to 0\\) by (9); hence \\(Q_{n}^{(1)}\\to 0\\).  \nFor the tail use (14):\n\\[\n|Q_{n}^{(2)}|\\le M^{2}\\sum_{d>D}\\frac{|\\,\\mu(d)|}{d^{2}}<\\eta. \\tag{15}\n\\]\n\nBecause \\(\\eta\\) is arbitrary, (15) forces \\(Q_{n}^{(2)}\\to 0\\), thus \\(Q_{n}\\to 0\\).\n\nStep 5. Conclusion.  \nCombining (10)–(12) and the above limit for \\(Q_{n}\\) we get  \n\\[\n\\lim_{n\\to\\infty}S_{n}\n=\\Bigl(\\frac{\\pi}{2}\\Bigr)^{2}\\cdot\\frac{1}{\\zeta(2)}\n=\\frac{\\pi^{2}}{4}\\cdot\\frac{6}{\\pi^{2}}\n=\\frac{3}{2}.\n\\]\n\\[\n\\boxed{L=\\dfrac{3}{2}}\n\\qquad\\qquad\\qquad\\qquad\\qquad\\qquad\\qquad\\qquad\\qquad\\qquad\\;\\;\\;\\;\\square\n\\]\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.441025",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension: the one–dimensional Riemann sum is replaced by a two–dimensional sum, drastically enlarging the combinatorial range.  \n• Additional constraint: the requirement \\(\\gcd(i,j)=1\\) introduces multiplicative number-theoretic structure, forcing use of Möbius inversion and the Riemann zeta function.  \n• Interaction of concepts: analysis (Riemann sums, dominated convergence) must be blended with analytic number theory (Dirichlet series, density of coprime pairs).  \n• Deeper theory: evaluating the limit hinges on understanding \\(\\sum\\mu(d)/d^{2}=1/\\zeta(2)\\), a non-elementary constant.  \n• More steps: the solution demands reformulation, inversion, asymptotic analysis of weighted sums, uniform error control, and identification of an Euler product—substantially beyond the single-integral estimate in the original problem."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}