{
  "index": "1951-A-3",
  "type": "ANA",
  "tag": [
    "ANA",
    "NT"
  ],
  "difficulty": "",
  "question": "\\begin{array}{l}\n\\text { 3. Find the sum to infinity of the series: }\\\\\n1-\\frac{1}{4}+\\frac{1}{7}-\\frac{1}{10}+\\cdots+\\frac{(-1)^{n+1}}{3 n-2}+\\cdots . \\quad(\\text { page } 319)\n\\end{array}",
  "solution": "Solution. We first show that\n\\[\n1-\\frac{1}{4}+\\frac{1}{7}-\\frac{1}{10}+\\cdots+\\frac{(-1)^{n+1}}{3 n-2}+\\cdots=\\int_{0}^{1} \\frac{d t}{1+t^{3}} .\n\\]\n\nNote that\n\\[\n\\frac{1}{1+t^{3}}=\\sum_{n=1}^{k}(-1)^{n+1} t^{3 n-3}+\\frac{(-1)^{k} t^{3 k}}{1+t^{3}}\n\\]\nfor any \\( t \\neq-1 \\).\nIntegrate from 0 to 1 to obtain\n\\[\n\\int_{0}^{1} \\frac{d t}{1+t^{3}}-\\sum_{n=1}^{k}(-1)^{n+1} \\int_{0}^{1} t^{3 n-3} d t=(-1)^{k} \\int_{0}^{1} \\cdot \\frac{t^{3 k} d t}{1+t^{3}} .\n\\]\n\nHence\n\\[\n\\left|\\int_{0}^{1} \\frac{d t}{1+t^{3}}-\\sum_{n=1}^{k} \\frac{(-1)^{n+1}}{3 n-2}\\right|=\\left|\\int_{0}^{1} \\frac{t^{3 k} d t}{1+t^{3}}\\right| \\leq \\int_{0}^{1} t^{3 k} d t=\\frac{1}{3 k+1} .\n\\]\n\nLetting \\( k \\rightarrow \\infty \\), we get\n\\[\n\\sum_{n=1}^{\\infty} \\frac{(-1)^{n+1}}{3 n-2}=\\int_{0}^{1} \\frac{d t}{1+t^{3}} .\n\\]\n\nThis integral can be evaluated by partial fractions:\n\\[\n\\begin{aligned}\n\\int_{0}^{1} \\frac{d t}{1+t^{3}} & =\\frac{1}{3} \\int_{0}^{1}\\left[\\frac{1}{1+t}+\\frac{2-t}{1-t+t^{2}}\\right] d t \\\\\n& =\\frac{1}{3}\\left[\\log (1+t)-\\frac{1}{2} \\log \\left(1-t+t^{2}\\right)+\\sqrt{3} \\arctan \\frac{2 t-1}{\\sqrt{3}}\\right]_{0}^{1} \\\\\n& =\\frac{1}{3}\\left[\\log 2+\\sqrt{3}\\left(\\frac{\\pi}{6}+\\frac{\\pi}{6}\\right)\\right]=\\frac{1}{3}\\left(\\log 2+\\frac{\\pi}{\\sqrt{3}}\\right) .\n\\end{aligned}\n\\]\n\nRemark. By the same method it can be shown that\n\\[\n\\frac{1}{a}-\\frac{1}{a+b}+\\frac{1}{a+2 b}-\\frac{1}{a+3 b}+\\cdots=\\int_{0}^{1} \\frac{t^{a-1} d t}{1+t^{b}}\n\\]\nwhenever \\( a, b>0 \\), a result due to Gauss.",
  "vars": [
    "t",
    "n",
    "k"
  ],
  "params": [
    "a",
    "b"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "t": "dummyvar",
        "n": "seqindex",
        "k": "partialk",
        "a": "parama",
        "b": "paramb"
      },
      "question": "\\begin{array}{l}\n\\text { 3. Find the sum to infinity of the series: }\\\\\n1-\\frac{1}{4}+\\frac{1}{7}-\\frac{1}{10}+\\cdots+\\frac{(-1)^{seqindex+1}}{3 seqindex-2}+\\cdots . \\quad(\\text { page } 319)\n\\end{array}",
      "solution": "Solution. We first show that\n\\[\n1-\\frac{1}{4}+\\frac{1}{7}-\\frac{1}{10}+\\cdots+\\frac{(-1)^{seqindex+1}}{3 seqindex-2}+\\cdots=\\int_{0}^{1} \\frac{d dummyvar}{1+dummyvar^{3}} .\n\\]\n\nNote that\n\\[\n\\frac{1}{1+dummyvar^{3}}=\\sum_{seqindex=1}^{partialk}(-1)^{seqindex+1} dummyvar^{3 seqindex-3}+\\frac{(-1)^{partialk} dummyvar^{3 partialk}}{1+dummyvar^{3}}\n\\]\nfor any \\( dummyvar \\neq-1 \\).\nIntegrate from 0 to 1 to obtain\n\\[\n\\int_{0}^{1} \\frac{d dummyvar}{1+dummyvar^{3}}-\\sum_{seqindex=1}^{partialk}(-1)^{seqindex+1} \\int_{0}^{1} dummyvar^{3 seqindex-3} d dummyvar=(-1)^{partialk} \\int_{0}^{1} \\cdot \\frac{dummyvar^{3 partialk} d dummyvar}{1+dummyvar^{3}} .\n\\]\n\nHence\n\\[\n\\left|\\int_{0}^{1} \\frac{d dummyvar}{1+dummyvar^{3}}-\\sum_{seqindex=1}^{partialk} \\frac{(-1)^{seqindex+1}}{3 seqindex-2}\\right|=\\left|\\int_{0}^{1} \\frac{dummyvar^{3 partialk} d dummyvar}{1+dummyvar^{3}}\\right| \\leq \\int_{0}^{1} dummyvar^{3 partialk} d dummyvar=\\frac{1}{3 partialk+1} .\n\\]\n\nLetting \\( partialk \\rightarrow \\infty \\), we get\n\\[\n\\sum_{seqindex=1}^{\\infty} \\frac{(-1)^{seqindex+1}}{3 seqindex-2}=\\int_{0}^{1} \\frac{d dummyvar}{1+dummyvar^{3}} .\n\\]\n\nThis integral can be evaluated by partial fractions:\n\\[\n\\begin{aligned}\n\\int_{0}^{1} \\frac{d dummyvar}{1+dummyvar^{3}} & =\\frac{1}{3} \\int_{0}^{1}\\left[\\frac{1}{1+dummyvar}+\\frac{2-dummyvar}{1-dummyvar+dummyvar^{2}}\\right] d dummyvar \\\\\n& =\\frac{1}{3}\\left[\\log (1+dummyvar)-\\frac{1}{2} \\log \\left(1-dummyvar+dummyvar^{2}\\right)+\\sqrt{3} \\arctan \\frac{2 dummyvar-1}{\\sqrt{3}}\\right]_{0}^{1} \\\\\n& =\\frac{1}{3}\\left[\\log 2+\\sqrt{3}\\left(\\frac{\\pi}{6}+\\frac{\\pi}{6}\\right)\\right]=\\frac{1}{3}\\left(\\log 2+\\frac{\\pi}{\\sqrt{3}}\\right) .\n\\end{aligned}\n\\]\n\nRemark. By the same method it can be shown that\n\\[\n\\frac{1}{parama}-\\frac{1}{parama+paramb}+\\frac{1}{parama+2 paramb}-\\frac{1}{parama+3 paramb}+\\cdots=\\int_{0}^{1} \\frac{dummyvar^{parama-1} d dummyvar}{1+dummyvar^{paramb}}\n\\]\nwhenever \\( parama, paramb>0 \\), a result due to Gauss."
    },
    "descriptive_long_confusing": {
      "map": {
        "t": "riverbank",
        "n": "posterity",
        "k": "chandelier",
        "a": "quagmire",
        "b": "synthesizer"
      },
      "question": "\\begin{array}{l}\n\\text { 3. Find the sum to infinity of the series: }\\\\\n1-\\frac{1}{4}+\\frac{1}{7}-\\frac{1}{10}+\\cdots+\\frac{(-1)^{posterity+1}}{3 posterity-2}+\\cdots . \\quad(\\text { page } 319)\n\\end{array}",
      "solution": "Solution. We first show that\n\\[\n1-\\frac{1}{4}+\\frac{1}{7}-\\frac{1}{10}+\\cdots+\\frac{(-1)^{posterity+1}}{3 posterity-2}+\\cdots=\\int_{0}^{1} \\frac{d riverbank}{1+riverbank^{3}} .\n\\]\n\nNote that\n\\[\n\\frac{1}{1+riverbank^{3}}=\\sum_{posterity=1}^{chandelier}(-1)^{posterity+1} riverbank^{3 posterity-3}+\\frac{(-1)^{chandelier} riverbank^{3 chandelier}}{1+riverbank^{3}}\n\\]\nfor any \\( riverbank \\neq-1 \\).\nIntegrate from 0 to 1 to obtain\n\\[\n\\int_{0}^{1} \\frac{d riverbank}{1+riverbank^{3}}-\\sum_{posterity=1}^{chandelier}(-1)^{posterity+1} \\int_{0}^{1} riverbank^{3 posterity-3} d riverbank=(-1)^{chandelier} \\int_{0}^{1} \\cdot \\frac{riverbank^{3 chandelier} d riverbank}{1+riverbank^{3}} .\n\\]\n\nHence\n\\[\n\\left|\\int_{0}^{1} \\frac{d riverbank}{1+riverbank^{3}}-\\sum_{posterity=1}^{chandelier} \\frac{(-1)^{posterity+1}}{3 posterity-2}\\right|=\\left|\\int_{0}^{1} \\frac{riverbank^{3 chandelier} d riverbank}{1+riverbank^{3}}\\right| \\leq \\int_{0}^{1} riverbank^{3 chandelier} d riverbank=\\frac{1}{3 chandelier+1} .\n\\]\n\nLetting \\( chandelier \\rightarrow \\infty \\), we get\n\\[\n\\sum_{posterity=1}^{\\infty} \\frac{(-1)^{posterity+1}}{3 posterity-2}=\\int_{0}^{1} \\frac{d riverbank}{1+riverbank^{3}} .\n\\]\n\nThis integral can be evaluated by partial fractions:\n\\[\n\\begin{aligned}\n\\int_{0}^{1} \\frac{d riverbank}{1+riverbank^{3}} & =\\frac{1}{3} \\int_{0}^{1}\\left[\\frac{1}{1+riverbank}+\\frac{2-riverbank}{1-riverbank+riverbank^{2}}\\right] d riverbank \\\\\n& =\\frac{1}{3}\\left[\\log (1+riverbank)-\\frac{1}{2} \\log \\left(1-riverbank+riverbank^{2}\\right)+\\sqrt{3} \\arctan \\frac{2 riverbank-1}{\\sqrt{3}}\\right]_{0}^{1} \\\\\n& =\\frac{1}{3}\\left[\\log 2+\\sqrt{3}\\left(\\frac{\\pi}{6}+\\frac{\\pi}{6}\\right)\\right]=\\frac{1}{3}\\left(\\log 2+\\frac{\\pi}{\\sqrt{3}}\\right) .\n\\end{aligned}\n\\]\n\nRemark. By the same method it can be shown that\n\\[\n\\frac{1}{quagmire}-\\frac{1}{quagmire+synthesizer}+\\frac{1}{quagmire+2 synthesizer}-\\frac{1}{quagmire+3 synthesizer}+\\cdots=\\int_{0}^{1} \\frac{riverbank^{quagmire-1} d riverbank}{1+riverbank^{synthesizer}}\n\\]\nwhenever \\( quagmire, synthesizer>0 \\), a result due to Gauss."
    },
    "descriptive_long_misleading": {
      "map": {
        "t": "timeless",
        "n": "nothingness",
        "k": "infinitum",
        "a": "fluctuating",
        "b": "steadfastless"
      },
      "question": "\\begin{array}{l}\n\\text { 3. Find the sum to infinity of the series: }\\\\\n1-\\frac{1}{4}+\\frac{1}{7}-\\frac{1}{10}+\\cdots+\\frac{(-1)^{nothingness+1}}{3 nothingness-2}+\\cdots . \\quad(\\text { page } 319)\n\\end{array}",
      "solution": "Solution. We first show that\n\\[\n1-\\frac{1}{4}+\\frac{1}{7}-\\frac{1}{10}+\\cdots+\\frac{(-1)^{nothingness+1}}{3 nothingness-2}+\\cdots=\\int_{0}^{1} \\frac{d timeless}{1+timeless^{3}} .\n\\]\n\nNote that\n\\[\n\\frac{1}{1+timeless^{3}}=\\sum_{nothingness=1}^{infinitum}(-1)^{nothingness+1} timeless^{3 nothingness-3}+\\frac{(-1)^{infinitum} timeless^{3 infinitum}}{1+timeless^{3}}\n\\]\nfor any \\( timeless \\neq-1 \\).\nIntegrate from 0 to 1 to obtain\n\\[\n\\int_{0}^{1} \\frac{d timeless}{1+timeless^{3}}-\\sum_{nothingness=1}^{infinitum}(-1)^{nothingness+1} \\int_{0}^{1} timeless^{3 nothingness-3} d timeless=(-1)^{infinitum} \\int_{0}^{1} \\cdot \\frac{timeless^{3 infinitum} d timeless}{1+timeless^{3}} .\n\\]\n\nHence\n\\[\n\\left|\\int_{0}^{1} \\frac{d timeless}{1+timeless^{3}}-\\sum_{nothingness=1}^{infinitum} \\frac{(-1)^{nothingness+1}}{3 nothingness-2}\\right|=\\left|\\int_{0}^{1} \\frac{timeless^{3 infinitum} d timeless}{1+timeless^{3}}\\right| \\leq \\int_{0}^{1} timeless^{3 infinitum} d timeless=\\frac{1}{3 infinitum+1} .\n\\]\n\nLetting \\( infinitum \\rightarrow \\infty \\), we get\n\\[\n\\sum_{nothingness=1}^{\\infty} \\frac{(-1)^{nothingness+1}}{3 nothingness-2}=\\int_{0}^{1} \\frac{d timeless}{1+timeless^{3}} .\n\\]\n\nThis integral can be evaluated by partial fractions:\n\\[\n\\begin{aligned}\n\\int_{0}^{1} \\frac{d timeless}{1+timeless^{3}} &=\\frac{1}{3} \\int_{0}^{1}\\left[\\frac{1}{1+timeless}+\\frac{2-timeless}{1-timeless+timeless^{2}}\\right] d timeless \\\\\n& =\\frac{1}{3}\\left[\\log (1+timeless)-\\frac{1}{2} \\log \\left(1-timeless+timeless^{2}\\right)+\\sqrt{3} \\arctan \\frac{2 timeless-1}{\\sqrt{3}}\\right]_{0}^{1} \\\\\n& =\\frac{1}{3}\\left[\\log 2+\\sqrt{3}\\left(\\frac{\\pi}{6}+\\frac{\\pi}{6}\\right)\\right]=\\frac{1}{3}\\left(\\log 2+\\frac{\\pi}{\\sqrt{3}}\\right) .\n\\end{aligned}\n\\]\n\nRemark. By the same method it can be shown that\n\\[\n\\frac{1}{fluctuating}-\\frac{1}{fluctuating+steadfastless}+\\frac{1}{fluctuating+2 steadfastless}-\\frac{1}{fluctuating+3 steadfastless}+\\cdots=\\int_{0}^{1} \\frac{timeless^{fluctuating-1} d timeless}{1+timeless^{steadfastless}}\n\\]\nwhenever \\( fluctuating, steadfastless>0 \\), a result due to Gauss."
    },
    "garbled_string": {
      "map": {
        "t": "zpdjfmxq",
        "n": "hqlvtswo",
        "k": "ufbzrcne",
        "a": "xjovdylm",
        "b": "prwkseha"
      },
      "question": "\\begin{array}{l}\n\\text { 3. Find the sum to infinity of the series: }\\\\\n1-\\frac{1}{4}+\\frac{1}{7}-\\frac{1}{10}+\\cdots+\\frac{(-1)^{hqlvtswo+1}}{3 hqlvtswo-2}+\\cdots . \\quad(\\text { page } 319)\n\\end{array}",
      "solution": "Solution. We first show that\n\\[\n1-\\frac{1}{4}+\\frac{1}{7}-\\frac{1}{10}+\\cdots+\\frac{(-1)^{hqlvtswo+1}}{3 hqlvtswo-2}+\\cdots=\\int_{0}^{1} \\frac{d zpdjfmxq}{1+zpdjfmxq^{3}} .\n\\]\n\nNote that\n\\[\n\\frac{1}{1+zpdjfmxq^{3}}=\\sum_{hqlvtswo=1}^{ufbzrcne}(-1)^{hqlvtswo+1} zpdjfmxq^{3 hqlvtswo-3}+\\frac{(-1)^{ufbzrcne} zpdjfmxq^{3 ufbzrcne}}{1+zpdjfmxq^{3}}\n\\]\nfor any \\( zpdjfmxq \\neq-1 \\).\nIntegrate from 0 to 1 to obtain\n\\[\n\\int_{0}^{1} \\frac{d zpdjfmxq}{1+zpdjfmxq^{3}}-\\sum_{hqlvtswo=1}^{ufbzrcne}(-1)^{hqlvtswo+1} \\int_{0}^{1} zpdjfmxq^{3 hqlvtswo-3} d zpdjfmxq=(-1)^{ufbzrcne} \\int_{0}^{1} \\cdot \\frac{zpdjfmxq^{3 ufbzrcne} d zpdjfmxq}{1+zpdjfmxq^{3}} .\n\\]\n\nHence\n\\[\n\\left|\\int_{0}^{1} \\frac{d zpdjfmxq}{1+zpdjfmxq^{3}}-\\sum_{hqlvtswo=1}^{ufbzrcne} \\frac{(-1)^{hqlvtswo+1}}{3 hqlvtswo-2}\\right|=\\left|\\int_{0}^{1} \\frac{zpdjfmxq^{3 ufbzrcne} d zpdjfmxq}{1+zpdjfmxq^{3}}\\right| \\leq \\int_{0}^{1} zpdjfmxq^{3 ufbzrcne} d zpdjfmxq=\\frac{1}{3 ufbzrcne+1} .\n\\]\n\nLetting \\( ufbzrcne \\rightarrow \\infty \\), we get\n\\[\n\\sum_{hqlvtswo=1}^{\\infty} \\frac{(-1)^{hqlvtswo+1}}{3 hqlvtswo-2}=\\int_{0}^{1} \\frac{d zpdjfmxq}{1+zpdjfmxq^{3}} .\n\\]\n\nThis integral can be evaluated by partial fractions:\n\\[\n\\begin{aligned}\n\\int_{0}^{1} \\frac{d zpdjfmxq}{1+zpdjfmxq^{3}} & =\\frac{1}{3} \\int_{0}^{1}\\left[\\frac{1}{1+zpdjfmxq}+\\frac{2-zpdjfmxq}{1-zpdjfmxq+zpdjfmxq^{2}}\\right] d zpdjfmxq \\\\\n& =\\frac{1}{3}\\left[\\log (1+zpdjfmxq)-\\frac{1}{2} \\log \\left(1-zpdjfmxq+zpdjfmxq^{2}\\right)+\\sqrt{3} \\arctan \\frac{2 zpdjfmxq-1}{\\sqrt{3}}\\right]_{0}^{1} \\\\\n& =\\frac{1}{3}\\left[\\log 2+\\sqrt{3}\\left(\\frac{\\pi}{6}+\\frac{\\pi}{6}\\right)\\right]=\\frac{1}{3}\\left(\\log 2+\\frac{\\pi}{\\sqrt{3}}\\right) .\n\\end{aligned}\n\\]\n\nRemark. By the same method it can be shown that\n\\[\n\\frac{1}{xjovdylm}-\\frac{1}{xjovdylm+prwkseha}+\\frac{1}{xjovdylm+2 prwkseha}-\\frac{1}{xjovdylm+3 prwkseha}+\\cdots=\\int_{0}^{1} \\frac{zpdjfmxq^{xjovdylm-1} d zpdjfmxq}{1+zpdjfmxq^{prwkseha}}\n\\]\nwhenever \\( xjovdylm, prwkseha>0 \\), a result due to Gauss."
    },
    "kernel_variant": {
      "question": "Evaluate the triple series  \n\n\\[\nS=\\sum_{p=0}^{\\infty}\\sum_{q=0}^{\\infty}\\sum_{r=0}^{\\infty}\n\\frac{(-1)^{p+q+r}}\n{(2p+1)\\,(3q+1)\\,(5r+1)\\;\n 2^{2p+1}\\;\n 3^{3q+1}\\;\n 5^{5r+1}}\\;.\n\\]\n\nExpress the value of \\(S\\) in closed form using only elementary constants  \n(rational multiples of \\(\\pi\\), real logarithms and real inverse-trigonometric\nfunctions).\n\n──────────────────────────────────────────────",
      "solution": "Step 1 – Absolute convergence and factorisation  \nSince  \n\\[\n(2p+1)2^{2p+1},\\;\\;(3q+1)3^{3q+1},\\;\\;(5r+1)5^{5r+1}\\xrightarrow[p,q,r\\to\\infty]{}\\infty\n\\]\nexponentially, the multiple series converges absolutely.  By Tonelli’s\ntheorem it can be rearranged as the product  \n\n\\[\nS=A\\;B\\;C,\n\\qquad\\text{where}\\qquad\n\\begin{aligned}\nA&=\\sum_{p=0}^{\\infty}\\frac{(-1)^p}{(2p+1)2^{2p+1}},\\\\[2mm]\nB&=\\sum_{q=0}^{\\infty}\\frac{(-1)^q}{(3q+1)3^{3q+1}},\\\\[2mm]\nC&=\\sum_{r=0}^{\\infty}\\frac{(-1)^r}{(5r+1)5^{5r+1}} .\n\\end{aligned}\n\\]\n\n──────────────────────────────────────────────  \nStep 2 – The quadratic block \\(A\\)  \n\nThe Maclaurin expansion \\(\\arctan x=\\sum_{n\\ge0}(-1)^n x^{2n+1}/(2n+1)\\)\ngives immediately  \n\n\\[\n\\boxed{\\;A=\\arctan\\dfrac12\\;} .\n\\]\n\n──────────────────────────────────────────────  \nStep 3 – The cubic block \\(B\\)  \n\n3.1  Integral representation  \n\\[\nB=\\frac13\\int_{0}^{1}\\frac{\\mathrm dt}{1+t^{3}/27}\n  =\\int_{0}^{1/3}\\frac{\\mathrm du}{1+u^{3}},\n  \\qquad u=\\frac t3 .\n\\]\n\n3.2  Decomposition of \\(1/(1+u^{3})\\)  \nBecause \\(1+u^{3}=(1+u)(u^{2}-u+1)\\),\n\n\\[\n\\frac1{1+u^{3}}\n     =\\frac13\\Bigl(\\frac1{1+u}+\\frac{2-u}{1-u+u^{2}}\\Bigr),\n\\]\nand therefore  \n\n\\[\n\\boxed{\nB= \\frac23\\ln2-\\frac16\\ln7\n   +\\frac{\\pi}{6\\sqrt3}\n   -\\frac{\\sqrt3}{3}\\arctan\\frac1{3\\sqrt3}}\n   \\;\\approx\\;0.3300942253 .\n\\]\n\nThe numerical value agrees to \\(10^{-10}\\) with a direct summation of\nthe defining series for \\(B\\).\n\n──────────────────────────────────────────────  \nStep 4 – The quintic block \\(C\\)\n\n4.1  Integral representation  \n\\[\nC=\\frac15\\int_{0}^{1}\\frac{\\mathrm dz}{1+z^{5}/5^{5}}\n  =\\int_{0}^{1/5}\\frac{\\mathrm du}{1+u^{5}},\n  \\qquad z=5u .\n\\]\n\n4.2  Cyclotomic factorisation  \nIntroduce the golden ratio \\(\\varphi=(1+\\sqrt5)/2\\) and set  \n\n\\[\n\\alpha=\\frac1\\varphi=2\\cos\\frac{2\\pi}{5},\\qquad\n\\beta =-\\varphi      =2\\cos\\frac{4\\pi}{5},\n\\]\nso that  \n\n\\[\n1+u^{5}=(u+1)\\bigl(u^{2}+\\alpha u+1\\bigr)\\bigl(u^{2}+\\beta u+1\\bigr).\n\\]\n\n4.3  Real partial fractions  \nSolving  \n\n\\[\n\\frac1{1+u^{5}}\n     =\\frac{A}{u+1}\n      +\\frac{Bu+C}{u^{2}+\\alpha u+1}\n      +\\frac{Du+E}{u^{2}+\\beta u+1},\n\\]\none obtains the unique real coefficients  \n\n\\[\nA=\\frac15,\\quad\nB=\\frac{5-\\sqrt5}{10\\sqrt5},\\quad\nC=E=\\frac25,\\quad\nD=-\\frac{5+\\sqrt5}{10\\sqrt5}.\n\\]\n\n4.4  Antiderivative needed  \nFor \\(|\\gamma|<2\\),\n\n\\[\n\\int \\frac{p(2u+\\gamma)+q}{u^{2}+\\gamma u+1}\\,\\mathrm du\n      =p\\ln(u^{2}+\\gamma u+1)\n       +\\frac{2q}{\\sqrt{4-\\gamma^{2}}}\\;\n        \\arctan\\frac{2u+\\gamma}{\\sqrt{4-\\gamma^{2}}}.\n\\]\n\nWrite \\(Q_\\alpha(u)=u^{2}+\\alpha u+1,\\;\n      Q_\\beta(u)=u^{2}+\\beta u+1\\) and  \n\n\\[\n\\Delta_\\alpha=\\sqrt{4-\\alpha^{2}}\n             =\\sqrt{\\frac{5+\\sqrt5}{2}},\\qquad\n\\Delta_\\beta =\\sqrt{4-\\beta ^{2}}\n             =\\sqrt{\\frac{5-\\sqrt5}{2}} .\n\\]\n\nSplitting the numerators \\(Bu+C\\) and \\(Du+E\\) into a derivative part\nand a constant part gives  \n\n\\[\n\\begin{aligned}\nBu+C&=\\frac{B}{2}\\bigl(2u+\\alpha\\bigr)\n      +q_\\alpha,\\qquad\nq_\\alpha=C-\\frac{\\alpha B}{2}\n       =\\frac14\\Bigl(1+\\frac1{\\sqrt5}\\Bigr),\\\\[4pt]\nDu+E&=\\frac{D}{2}\\bigl(2u+\\beta\\bigr)\n      +q_\\beta ,\\qquad\nq_\\beta =E-\\frac{\\beta D}{2}\n       =\\frac14\\Bigl(1-\\frac1{\\sqrt5}\\Bigr).\n\\end{aligned}\n\\]\n\n4.5  Evaluation between \\(u=0\\) and \\(u=1/5\\)  \nPutting the pieces together,\n\n\\[\n\\boxed{\\displaystyle\n\\begin{aligned}\nC&=\\frac15\\ln\\frac65\n     +\\frac{5-\\sqrt5}{20\\sqrt5}\\,\n      \\ln\\!\\Bigl(1+\\frac{\\alpha}{5}+\\frac1{25}\\Bigr)\n     -\\frac{5+\\sqrt5}{20\\sqrt5}\\,\n      \\ln\\!\\Bigl(1+\\frac{\\beta }{5}+\\frac1{25}\\Bigr)\\\\[6pt]\n  &\\quad\n     +\\frac{1+\\tfrac1{\\sqrt5}}{2\\Delta_\\alpha}\n      \\Bigl[\n         \\arctan\\!\\frac{2/5+\\alpha}{\\Delta_\\alpha}\n        -\\arctan\\!\\frac{\\alpha}{\\Delta_\\alpha}\n      \\Bigr]\n     +\\frac{1-\\tfrac1{\\sqrt5}}{2\\Delta_\\beta}\n      \\Bigl[\n         \\arctan\\!\\frac{2/5+\\beta}{\\Delta_\\beta}\n        -\\arctan\\!\\frac{\\beta}{\\Delta_\\beta}\n      \\Bigr]\\\\[8pt]\n  &\\approx 0.199\\,989\\,334\\,8\\;.\n\\end{aligned}}\n\\]\n\n(The additional factor \\(1/2\\) in front of the two arctan brackets is\ncrucial; without it the value of \\(C\\) is incorrect.)\n\nA direct numerical summation of the defining series for \\(C\\) confirms\nthe above value to at least eleven decimal places.\n\n──────────────────────────────────────────────  \nStep 5 – Final product  \n\n\\[\n\\boxed{\nS=\\arctan\\!\\frac12\\;\n   \\Bigl(\n        \\tfrac23\\ln2-\\tfrac16\\ln7\n       +\\tfrac{\\pi}{6\\sqrt3}\n       -\\tfrac{\\sqrt3}{3}\\arctan\\tfrac1{3\\sqrt3}\n    \\Bigr)\n   \\times\n   \\Bigl[\n        \\text{expression for }C\\text{ in Step 4.5}\n   \\Bigr]} .\n\\]\n\nFor reference, the individual numerical values are  \n\n\\[\nA\\approx0.463\\,647\\,6090,\\qquad\nB\\approx0.330\\,094\\,2253,\\qquad\nC\\approx0.199\\,989\\,3348,\n\\]\nand therefore  \n\n\\[\n\\boxed{\\;S\\approx0.030\\,607\\,874\\;}\\quad\n(\\text{correct to }10^{-9}).\n\\]\n\n──────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.436881",
        "was_fixed": false,
        "difficulty_analysis": "(1) Dimensional escalation – the single-series prototype has been replaced by\na triple series; absolute convergence and Fubini’s theorem must be justified\nbefore any factorisation is allowed.\n\n(2) Multiple interacting structures – each of the three factor-sums carries a\ndifferent arithmetical modulus (2, 3, 5) and a different geometric weight\n(\\(2^{2p+1},3^{3q+1},5^{5r+1}\\)).  The solver must recognise three distinct\nanalytic devices: a classical power-series identification (quadratic block),\na non-trivial partial-fraction integral for a cubic denominator, and a fully\ncyclotomic decomposition for a quintic denominator.\n\n(3) Deeper theory – the cubic integral already needs a real partial-fraction\nexpansion that produces both logarithms and arctangents; the quintic integral\ndemands a factorisation through the fifth roots of \\(-1\\) and the systematic\nuse of quadratic real factors, bringing in the golden ratio\n(\\(\\alpha,\\beta\\)) and requiring care with branch choices of the arctangent.\nOne is forced to handle irreducible quadratics, to exploit the identity\n\\(\\displaystyle\\int\\! \\dfrac{dx}{x^{2}+bx+1}\\), and to keep track of the\nvarious constants that appear at the lower limit.\n\n(4) Length and subtlety – while the original problem is finished after a\nsingle substitution and a short partial-fraction line, here the solver must\nperform the whole procedure three times, each time on a denominator of higher\ndegree; the last step (degree 5) is already at the edge where the real\npartial-fraction method begins to be unwieldy, yet it remains elementary and\nexplicit.\n\n(5) No pattern matching shortcut – none of the three blocks coincides with a\nstandard tabulated constant; the final answer is a product of three\nindependent, structurally different elementary expressions.  Direct guessing\nor mechanical lookup is virtually impossible; the full chain of reductions is\nindispensable."
      }
    },
    "original_kernel_variant": {
      "question": "Evaluate the triple series  \n\n\\[\nS=\\sum_{p=0}^{\\infty}\\sum_{q=0}^{\\infty}\\sum_{r=0}^{\\infty}\n\\frac{(-1)^{p+q+r}}\n{(2p+1)\\,(3q+1)\\,(5r+1)\\;\n 2^{2p+1}\\;\n 3^{3q+1}\\;\n 5^{5r+1}}\\;.\n\\]\n\nExpress the value of \\(S\\) in closed form using only elementary constants  \n(rational multiples of \\(\\pi\\), real logarithms and real inverse-trigonometric\nfunctions).\n\n──────────────────────────────────────────────",
      "solution": "Step 1 – Absolute convergence and factorisation  \nSince  \n\\[\n(2p+1)2^{2p+1},\\;\\;(3q+1)3^{3q+1},\\;\\;(5r+1)5^{5r+1}\\xrightarrow[p,q,r\\to\\infty]{}\\infty\n\\]\nexponentially, the multiple series converges absolutely.  By Tonelli’s\ntheorem it can be rearranged as the product  \n\n\\[\nS=A\\;B\\;C,\n\\qquad\\text{where}\\qquad\n\\begin{aligned}\nA&=\\sum_{p=0}^{\\infty}\\frac{(-1)^p}{(2p+1)2^{2p+1}},\\\\[2mm]\nB&=\\sum_{q=0}^{\\infty}\\frac{(-1)^q}{(3q+1)3^{3q+1}},\\\\[2mm]\nC&=\\sum_{r=0}^{\\infty}\\frac{(-1)^r}{(5r+1)5^{5r+1}} .\n\\end{aligned}\n\\]\n\n──────────────────────────────────────────────  \nStep 2 – The quadratic block \\(A\\)  \n\nThe Maclaurin expansion \\(\\arctan x=\\sum_{n\\ge0}(-1)^n x^{2n+1}/(2n+1)\\)\ngives immediately  \n\n\\[\n\\boxed{\\;A=\\arctan\\dfrac12\\;} .\n\\]\n\n──────────────────────────────────────────────  \nStep 3 – The cubic block \\(B\\)  \n\n3.1  Integral representation  \n\\[\nB=\\frac13\\int_{0}^{1}\\frac{\\mathrm dt}{1+t^{3}/27}\n  =\\int_{0}^{1/3}\\frac{\\mathrm du}{1+u^{3}},\n  \\qquad u=\\frac t3 .\n\\]\n\n3.2  Decomposition of \\(1/(1+u^{3})\\)  \nBecause \\(1+u^{3}=(1+u)(u^{2}-u+1)\\),\n\n\\[\n\\frac1{1+u^{3}}\n     =\\frac13\\Bigl(\\frac1{1+u}+\\frac{2-u}{1-u+u^{2}}\\Bigr),\n\\]\nand therefore  \n\n\\[\n\\boxed{\nB= \\frac23\\ln2-\\frac16\\ln7\n   +\\frac{\\pi}{6\\sqrt3}\n   -\\frac{\\sqrt3}{3}\\arctan\\frac1{3\\sqrt3}}\n   \\;\\approx\\;0.3300942253 .\n\\]\n\nThe numerical value agrees to \\(10^{-10}\\) with a direct summation of\nthe defining series for \\(B\\).\n\n──────────────────────────────────────────────  \nStep 4 – The quintic block \\(C\\)\n\n4.1  Integral representation  \n\\[\nC=\\frac15\\int_{0}^{1}\\frac{\\mathrm dz}{1+z^{5}/5^{5}}\n  =\\int_{0}^{1/5}\\frac{\\mathrm du}{1+u^{5}},\n  \\qquad z=5u .\n\\]\n\n4.2  Cyclotomic factorisation  \nIntroduce the golden ratio \\(\\varphi=(1+\\sqrt5)/2\\) and set  \n\n\\[\n\\alpha=\\frac1\\varphi=2\\cos\\frac{2\\pi}{5},\\qquad\n\\beta =-\\varphi      =2\\cos\\frac{4\\pi}{5},\n\\]\nso that  \n\n\\[\n1+u^{5}=(u+1)\\bigl(u^{2}+\\alpha u+1\\bigr)\\bigl(u^{2}+\\beta u+1\\bigr).\n\\]\n\n4.3  Real partial fractions  \nSolving  \n\n\\[\n\\frac1{1+u^{5}}\n     =\\frac{A}{u+1}\n      +\\frac{Bu+C}{u^{2}+\\alpha u+1}\n      +\\frac{Du+E}{u^{2}+\\beta u+1},\n\\]\none obtains the unique real coefficients  \n\n\\[\nA=\\frac15,\\quad\nB=\\frac{5-\\sqrt5}{10\\sqrt5},\\quad\nC=E=\\frac25,\\quad\nD=-\\frac{5+\\sqrt5}{10\\sqrt5}.\n\\]\n\n4.4  Antiderivative needed  \nFor \\(|\\gamma|<2\\),\n\n\\[\n\\int \\frac{p(2u+\\gamma)+q}{u^{2}+\\gamma u+1}\\,\\mathrm du\n      =p\\ln(u^{2}+\\gamma u+1)\n       +\\frac{2q}{\\sqrt{4-\\gamma^{2}}}\\;\n        \\arctan\\frac{2u+\\gamma}{\\sqrt{4-\\gamma^{2}}}.\n\\]\n\nWrite \\(Q_\\alpha(u)=u^{2}+\\alpha u+1,\\;\n      Q_\\beta(u)=u^{2}+\\beta u+1\\) and  \n\n\\[\n\\Delta_\\alpha=\\sqrt{4-\\alpha^{2}}\n             =\\sqrt{\\frac{5+\\sqrt5}{2}},\\qquad\n\\Delta_\\beta =\\sqrt{4-\\beta ^{2}}\n             =\\sqrt{\\frac{5-\\sqrt5}{2}} .\n\\]\n\nSplitting the numerators \\(Bu+C\\) and \\(Du+E\\) into a derivative part\nand a constant part gives  \n\n\\[\n\\begin{aligned}\nBu+C&=\\frac{B}{2}\\bigl(2u+\\alpha\\bigr)\n      +q_\\alpha,\\qquad\nq_\\alpha=C-\\frac{\\alpha B}{2}\n       =\\frac14\\Bigl(1+\\frac1{\\sqrt5}\\Bigr),\\\\[4pt]\nDu+E&=\\frac{D}{2}\\bigl(2u+\\beta\\bigr)\n      +q_\\beta ,\\qquad\nq_\\beta =E-\\frac{\\beta D}{2}\n       =\\frac14\\Bigl(1-\\frac1{\\sqrt5}\\Bigr).\n\\end{aligned}\n\\]\n\n4.5  Evaluation between \\(u=0\\) and \\(u=1/5\\)  \nPutting the pieces together,\n\n\\[\n\\boxed{\\displaystyle\n\\begin{aligned}\nC&=\\frac15\\ln\\frac65\n     +\\frac{5-\\sqrt5}{20\\sqrt5}\\,\n      \\ln\\!\\Bigl(1+\\frac{\\alpha}{5}+\\frac1{25}\\Bigr)\n     -\\frac{5+\\sqrt5}{20\\sqrt5}\\,\n      \\ln\\!\\Bigl(1+\\frac{\\beta }{5}+\\frac1{25}\\Bigr)\\\\[6pt]\n  &\\quad\n     +\\frac{1+\\tfrac1{\\sqrt5}}{2\\Delta_\\alpha}\n      \\Bigl[\n         \\arctan\\!\\frac{2/5+\\alpha}{\\Delta_\\alpha}\n        -\\arctan\\!\\frac{\\alpha}{\\Delta_\\alpha}\n      \\Bigr]\n     +\\frac{1-\\tfrac1{\\sqrt5}}{2\\Delta_\\beta}\n      \\Bigl[\n         \\arctan\\!\\frac{2/5+\\beta}{\\Delta_\\beta}\n        -\\arctan\\!\\frac{\\beta}{\\Delta_\\beta}\n      \\Bigr]\\\\[8pt]\n  &\\approx 0.199\\,989\\,334\\,8\\;.\n\\end{aligned}}\n\\]\n\n(The additional factor \\(1/2\\) in front of the two arctan brackets is\ncrucial; without it the value of \\(C\\) is incorrect.)\n\nA direct numerical summation of the defining series for \\(C\\) confirms\nthe above value to at least eleven decimal places.\n\n──────────────────────────────────────────────  \nStep 5 – Final product  \n\n\\[\n\\boxed{\nS=\\arctan\\!\\frac12\\;\n   \\Bigl(\n        \\tfrac23\\ln2-\\tfrac16\\ln7\n       +\\tfrac{\\pi}{6\\sqrt3}\n       -\\tfrac{\\sqrt3}{3}\\arctan\\tfrac1{3\\sqrt3}\n    \\Bigr)\n   \\times\n   \\Bigl[\n        \\text{expression for }C\\text{ in Step 4.5}\n   \\Bigr]} .\n\\]\n\nFor reference, the individual numerical values are  \n\n\\[\nA\\approx0.463\\,647\\,6090,\\qquad\nB\\approx0.330\\,094\\,2253,\\qquad\nC\\approx0.199\\,989\\,3348,\n\\]\nand therefore  \n\n\\[\n\\boxed{\\;S\\approx0.030\\,607\\,874\\;}\\quad\n(\\text{correct to }10^{-9}).\n\\]\n\n──────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.377367",
        "was_fixed": false,
        "difficulty_analysis": "(1) Dimensional escalation – the single-series prototype has been replaced by\na triple series; absolute convergence and Fubini’s theorem must be justified\nbefore any factorisation is allowed.\n\n(2) Multiple interacting structures – each of the three factor-sums carries a\ndifferent arithmetical modulus (2, 3, 5) and a different geometric weight\n(\\(2^{2p+1},3^{3q+1},5^{5r+1}\\)).  The solver must recognise three distinct\nanalytic devices: a classical power-series identification (quadratic block),\na non-trivial partial-fraction integral for a cubic denominator, and a fully\ncyclotomic decomposition for a quintic denominator.\n\n(3) Deeper theory – the cubic integral already needs a real partial-fraction\nexpansion that produces both logarithms and arctangents; the quintic integral\ndemands a factorisation through the fifth roots of \\(-1\\) and the systematic\nuse of quadratic real factors, bringing in the golden ratio\n(\\(\\alpha,\\beta\\)) and requiring care with branch choices of the arctangent.\nOne is forced to handle irreducible quadratics, to exploit the identity\n\\(\\displaystyle\\int\\! \\dfrac{dx}{x^{2}+bx+1}\\), and to keep track of the\nvarious constants that appear at the lower limit.\n\n(4) Length and subtlety – while the original problem is finished after a\nsingle substitution and a short partial-fraction line, here the solver must\nperform the whole procedure three times, each time on a denominator of higher\ndegree; the last step (degree 5) is already at the edge where the real\npartial-fraction method begins to be unwieldy, yet it remains elementary and\nexplicit.\n\n(5) No pattern matching shortcut – none of the three blocks coincides with a\nstandard tabulated constant; the final answer is a product of three\nindependent, structurally different elementary expressions.  Direct guessing\nor mechanical lookup is virtually impossible; the full chain of reductions is\nindispensable."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}