{
  "index": "2002-A-6",
  "type": "ANA",
  "tag": [
    "ANA",
    "NT"
  ],
  "difficulty": "",
  "question": "Fix an integer $b \\geq 2$. Let $f(1) = 1$, $f(2) = 2$, and for each\n$n \\geq 3$, define $f(n) = n f(d)$, where $d$ is the number of\nbase-$b$ digits of $n$. For which values of $b$ does\n\\[\n\\sum_{n=1}^\\infty \\frac{1}{f(n)}\n\\]\nconverge?",
  "solution": "The sum converges for $b=2$ and diverges for $b \\geq 3$.\nWe first consider $b \\geq 3$. Suppose the sum converges;\nthen the fact\nthat $f(n) = n f(d)$ whenever $b^{d-1} \\leq n \\leq b^{d} - 1$ yields\n\\begin{equation} \\label{a6eq1}\n\\sum_{n=1}^\\infty \\frac{1}{f(n)}\n= \\sum_{d=1}^\\infty \\frac{1}{f(d)} \\sum_{n=b^{d-1}}^{b^d - 1} \\frac{1}{n}.\n\\end{equation}\nHowever, by comparing the integral of $1/x$ with a Riemann sum,\nwe see that\n\\begin{align*}\n\\sum_{n=b^{d-1}}^{b^d - 1} \\frac{1}{n}\n&> \\int_{b^{d-1}}^{b^d} \\frac{dx}{x} \\\\\n&= \\log (b^d) - \\log (b^{d-1}) = \\log b,\n\\end{align*}\nwhere $\\log$ denotes the natural logarithm. Thus \\eqref{a6eq1} yields\n\\[\n\\sum_{n=1}^\\infty \\frac{1}{f(n)}\n> (\\log b) \\sum_{n=1}^\\infty \\frac{1}{f(n)},\n\\]\na contradiction since $\\log b > 1$ for $b \\geq 3$. Therefore the\nsum diverges.\n\nFor $b=2$, we have a slightly different identity because $f(2) \\neq\n2 f(2)$. Instead, for any positive integer $i$, we have\n\\begin{equation} \\label{a6eq2}\n\\sum_{n=1}^{2^i-1} \\frac{1}{f(n)}\n= 1 + \\frac{1}{2} + \\frac{1}{6} +\n\\sum_{d=3}^i \\frac{1}{f(d)} \\sum_{n=2^{d-1}}^{2^d - 1} \\frac{1}{n}.\n\\end{equation}\nAgain comparing an integral to a Riemann sum, we see that for $d\\geq 3$,\n\\begin{align*}\n\\sum_{n=2^{d-1}}^{2^d - 1} \\frac{1}{n} &<\n\\frac{1}{2^{d-1}} - \\frac{1}{2^d} + \\int_{2^{d-1}}^{2^d} \\frac{dx}{x}\n\\\\\n&= \\frac{1}{2^d} + \\log 2 \\\\\n&\\leq \\frac{1}{8} + \\log 2 < 0.125 + 0.7 < 1.\n\\end{align*}\nPut $c = \\frac{1}{8} + \\log 2$ and $L = 1+\\frac{1}{2} +\n\\frac{1}{6(1-c)}$. Then we can prove that $\\sum_{n=1}^{2^i-1}\n\\frac{1}{f(n)} < L$ for all $i \\geq 2$ by induction on $i$. The case\n$i=2$ is clear. For the induction, note that by \\eqref{a6eq2},\n\\begin{align*}\n\\sum_{n=1}^{2^i-1} \\frac{1}{f(n)}\n&< 1 + \\frac{1}{2} + \\frac{1}{6} + c \\sum_{d=3}^i \\frac{1}{f(d)} \\\\\n&< 1 + \\frac{1}{2} + \\frac{1}{6} + c \\frac{1}{6(1-c)} \\\\\n&= 1 + \\frac{1}{2} + \\frac{1}{6(1-c)} = L,\n\\end{align*}\nas desired. We conclude that $\\sum_{n=1}^\\infty \\frac{1}{f(n)}$\nconverges to a limit less than or equal to $L$.\n\nNote: the above argument proves that the sum for $b=2$ is at most\n$L < 2.417$. One can also obtain a lower bound by the same technique,\nnamely $1 + \\frac{1}{2} + \\frac{1}{6(1 - c')}$ with $c' = \\log 2$.\nThis bound exceeds $2.043$. (By contrast, summing the first 100000 terms of\nthe series only yields a lower bound of $1.906$.)\nRepeating the same arguments with $d \\geq 4$\nas the cutoff yields the upper bound $2.185$ and the lower bound $2.079$.",
  "vars": [
    "n",
    "d",
    "i",
    "x"
  ],
  "params": [
    "b",
    "f",
    "c",
    "L"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "n": "indexval",
        "d": "digitnum",
        "i": "iteridx",
        "x": "realvar",
        "b": "baseint",
        "f": "funcmap",
        "c": "boundval",
        "L": "limupper"
      },
      "question": "Fix an integer $baseint \\geq 2$. Let $funcmap(1) = 1$, $funcmap(2) = 2$, and for each\n$indexval \\geq 3$, define $funcmap(indexval) = indexval\\, funcmap(digitnum)$, where $digitnum$ is the number of\nbase-$baseint$ digits of $indexval$. For which values of $baseint$ does\n\\[\n\\sum_{indexval=1}^\\infty \\frac{1}{funcmap(indexval)}\n\\]\nconverge?",
      "solution": "The sum converges for $baseint=2$ and diverges for $baseint \\geq 3$.\nWe first consider $baseint \\geq 3$. Suppose the sum converges; then the fact that $funcmap(indexval)= indexval\\, funcmap(digitnum)$ whenever $baseint^{digitnum-1}\\leq indexval \\leq baseint^{digitnum}-1$ yields\n\\begin{equation} \\label{a6eq1}\n\\sum_{indexval=1}^\\infty \\frac{1}{funcmap(indexval)}\n= \\sum_{digitnum=1}^\\infty \\frac{1}{funcmap(digitnum)} \\sum_{indexval=baseint^{digitnum-1}}^{baseint^{digitnum}-1} \\frac{1}{indexval}.\n\\end{equation}\nHowever, by comparing the integral of $1/realvar$ with a Riemann sum,\nwe see that\n\\begin{align*}\n\\sum_{indexval=baseint^{digitnum-1}}^{baseint^{digitnum}-1} \\frac{1}{indexval}\n&> \\int_{baseint^{digitnum-1}}^{baseint^{digitnum}} \\frac{\\mathrm{d}realvar}{realvar} \\\\\n&= \\log (baseint^{digitnum}) - \\log (baseint^{digitnum-1}) = \\log baseint,\n\\end{align*}\nwhere $\\log$ denotes the natural logarithm. Thus \\eqref{a6eq1} yields\n\\[\n\\sum_{indexval=1}^\\infty \\frac{1}{funcmap(indexval)}\n> (\\log baseint) \\sum_{indexval=1}^\\infty \\frac{1}{funcmap(indexval)},\n\\]\na contradiction since $\\log baseint > 1$ for $baseint \\geq 3$. Therefore the\nsum diverges.\n\nFor $baseint=2$, we have a slightly different identity because $funcmap(2) \\neq 2\\, funcmap(2)$. Instead, for any positive integer $iteridx$, we have\n\\begin{equation} \\label{a6eq2}\n\\sum_{indexval=1}^{2^{iteridx}-1} \\frac{1}{funcmap(indexval)}\n= 1 + \\frac{1}{2} + \\frac{1}{6} +\n\\sum_{digitnum=3}^{iteridx} \\frac{1}{funcmap(digitnum)} \\sum_{indexval=2^{digitnum-1}}^{2^{digitnum}-1} \\frac{1}{indexval}.\n\\end{equation}\nAgain comparing an integral to a Riemann sum, we see that for $digitnum \\geq 3$,\n\\begin{align*}\n\\sum_{indexval=2^{digitnum-1}}^{2^{digitnum}-1} \\frac{1}{indexval}\n&< \\frac{1}{2^{digitnum-1}} - \\frac{1}{2^{digitnum}} + \\int_{2^{digitnum-1}}^{2^{digitnum}} \\frac{\\mathrm{d}realvar}{realvar} \\\\\n&= \\frac{1}{2^{digitnum}} + \\log 2 \\\\\n&\\leq \\frac{1}{8} + \\log 2 < 0.125 + 0.7 < 1.\n\\end{align*}\nPut $boundval = \\frac{1}{8} + \\log 2$ and $limupper = 1 + \\frac{1}{2} + \\frac{1}{6(1-boundval)}$. Then we can prove that $\\sum_{indexval=1}^{2^{iteridx}-1} \\frac{1}{funcmap(indexval)} < limupper$ for all $iteridx \\geq 2$ by induction on $iteridx$. The case $iteridx = 2$ is clear. For the induction, note that by \\eqref{a6eq2},\n\\begin{align*}\n\\sum_{indexval=1}^{2^{iteridx}-1} \\frac{1}{funcmap(indexval)}\n&< 1 + \\frac{1}{2} + \\frac{1}{6} + boundval \\sum_{digitnum=3}^{iteridx} \\frac{1}{funcmap(digitnum)} \\\\\n&< 1 + \\frac{1}{2} + \\frac{1}{6} + boundval \\frac{1}{6(1-boundval)} \\\\\n&= 1 + \\frac{1}{2} + \\frac{1}{6(1-boundval)} = limupper,\n\\end{align*}\nas desired. We conclude that $\\sum_{indexval=1}^\\infty \\frac{1}{funcmap(indexval)}$\nconverges to a limit less than or equal to $limupper$.\n\nNote: the above argument proves that the sum for $baseint=2$ is at most $limupper < 2.417$. One can also obtain a lower bound by the same technique, namely $1 + \\frac{1}{2} + \\frac{1}{6(1 - boundval')}$ with $boundval' = \\log 2$. This bound exceeds $2.043$. (By contrast, summing the first 100000 terms of the series only yields a lower bound of $1.906$.) Repeating the same arguments with $digitnum \\geq 4$ as the cutoff yields the upper bound $2.185$ and the lower bound $2.079$.}",
      "confidence": 0.13
    },
    "descriptive_long_confusing": {
      "map": {
        "n": "umbrella",
        "d": "lighthouse",
        "i": "cinnamon",
        "x": "waterfall",
        "b": "tornadoes",
        "f": "quasarbeam",
        "c": "sandstone",
        "L": "jellyfish"
      },
      "question": "Fix an integer $tornadoes \\geq 2$. Let $quasarbeam(1) = 1$, $quasarbeam(2) = 2$, and for each\n$umbrella \\geq 3$, define $quasarbeam(umbrella) = umbrella \\, quasarbeam(lighthouse)$, where $lighthouse$ is the number of\nbase-$tornadoes$ digits of $umbrella$. For which values of $tornadoes$ does\n\\[\n\\sum_{umbrella=1}^\\infty \\frac{1}{quasarbeam(umbrella)}\n\\]\nconverge?",
      "solution": "The sum converges for $tornadoes=2$ and diverges for $tornadoes \\geq 3$.\nWe first consider $tornadoes \\geq 3$. Suppose the sum converges;\nthen the fact\nthat $quasarbeam(umbrella) = umbrella \\, quasarbeam(lighthouse)$ whenever $tornadoes^{lighthouse-1} \\leq umbrella \\leq tornadoes^{lighthouse} - 1$ yields\n\\begin{equation} \\label{a6eq1}\n\\sum_{umbrella=1}^\\infty \\frac{1}{quasarbeam(umbrella)}\n= \\sum_{lighthouse=1}^\\infty \\frac{1}{quasarbeam(lighthouse)} \\sum_{umbrella=tornadoes^{lighthouse-1}}^{tornadoes^{lighthouse} - 1} \\frac{1}{umbrella}.\n\\end{equation}\nHowever, by comparing the integral of $1/waterfall$ with a Riemann sum,\nwe see that\n\\begin{align*}\n\\sum_{umbrella=tornadoes^{lighthouse-1}}^{tornadoes^{lighthouse} - 1} \\frac{1}{umbrella}\n&> \\int_{tornadoes^{lighthouse-1}}^{tornadoes^{lighthouse}} \\frac{d\\,waterfall}{waterfall} \\\\\n&= \\log (tornadoes^{lighthouse}) - \\log (tornadoes^{lighthouse-1}) = \\log tornadoes,\n\\end{align*}\nwhere $\\log$ denotes the natural logarithm. Thus \\eqref{a6eq1} yields\n\\[\n\\sum_{umbrella=1}^\\infty \\frac{1}{quasarbeam(umbrella)}\n> (\\log tornadoes) \\sum_{umbrella=1}^\\infty \\frac{1}{quasarbeam(umbrella)},\n\\]\na contradiction since $\\log tornadoes > 1$ for $tornadoes \\geq 3$. Therefore the\nsum diverges.\n\nFor $tornadoes=2$, we have a slightly different identity because $quasarbeam(2) \\neq\n2 \\, quasarbeam(2)$. Instead, for any positive integer $cinnamon$, we have\n\\begin{equation} \\label{a6eq2}\n\\sum_{umbrella=1}^{2^{cinnamon}-1} \\frac{1}{quasarbeam(umbrella)}\n= 1 + \\frac{1}{2} + \\frac{1}{6} +\n\\sum_{lighthouse=3}^{cinnamon} \\frac{1}{quasarbeam(lighthouse)} \\sum_{umbrella=2^{lighthouse-1}}^{2^{lighthouse} - 1} \\frac{1}{umbrella}.\n\\end{equation}\nAgain comparing an integral to a Riemann sum, we see that for $lighthouse\\geq 3$,\n\\begin{align*}\n\\sum_{umbrella=2^{lighthouse-1}}^{2^{lighthouse} - 1} \\frac{1}{umbrella} &<\n\\frac{1}{2^{lighthouse-1}} - \\frac{1}{2^{lighthouse}} + \\int_{2^{lighthouse-1}}^{2^{lighthouse}} \\frac{d\\,waterfall}{waterfall}\n\\\\\n&= \\frac{1}{2^{lighthouse}} + \\log 2 \\\\\n&\\leq \\frac{1}{8} + \\log 2 < 0.125 + 0.7 < 1.\n\\end{align*}\nPut $sandstone = \\frac{1}{8} + \\log 2$ and $jellyfish = 1+\\frac{1}{2} +\n\\frac{1}{6(1-sandstone)}$. Then we can prove that $\\sum_{umbrella=1}^{2^{cinnamon}-1}\n\\frac{1}{quasarbeam(umbrella)} < jellyfish$ for all $cinnamon \\geq 2$ by induction on $cinnamon$. The case\n$cinnamon=2$ is clear. For the induction, note that by \\eqref{a6eq2},\n\\begin{align*}\n\\sum_{umbrella=1}^{2^{cinnamon}-1} \\frac{1}{quasarbeam(umbrella)}\n&< 1 + \\frac{1}{2} + \\frac{1}{6} + sandstone \\sum_{lighthouse=3}^{cinnamon} \\frac{1}{quasarbeam(lighthouse)} \\\\\n&< 1 + \\frac{1}{2} + \\frac{1}{6} + sandstone \\frac{1}{6(1-sandstone)} \\\\\n&= 1 + \\frac{1}{2} + \\frac{1}{6(1-sandstone)} = jellyfish,\n\\end{align*}\nas desired. We conclude that $\\sum_{umbrella=1}^\\infty \\frac{1}{quasarbeam(umbrella)}$\nconverges to a limit less than or equal to $jellyfish$.\n\nNote: the above argument proves that the sum for $tornadoes=2$ is at most\n$jellyfish < 2.417$. One can also obtain a lower bound by the same technique,\nnamely $1 + \\frac{1}{2} + \\frac{1}{6(1 - sandstone')} $ with $sandstone' = \\log 2$.\nThis bound exceeds $2.043$. (By contrast, summing the first 100000 terms of\nthe series only yields a lower bound of $1.906$.)\nRepeating the same arguments with $lighthouse \\geq 4$\nas the cutoff yields the upper bound $2.185$ and the lower bound $2.079$. "
    },
    "descriptive_long_misleading": {
      "map": {
        "n": "continuousvar",
        "d": "analogindex",
        "i": "wholevalue",
        "x": "constantval",
        "b": "apexpower",
        "f": "malfunction",
        "c": "variable",
        "L": "lowerbound"
      },
      "question": "Fix an integer $apexpower \\geq 2$. Let $malfunction(1) = 1$, $malfunction(2) = 2$, and for each\n$continuousvar \\geq 3$, define $malfunction(continuousvar) = continuousvar\\,malfunction(analogindex)$, where $analogindex$ is the number of\nbase-$apexpower$ digits of $continuousvar$. For which values of $apexpower$ does\n\\[\n\\sum_{continuousvar=1}^\\infty \\frac{1}{malfunction(continuousvar)}\n\\]\nconverge?",
      "solution": "The sum converges for $apexpower=2$ and diverges for $apexpower \\geq 3$.\nWe first consider $apexpower \\geq 3$. Suppose the sum converges;\nthen the fact\nthat $malfunction(continuousvar) = continuousvar \\, malfunction(analogindex)$ whenever $apexpower^{analogindex-1} \\leq continuousvar \\leq apexpower^{analogindex} - 1$ yields\n\\begin{equation} \\label{a6eq1}\n\\sum_{continuousvar=1}^\\infty \\frac{1}{malfunction(continuousvar)}\n= \\sum_{analogindex=1}^\\infty \\frac{1}{malfunction(analogindex)} \\sum_{continuousvar=apexpower^{analogindex-1}}^{apexpower^{analogindex} - 1} \\frac{1}{continuousvar}.\n\\end{equation}\nHowever, by comparing the integral of $1/constantval$ with a Riemann sum,\nwe see that\n\\begin{align*}\n\\sum_{continuousvar=apexpower^{analogindex-1}}^{apexpower^{analogindex} - 1} \\frac{1}{continuousvar}\n&> \\int_{apexpower^{analogindex-1}}^{apexpower^{analogindex}} \\frac{dconstantval}{constantval} \\\\\n&= \\log (apexpower^{analogindex}) - \\log (apexpower^{analogindex-1}) = \\log apexpower,\n\\end{align*}\nwhere $\\log$ denotes the natural logarithm. Thus \\eqref{a6eq1} yields\n\\[\n\\sum_{continuousvar=1}^\\infty \\frac{1}{malfunction(continuousvar)}\n> (\\log apexpower) \\sum_{continuousvar=1}^\\infty \\frac{1}{malfunction(continuousvar)},\n\\]\na contradiction since $\\log apexpower > 1$ for $apexpower \\geq 3$. Therefore the\nsum diverges.\n\nFor $apexpower=2$, we have a slightly different identity because $malfunction(2) \\neq\n2 \\, malfunction(2)$. Instead, for any positive integer $wholevalue$, we have\n\\begin{equation} \\label{a6eq2}\n\\sum_{continuousvar=1}^{2^{wholevalue}-1} \\frac{1}{malfunction(continuousvar)}\n= 1 + \\frac{1}{2} + \\frac{1}{6} +\n\\sum_{analogindex=3}^{wholevalue} \\frac{1}{malfunction(analogindex)} \\sum_{continuousvar=2^{analogindex-1}}^{2^{analogindex} - 1} \\frac{1}{continuousvar}.\n\\end{equation}\nAgain comparing an integral to a Riemann sum, we see that for $analogindex\\geq 3$,\n\\begin{align*}\n\\sum_{continuousvar=2^{analogindex-1}}^{2^{analogindex} - 1} \\frac{1}{continuousvar} &<\n\\frac{1}{2^{analogindex-1}} - \\frac{1}{2^{analogindex}} + \\int_{2^{analogindex-1}}^{2^{analogindex}} \\frac{dconstantval}{constantval}\n\\\\\n&= \\frac{1}{2^{analogindex}} + \\log 2 \\\\\n&\\leq \\frac{1}{8} + \\log 2 < 0.125 + 0.7 < 1.\n\\end{align*}\nPut $variable = \\frac{1}{8} + \\log 2$ and $lowerbound = 1+\\frac{1}{2} +\n\\frac{1}{6(1-variable)}$. Then we can prove that $\\sum_{continuousvar=1}^{2^{wholevalue}-1}\n\\frac{1}{malfunction(continuousvar)} < lowerbound$ for all $wholevalue \\geq 2$ by induction on $wholevalue$. The case\n$wholevalue=2$ is clear. For the induction, note that by \\eqref{a6eq2},\n\\begin{align*}\n\\sum_{continuousvar=1}^{2^{wholevalue}-1} \\frac{1}{malfunction(continuousvar)}\n&< 1 + \\frac{1}{2} + \\frac{1}{6} + variable \\sum_{analogindex=3}^{wholevalue} \\frac{1}{malfunction(analogindex)} \\\\\n&< 1 + \\frac{1}{2} + \\frac{1}{6} + variable \\frac{1}{6(1-variable)} \\\\\n&= 1 + \\frac{1}{2} + \\frac{1}{6(1-variable)} = lowerbound,\n\\end{align*}\nas desired. We conclude that $\\sum_{continuousvar=1}^\\infty \\frac{1}{malfunction(continuousvar)}$\nconverges to a limit less than or equal to $lowerbound$.\n\nNote: the above argument proves that the sum for $apexpower=2$ is at most\n$lowerbound < 2.417$. One can also obtain a lower bound by the same technique,\nnamely $1 + \\frac{1}{2} + \\frac{1}{6(1 - variable')} $ with $variable' = \\log 2$.\nThis bound exceeds $2.043$. (By contrast, summing the first 100000 terms of\nthe series only yields a lower bound of $1.906$.)\nRepeating the same arguments with $analogindex \\geq 4$\nas the cutoff yields the upper bound $2.185$ and the lower bound $2.079$. "
    },
    "garbled_string": {
      "map": {
        "n": "qzxwvtnp",
        "d": "hjgrksla",
        "i": "mfldqwer",
        "x": "trbslmno",
        "b": "ykcpdgiu",
        "f": "asvrplke",
        "c": "vhnqptui",
        "L": "rpsdqkzm"
      },
      "question": "Fix an integer $ykcpdgiu \\geq 2$. Let $asvrplke(1) = 1$, $asvrplke(2) = 2$, and for each\n$qzxwvtnp \\geq 3$, define $asvrplke(qzxwvtnp) = qzxwvtnp asvrplke(hjgrksla)$, where $hjgrksla$ is the number of\nbase-$ykcpdgiu$ digits of $qzxwvtnp$. For which values of $ykcpdgiu$ does\n\\[\n\\sum_{qzxwvtnp=1}^\\infty \\frac{1}{asvrplke(qzxwvtnp)}\n\\]\nconverge?",
      "solution": "The sum converges for $ykcpdgiu=2$ and diverges for $ykcpdgiu \\geq 3$.\nWe first consider $ykcpdgiu \\geq 3$. Suppose the sum converges;\nthen the fact\nthat $asvrplke(qzxwvtnp) = qzxwvtnp asvrplke(hjgrksla)$ whenever $ykcpdgiu^{hjgrksla-1} \\leq qzxwvtnp \\leq ykcpdgiu^{hjgrksla} - 1$ yields\n\\begin{equation} \\label{a6eq1}\n\\sum_{qzxwvtnp=1}^\\infty \\frac{1}{asvrplke(qzxwvtnp)}\n= \\sum_{hjgrksla=1}^\\infty \\frac{1}{asvrplke(hjgrksla)} \\sum_{qzxwvtnp=ykcpdgiu^{hjgrksla-1}}^{ykcpdgiu^{hjgrksla} - 1} \\frac{1}{qzxwvtnp}.\n\\end{equation}\nHowever, by comparing the integral of $1/trbslmno$ with a Riemann sum,\nwe see that\n\\begin{align*}\n\\sum_{qzxwvtnp=ykcpdgiu^{hjgrksla-1}}^{ykcpdgiu^{hjgrksla} - 1} \\frac{1}{qzxwvtnp}\n&> \\int_{ykcpdgiu^{hjgrksla-1}}^{ykcpdgiu^{hjgrksla}} \\frac{d trbslmno}{trbslmno} \\\\\n&= \\log (ykcpdgiu^{hjgrksla}) - \\log (ykcpdgiu^{hjgrksla-1}) = \\log ykcpdgiu,\n\\end{align*}\nwhere $\\log$ denotes the natural logarithm. Thus \\eqref{a6eq1} yields\n\\[\n\\sum_{qzxwvtnp=1}^\\infty \\frac{1}{asvrplke(qzxwvtnp)}\n> (\\log ykcpdgiu) \\sum_{qzxwvtnp=1}^\\infty \\frac{1}{asvrplke(qzxwvtnp)},\n\\]\na contradiction since $\\log ykcpdgiu > 1$ for $ykcpdgiu \\geq 3$. Therefore the\nsum diverges.\n\nFor $ykcpdgiu=2$, we have a slightly different identity because $asvrplke(2) \\neq\n2 asvrplke(2)$. Instead, for any positive integer $mfldqwer$, we have\n\\begin{equation} \\label{a6eq2}\n\\sum_{qzxwvtnp=1}^{2^{mfldqwer}-1} \\frac{1}{asvrplke(qzxwvtnp)}\n= 1 + \\frac{1}{2} + \\frac{1}{6} +\n\\sum_{hjgrksla=3}^{mfldqwer} \\frac{1}{asvrplke(hjgrksla)} \\sum_{qzxwvtnp=2^{hjgrksla-1}}^{2^{hjgrksla} - 1} \\frac{1}{qzxwvtnp}.\n\\end{equation}\nAgain comparing an integral to a Riemann sum, we see that for $hjgrksla\\geq 3$,\n\\begin{align*}\n\\sum_{qzxwvtnp=2^{hjgrksla-1}}^{2^{hjgrksla} - 1} \\frac{1}{qzxwvtnp} &<\n\\frac{1}{2^{hjgrksla-1}} - \\frac{1}{2^{hjgrksla}} + \\int_{2^{hjgrksla-1}}^{2^{hjgrksla}} \\frac{d trbslmno}{trbslmno}\n\\\\\n&= \\frac{1}{2^{hjgrksla}} + \\log 2 \\\\\n&\\leq \\frac{1}{8} + \\log 2 < 0.125 + 0.7 < 1.\n\\end{align*}\nPut $vhnqptui = \\frac{1}{8} + \\log 2$ and $rpsdqkzm = 1+\\frac{1}{2} +\n\\frac{1}{6(1-vhnqptui)}$. Then we can prove that $\\sum_{qzxwvtnp=1}^{2^{mfldqwer}-1}\n\\frac{1}{asvrplke(qzxwvtnp)} < rpsdqkzm$ for all $mfldqwer \\geq 2$ by induction on $mfldqwer$. The case\n$mfldqwer=2$ is clear. For the induction, note that by \\eqref{a6eq2},\n\\begin{align*}\n\\sum_{qzxwvtnp=1}^{2^{mfldqwer}-1} \\frac{1}{asvrplke(qzxwvtnp)}\n&< 1 + \\frac{1}{2} + \\frac{1}{6} + vhnqptui \\sum_{hjgrksla=3}^{mfldqwer} \\frac{1}{asvrplke(hjgrksla)} \\\\\n&< 1 + \\frac{1}{2} + \\frac{1}{6} + vhnqptui \\frac{1}{6(1-vhnqptui)} \\\\\n&= 1 + \\frac{1}{2} + \\frac{1}{6(1-vhnqptui)} = rpsdqkzm,\n\\end{align*}\nas desired. We conclude that $\\sum_{qzxwvtnp=1}^\\infty \\frac{1}{asvrplke(qzxwvtnp)}$\nconverges to a limit less than or equal to $rpsdqkzm$.\n\nNote: the above argument proves that the sum for $ykcpdgiu=2$ is at most\n$rpsdqkzm < 2.417$. One can also obtain a lower bound by the same technique,\nnamely $1 + \\frac{1}{2} + \\frac{1}{6(1 - vhnqptui')}$ with $vhnqptui' = \\log 2$.\nThis bound exceeds $2.043$. (By contrast, summing the first 100000 terms of\nthe series only yields a lower bound of $1.906$.)\nRepeating the same arguments with $hjgrksla \\geq 4$\nas the cutoff yields the upper bound $2.185$ and the lower bound $2.079$. "
    },
    "kernel_variant": {
      "question": "Fix an integer b ≥ 2.\nFor every positive integer n let d(n) denote the number of base-b digits of n (so b^{d(n)-1} ≤ n ≤ b^{d(n)}-1).\nDefine the function f: ℕ → ℝ_{>0} by\n   f(1) = 3, f(2) = 5, and f(n) = n · f(d(n))  for all n ≥ 3.\nFor which values of b does the series\n            ∑_{n=1}^{∞} 1 / f(n)\nconverge?",
      "solution": "Throughout, d (without an argument) will always mean “number of base-b digits”.  Recall that an integer n has exactly d digits iff b^{d-1} ≤ n ≤ b^{d}-1.\n\n1.  A block decomposition.\n   ------------------------------------------------\n   Separate the finitely many terms that do not satisfy the recursion.\n        A := 1/f(1) + 1/f(2) = 1/3 + 1/5.\n   When b ≥ 4 there are additional one-digit numbers 3,4,…,b−1; put\n        B := Σ_{n=3}^{b-1} 1/f(n)  (take B = 0 if b = 2 or 3).\n   \n   For every integer d ≥ 2 define\n        L_d := max{3 , b^{d-1}}, U_d := b^{d}-1,\n        H_d := Σ_{n=L_d}^{U_d} 1/n  (empty sum = 0 if L_d>U_d).\n   Whenever n ∈ [L_d,U_d] we have n ≥ 3, hence f(n) = n f(d).  Therefore\n        Σ_{n=L_d}^{U_d} 1/f(n) = H_d / f(d).\n   Consequently the whole series can be written as\n        S := Σ_{n=1}^{∞} 1/f(n) = A + B + Σ_{d=2}^{∞} H_d / f(d).  (★)\n\n2.  Estimates for the harmonic blocks.\n   ----------------------------------\n   Because 1/x is decreasing one has, for every d ≥ 2,\n        ∫_{b^{d-1}}^{b^{d}} dx/x = ln b ≤ H_d ≤ ln b + 1/b^{d-1}  (b ≥ 3),\n        ∫_{2^{d-1}}^{2^{d}} dx/x = ln 2 ≤ H_d ≤ ln 2 + 1/2^{d-1}  (b = 2).\n   In particular\n        H_d ≥ ln b  for all d ≥ 2 if b ≥ 3,                      (1)\n        H_d ≤ ln 2 + 1/2^{d-1} ≤ ln 2 + 1/8 for all d ≥ 4 if b = 2. (2)\n   Set  c := ln 2 + 1/8 ≈ 0.818 < 1.\n\n3.  Divergence when b ≥ 3.\n   -----------------------\n   Put   T := Σ_{d=2}^{∞} 1/f(d).\n   From (★) and the lower bound (1) we obtain\n        S ≥ A + B + (ln b) · T.                                            (3)\n\n   Next group the indices d by their own number k ≥ 1 of digits:\n        D_k := { d : b^{k-1} ≤ d ≤ b^{k}-1 }.\n   For every d ∈ D_k we have f(d) = d f(k), so\n        Σ_{d∈D_k} 1/f(d) = (1/f(k)) Σ_{d=b^{k-1}}^{b^{k}-1} 1/d ≥ (ln b)/f(k).\n   Summing over k ≥ 2 gives\n        T = 1/f(2) + Σ_{k=2}^{∞} Σ_{d∈D_k} 1/f(d)\n          ≥ 1/f(2) + (ln b) Σ_{k=2}^{∞} 1/f(k)\n          = 1/f(2) + (ln b) T.                                              (4)\n   If T were finite, inequality (4) would imply\n        T ≥ T + 1/f(2) + (ln b − 1)T, which is impossible because ln b > 1.\n   Hence T = ∞, and by (3) the main series S diverges.  Therefore\n        Σ_{n=1}^{∞} 1/f(n) diverges for every base b ≥ 3.\n\n4.  Convergence when b = 2.\n   ------------------------\n   In this case B = 0.  Fix D = 4 and define the partial sums\n        S_i := Σ_{n=1}^{2^{i}-1} 1/f(n)  (i ≥ D).\n   Split the series in (★) after the first D−1 blocks:\n        A_D := Σ_{n=1}^{2^{D-1}-1} 1/f(n)   (a finite constant).\n   Using (2) we get, for every i ≥ D,\n        S_i = A_D + Σ_{d=D}^{i} H_d / f(d)\n             ≤ A_D + c Σ_{d=D}^{i} 1/f(d) ≤ A_D + c S_i.\n   Because 0 < c < 1 this yields\n        (1 − c) S_i ≤ A_D ⇒ S_i ≤ A_D /(1 − c) for all i ≥ D.\n   Thus the non-decreasing sequence (S_i) is bounded, so the limit\n        S = lim_{i→∞} S_i\n   exists and is finite.  Therefore the series converges when b = 2.\n\n5.  Final classification.\n   ----------------------\n   The series  Σ_{n=1}^{∞} 1 / f(n)  converges precisely for b = 2 and\n   diverges for every integer base b ≥ 3.",
      "_meta": {
        "core_steps": [
          "Partition the series by the number d of base-b digits of n and use f(n)=n·f(d).",
          "Rewrite the series as Σ_d (1/f(d)) · Σ_{n=b^{d-1}}^{b^{d}-1} (1/n).",
          "Estimate each inner harmonic block with a ∫(1/x)dx comparison (lower bound > log b, upper bound < constant when b=2).",
          "For b≥3: lower bound >1 multiplies the whole series, forcing a contradiction ⇒ divergence.",
          "For b=2: choose a digit-length cutoff where the upper bound c<1, then apply induction to keep partial sums bounded ⇒ convergence."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "The preset values of f on finitely many small n; they only affect finitely many summands and hence do not influence convergence.",
            "original": "f(1)=1, f(2)=2"
          },
          "slot2": {
            "description": "The exact digit-length cutoff used for the ‘c<1’ bound in the b=2 case (and the resulting numerical constant c). Any larger cutoff giving c<1 would serve the same purpose.",
            "original": "cutoff d≥3, yielding c = 1/8 + log 2 ≈ 0.825..."
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}