{
  "index": "1988-A-3",
  "type": "ANA",
  "tag": [
    "ANA",
    "NT"
  ],
  "difficulty": "",
  "question": "Determine, with proof, the set of real numbers $x$ for which\n\\[\n\\sum_{n=1}^\\infty \\left( \\frac{1}{n} \\csc \\frac{1}{n} - 1 \\right)^x\n\\]\nconverges.",
  "solution": "Solution. Define\n\\[\na_{n}=\\frac{1}{n} \\csc \\frac{1}{n}-1=\\frac{1}{n \\sin \\frac{1}{n}}-1 .\n\\]\n\nTaking \\( t=1 / n \\) in the inequality \\( 0<\\sin t<t \\) for \\( t \\in(0, \\pi) \\), we obtain \\( a_{n}>0 \\), so each term \\( a_{n}^{x} \\) of the series is defined for any real \\( x \\). Using \\( \\sin t=t-t^{3} / 3!+O\\left(t^{5}\\right) \\) as \\( t \\rightarrow 0 \\), we have, as \\( n \\rightarrow \\infty \\),\n\\[\n\\begin{aligned}\na_{n} & =\\frac{1}{n\\left(\\frac{1}{n}-\\frac{1}{6 n^{3}}+O\\left(\\frac{1}{n^{5}}\\right)\\right)}-1 \\\\\n& =\\frac{1}{1-\\frac{1}{6 n^{2}}+O\\left(\\frac{1}{n^{4}}\\right)}-1 \\\\\n& =\\frac{1}{6 n^{2}}+O\\left(\\frac{1}{n^{4}}\\right) .\n\\end{aligned}\n\\]\n\nIn particular, if \\( b_{n}=1 / n^{2} \\), then \\( a_{n}^{x} / b_{n}^{x} \\) has a finite limit as \\( n \\rightarrow \\infty \\), so by the Limit Comparison Test [Spv, Ch. 22, Theorem 2], \\( \\sum_{n=1}^{\\infty} a_{n}^{x} \\) converges if and only if \\( \\sum_{n=1}^{\\infty} b_{n}^{x}=\\sum_{n=1}^{\\infty} n^{-2 x} \\) converges, which by the Integral Comparison Test [Spv, Ch. 22, Theorem 4] holds if and only if \\( 2 x>1 \\), i.e., \\( x>1 / 2 \\).\n\nRemark (big- \\( O \\) and little-o notation). Recall that \\( O(g(n)) \\) is a stand-in for a function \\( f(n) \\) for which there exists a constant \\( C \\) such that \\( |f(n)| \\leq C|g(n)| \\) for all sufficiently large \\( n \\). (This does not necessarily imply that \\( \\lim _{n \\rightarrow \\infty} f(n) / g(n) \\) exists.)\nSimilarly \" \\( f(t)=O(g(t)) \\) as \\( t \\rightarrow 0 \\) \" means that there exists a constant \\( C \\) such that \\( |f(t)| \\leq C|g(t)| \\) for sufficiently small nonzero \\( t \\).\n\nOn the other hand, \\( o(g(n)) \\) is a stand-in for a function \\( f(n) \\) such that\n\\[\n\\lim _{n \\rightarrow \\infty} \\frac{f(n)}{g(n)}=0\n\\]\n\nOne can similarly define \" \\( f(t)=o(g(t)) \\) as \\( t \\rightarrow 0 \" \\).",
  "vars": [
    "x",
    "n",
    "t"
  ],
  "params": [
    "a_n",
    "b_n",
    "C"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "exponent",
        "n": "indexvar",
        "t": "smallvar",
        "a_n": "seriesterm",
        "b_n": "compterm",
        "C": "boundconstant"
      },
      "question": "Determine, with proof, the set of real numbers exponent for which\n\\[\n\\sum_{indexvar=1}^\\infty \\left( \\frac{1}{indexvar} \\csc \\frac{1}{indexvar} - 1 \\right)^{exponent}\n\\]\nconverges.",
      "solution": "Solution. Define\n\\[\nseriesterm=\\frac{1}{indexvar} \\csc \\frac{1}{indexvar}-1=\\frac{1}{indexvar \\sin \\frac{1}{indexvar}}-1 .\n\\]\n\nTaking \\( smallvar=1 / indexvar \\) in the inequality \\( 0<\\sin smallvar<smallvar \\) for \\( smallvar \\in(0, \\pi) \\), we obtain \\( seriesterm>0 \\), so each term \\( seriesterm^{exponent} \\) of the series is defined for any real exponent. Using \\( \\sin smallvar=smallvar-smallvar^{3} / 3!+O\\left(smallvar^{5}\\right) \\) as \\( smallvar \\rightarrow 0 \\), we have, as \\( indexvar \\rightarrow \\infty \\),\n\\[\n\\begin{aligned}\nseriesterm & =\\frac{1}{indexvar\\left(\\frac{1}{indexvar}-\\frac{1}{6 indexvar^{2}}+O\\left(\\frac{1}{indexvar^{4}}\\right)\\right)}-1 \\\\\n& =\\frac{1}{1-\\frac{1}{6 indexvar^{2}}+O\\left(\\frac{1}{indexvar^{4}}\\right)}-1 \\\\\n& =\\frac{1}{6 indexvar^{2}}+O\\left(\\frac{1}{indexvar^{4}}\\right) .\n\\end{aligned}\n\\]\n\nIn particular, if \\( compterm=1 / indexvar^{2} \\), then \\( seriesterm^{exponent} / compterm^{exponent} \\) has a finite limit as \\( indexvar \\rightarrow \\infty \\), so by the Limit Comparison Test [Spv, Ch. 22, Theorem 2], \\( \\sum_{indexvar=1}^{\\infty} seriesterm^{exponent} \\) converges if and only if \\( \\sum_{indexvar=1}^{\\infty} compterm^{exponent}=\\sum_{indexvar=1}^{\\infty} indexvar^{-2 exponent} \\) converges, which by the Integral Comparison Test [Spv, Ch. 22, Theorem 4] holds if and only if \\( 2 exponent>1 \\), i.e., \\( exponent>1 / 2 \\).\n\nRemark (big- \\( O \\) and little-o notation). Recall that \\( O(g(indexvar)) \\) is a stand-in for a function \\( f(indexvar) \\) for which there exists a boundconstant such that \\( |f(indexvar)| \\leq boundconstant|g(indexvar)| \\) for all sufficiently large indexvar. (This does not necessarily imply that \\( \\lim _{indexvar \\rightarrow \\infty} f(indexvar) / g(indexvar) \\) exists.)\nSimilarly \" \\( f(smallvar)=O(g(smallvar)) \\) as \\( smallvar \\rightarrow 0 \\) \" means that there exists a boundconstant such that \\( |f(smallvar)| \\leq boundconstant|g(smallvar)| \\) for sufficiently small nonzero smallvar.\n\nOn the other hand, \\( o(g(indexvar)) \\) is a stand-in for a function \\( f(indexvar) \\) such that\n\\[\n\\lim _{indexvar \\rightarrow \\infty} \\frac{f(indexvar)}{g(indexvar)}=0\n\\]\n\nOne can similarly define \" \\( f(smallvar)=o(g(smallvar)) \\) as \\( smallvar \\rightarrow 0 \" ."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "riverbank",
        "n": "honeycomb",
        "t": "blueberry",
        "a_n": "moonlight",
        "b_n": "sunflower",
        "C": "raincloud"
      },
      "question": "Determine, with proof, the set of real numbers $riverbank$ for which\n\\[\n\\sum_{honeycomb=1}^\\infty \\left( \\frac{1}{honeycomb} \\csc \\frac{1}{honeycomb} - 1 \\right)^{riverbank}\n\\]\nconverges.",
      "solution": "Solution. Define\n\\[\nmoonlight=\\frac{1}{honeycomb} \\csc \\frac{1}{honeycomb}-1=\\frac{1}{honeycomb \\sin \\frac{1}{honeycomb}}-1 .\n\\]\n\nTaking \\( blueberry=1 / honeycomb \\) in the inequality \\( 0<\\sin blueberry<blueberry \\) for \\( blueberry \\in(0, \\pi) \\), we obtain \\( moonlight>0 \\), so each term \\( moonlight^{riverbank} \\) of the series is defined for any real \\( riverbank \\). Using \\( \\sin blueberry=blueberry-blueberry^{3} / 3!+O\\left(blueberry^{5}\\right) \\) as \\( blueberry \\rightarrow 0 \\), we have, as \\( honeycomb \\rightarrow \\infty \\),\n\\[\n\\begin{aligned}\nmoonlight & =\\frac{1}{honeycomb\\left(\\frac{1}{honeycomb}-\\frac{1}{6 honeycomb^{3}}+O\\left(\\frac{1}{honeycomb^{5}}\\right)\\right)}-1 \\\\\n& =\\frac{1}{1-\\frac{1}{6 honeycomb^{2}}+O\\left(\\frac{1}{honeycomb^{4}}\\right)}-1 \\\\\n& =\\frac{1}{6 honeycomb^{2}}+O\\left(\\frac{1}{honeycomb^{4}}\\right) .\n\\end{aligned}\n\\]\n\nIn particular, if \\( sunflower=1 / honeycomb^{2} \\), then \\( moonlight^{riverbank} / sunflower^{riverbank} \\) has a finite limit as \\( honeycomb \\rightarrow \\infty \\), so by the Limit Comparison Test [Spv, Ch. 22, Theorem 2], \\( \\sum_{honeycomb=1}^{\\infty} moonlight^{riverbank} \\) converges if and only if \\( \\sum_{honeycomb=1}^{\\infty} sunflower^{riverbank}=\\sum_{honeycomb=1}^{\\infty} honeycomb^{-2 riverbank} \\) converges, which by the Integral Comparison Test [Spv, Ch. 22, Theorem 4] holds if and only if \\( 2 riverbank>1 \\), i.e., \\( riverbank>1 / 2 \\).\n\nRemark (big- \\( O \\) and little-o notation). Recall that \\( O(g(honeycomb)) \\) is a stand-in for a function \\( f(honeycomb) \\) for which there exists a constant \\( raincloud \\) such that \\( |f(honeycomb)| \\leq raincloud|g(honeycomb)| \\) for all sufficiently large \\( honeycomb \\). (This does not necessarily imply that \\( \\lim _{honeycomb \\rightarrow \\infty} f(honeycomb) / g(honeycomb) \\) exists.)\nSimilarly \" \\( f(blueberry)=O(g(blueberry)) \\) as \\( blueberry \\rightarrow 0 \\) \" means that there exists a constant \\( raincloud \\) such that \\( |f(blueberry)| \\leq raincloud|g(blueberry)| \\) for sufficiently small nonzero \\( blueberry \\).\n\nOn the other hand, \\( o(g(honeycomb)) \\) is a stand-in for a function \\( f(honeycomb) \\) such that\n\\[\n\\lim _{honeycomb \\rightarrow \\infty} \\frac{f(honeycomb)}{g(honeycomb)}=0\n\\]\n\nOne can similarly define \" \\( f(blueberry)=o(g(blueberry)) \\) as \\( blueberry \\rightarrow 0 \" ."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "knownvalue",
        "n": "boundless",
        "t": "timeless",
        "a_n": "steadystate",
        "b_n": "disorderly",
        "C": "variable"
      },
      "question": "Determine, with proof, the set of real numbers $knownvalue$ for which\n\\[\n\\sum_{boundless=1}^\\infty \\left( \\frac{1}{boundless} \\csc \\frac{1}{boundless} - 1 \\right)^{knownvalue}\n\\]\nconverges.",
      "solution": "Solution. Define\n\\[\nsteadystate=\\frac{1}{boundless} \\csc \\frac{1}{boundless}-1=\\frac{1}{boundless \\sin \\frac{1}{boundless}}-1 .\n\\]\n\nTaking \\( timeless=1 / boundless \\) in the inequality \\( 0<\\sin timeless<timeless \\) for \\( timeless \\in(0, \\pi) \\), we obtain \\( steadystate>0 \\), so each term \\( steadystate^{knownvalue} \\) of the series is defined for any real \\( knownvalue \\). Using \\( \\sin timeless=timeless-timeless^{3} / 3!+O\\left(timeless^{5}\\right) \\) as \\( timeless \\rightarrow 0 \\), we have, as \\( boundless \\rightarrow \\infty \\),\n\\[\n\\begin{aligned}\nsteadystate & =\\frac{1}{boundless\\left(\\frac{1}{boundless}-\\frac{1}{6 boundless^{3}}+O\\left(\\frac{1}{boundless^{5}}\\right)\\right)}-1 \\\\\n& =\\frac{1}{1-\\frac{1}{6 boundless^{2}}+O\\left(\\frac{1}{boundless^{4}}\\right)}-1 \\\\\n& =\\frac{1}{6 boundless^{2}}+O\\left(\\frac{1}{boundless^{4}}\\right) .\n\\end{aligned}\n\\]\n\nIn particular, if \\( disorderly=1 / boundless^{2} \\), then \\( steadystate^{knownvalue} / disorderly^{knownvalue} \\) has a finite limit as \\( boundless \\rightarrow \\infty \\), so by the Limit Comparison Test [Spv, Ch. 22, Theorem 2], \\( \\sum_{boundless=1}^{\\infty} steadystate^{knownvalue} \\) converges if and only if \\( \\sum_{boundless=1}^{\\infty} disorderly^{knownvalue}=\\sum_{boundless=1}^{\\infty} boundless^{-2 knownvalue} \\) converges, which by the Integral Comparison Test [Spv, Ch. 22, Theorem 4] holds if and only if \\( 2 knownvalue>1 \\), i.e., \\( knownvalue>1 / 2 \\).\n\nRemark (big- \\( O \\) and little-o notation). Recall that \\( O(g(boundless)) \\) is a stand-in for a function \\( f(boundless) \\) for which there exists a constant \\( variable \\) such that \\( |f(boundless)| \\leq variable|g(boundless)| \\) for all sufficiently large \\( boundless \\). (This does not necessarily imply that \\( \\lim _{boundless \\rightarrow \\infty} f(boundless) / g(boundless) \\) exists.)\nSimilarly \" \\( f(timeless)=O(g(timeless)) \\) as \\( timeless \\rightarrow 0 \\) \" means that there exists a constant \\( variable \\) such that \\( |f(timeless)| \\leq variable|g(timeless)| \\) for sufficiently small nonzero \\( timeless \\).\n\nOn the other hand, \\( o(g(boundless)) \\) is a stand-in for a function \\( f(boundless) \\) such that\n\\[\n\\lim _{boundless \\rightarrow \\infty} \\frac{f(boundless)}{g(boundless)}=0\n\\]\n\nOne can similarly define \" \\( f(timeless)=o(g(timeless)) \\) as \\( timeless \\rightarrow 0 \" ."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "n": "hjgrksla",
        "t": "plmoknij",
        "a_n": "fdertyui",
        "b_n": "vbnmlkjh",
        "C": "asdfghjk"
      },
      "question": "Determine, with proof, the set of real numbers qzxwvtnp for which\n\\[\n\\sum_{hjgrksla=1}^\\infty \\left( \\frac{1}{hjgrksla} \\csc \\frac{1}{hjgrksla} - 1 \\right)^{qzxwvtnp}\n\\]\nconverges.",
      "solution": "Solution. Define\n\\[\nfdertyui=\\frac{1}{hjgrksla} \\csc \\frac{1}{hjgrksla}-1=\\frac{1}{hjgrksla \\sin \\frac{1}{hjgrksla}}-1 .\n\\]\n\nTaking \\( plmoknij=1 / hjgrksla \\) in the inequality \\( 0<\\sin plmoknij<plmoknij \\) for \\( plmoknij \\in(0, \\pi) \\), we obtain \\( fdertyui>0 \\), so each term \\( fdertyui^{qzxwvtnp} \\) of the series is defined for any real \\( qzxwvtnp \\). Using \\( \\sin plmoknij=plmoknij-plmoknij^{3} / 3!+O\\left(plmoknij^{5}\\right) \\) as \\( plmoknij \\rightarrow 0 \\), we have, as \\( hjgrksla \\rightarrow \\infty \\),\n\\[\n\\begin{aligned}\nfdertyui & =\\frac{1}{hjgrksla\\left(\\frac{1}{hjgrksla}-\\frac{1}{6 hjgrksla^{3}}+O\\left(\\frac{1}{hjgrksla^{5}}\\right)\\right)}-1 \\\\\n& =\\frac{1}{1-\\frac{1}{6 hjgrksla^{2}}+O\\left(\\frac{1}{hjgrksla^{4}}\\right)}-1 \\\\\n& =\\frac{1}{6 hjgrksla^{2}}+O\\left(\\frac{1}{hjgrksla^{4}}\\right) .\n\\end{aligned}\n\\]\n\nIn particular, if \\( vbnmlkjh=1 / hjgrksla^{2} \\), then \\( fdertyui^{qzxwvtnp} / vbnmlkjh^{qzxwvtnp} \\) has a finite limit as \\( hjgrksla \\rightarrow \\infty \\), so by the Limit Comparison Test [Spv, Ch. 22, Theorem 2], \\( \\sum_{hjgrksla=1}^{\\infty} fdertyui^{qzxwvtnp} \\) converges if and only if \\( \\sum_{hjgrksla=1}^{\\infty} vbnmlkjh^{qzxwvtnp}=\\sum_{hjgrksla=1}^{\\infty} hjgrksla^{-2 qzxwvtnp} \\) converges, which by the Integral Comparison Test [Spv, Ch. 22, Theorem 4] holds if and only if \\( 2 qzxwvtnp>1 \\), i.e., \\( qzxwvtnp>1 / 2 \\).\n\nRemark (big- \\( O \\) and little-o notation). Recall that \\( O(g(hjgrksla)) \\) is a stand-in for a function \\( f(hjgrksla) \\) for which there exists a constant \\( asdfghjk \\) such that \\( |f(hjgrksla)| \\leq asdfghjk|g(hjgrksla)| \\) for all sufficiently large \\( hjgrksla \\). (This does not necessarily imply that \\( \\lim _{hjgrksla \\rightarrow \\infty} f(hjgrksla) / g(hjgrksla) \\) exists.)\nSimilarly \" \\( f(plmoknij)=O(g(plmoknij)) \\) as \\( plmoknij \\rightarrow 0 \\) \" means that there exists a constant \\( asdfghjk \\) such that \\( |f(plmoknij)| \\leq asdfghjk|g(plmoknij)| \\) for sufficiently small nonzero \\( plmoknij \\).\n\nOn the other hand, \\( o(g(hjgrksla)) \\) is a stand-in for a function \\( f(hjgrksla) \\) such that\n\\[\n\\lim _{hjgrksla \\rightarrow \\infty} \\frac{f(hjgrksla)}{g(hjgrksla)}=0\n\\]\n\nOne can similarly define \" \\( f(plmoknij)=o(g(plmoknij)) \\) as \\( plmoknij \\rightarrow 0 \" ."
    },
    "kernel_variant": {
      "question": "Determine, with proof, the complete set of real triples \n\\[\n(x,y,z)\\in\\mathbb R^{3}\n\\]\nfor which the alternating series  \n\\[\n\\boxed{\\,S(x,y,z)\\;=\\;\n\\sum_{n=3}^{\\infty}(-1)^{n}\\,\n\\frac{\\Bigl[\n\\,3\\sec\\!\\bigl(\\tfrac{3}{n^{5/2}}\\bigr)-3\n\\;+\\;\n2\\bigl(\\csc\\!\\bigl(\\tfrac{2}{n^{3}}\\bigr)-\\tfrac{n^{3}}{2}\\bigr)\n\\Bigr]^{\\,x}}\n{\\,n^{\\,y}\\,(\\log n)^{\\,z}}}\\qquad(\\log=\\text{natural logarithm})\n\\]\n\n(a) converges absolutely;  \n(b) converges conditionally, i.e. converges but not absolutely. \n\n(Real powers are understood through the principal branch; the bracket is strictly positive for every \\(n\\ge 3\\), so all terms are real for every real \\(x\\).)\n\n————————————————————————————————————",
      "solution": "Notation.  Put  \n\\[\na_{n}:=\n3\\sec\\!\\left(\\frac{3}{n^{5/2}}\\right)-3+\n2\\left(\\csc\\!\\left(\\frac{2}{n^{3}}\\right)-\\frac{n^{3}}{2}\\right),\n\\qquad n\\ge 3 ,\n\\]\n\\[\nb_{n}:=\\frac{a_{n}^{\\,x}}{n^{y}(\\log n)^{z}}, \n\\qquad \nS(x,y,z)=\\sum_{n=3}^{\\infty}(-1)^{n} b_{n},\n\\]\nand  \n\\[\np:=3x+y\\quad(\\text{the effective power of }n).\n\\]\n\nStep 1.  Precise asymptotics of \\(a_{n}\\).\n\nUsing the standard small–angle expansions  \n\\[\n\\sec t=1+\\frac{t^{2}}{2}+O(t^{4}),\\qquad \n\\csc t=t^{-1}+\\frac{t}{6}+O(t^{3})\\quad(t\\to0),\n\\]\none finds  \n\\[\n\\begin{aligned}\n3\\sec\\!\\bigl(\\tfrac{3}{n^{5/2}}\\bigr)-3\n   &=\\frac{27}{2n^{5}}+O(n^{-10}),\\\\\n2\\Bigl(\\csc\\!\\bigl(\\tfrac{2}{n^{3}}\\bigr)-\\tfrac{n^{3}}{2}\\Bigr)\n   &=\\frac{2}{3n^{3}}+O(n^{-9}),\n\\end{aligned}\n\\]\nhence  \n\\[\na_{n}= \\frac{2}{3n^{3}}+\\frac{27}{2n^{5}}+O(n^{-7})\n     =\\kappa n^{-3}\\!\\Bigl(1+\\rho n^{-2}+O(n^{-4})\\Bigr),\n\\qquad \n\\kappa:=\\frac{2}{3},\\;\\; \\rho:=\\frac{81}{4}.\n\\]\nIn particular \\(a_{n}>0\\) for every \\(n\\ge3\\), so \\(a_{n}^{x}\\) is well defined for all real \\(x\\).\n\nStep 2.  Leading form of \\(b_{n}\\).\n\nBecause \\(a_{n}\\sim\\kappa n^{-3}\\) with \\(\\kappa>0\\),\n\\[\na_{n}^{\\,x}= \\kappa^{\\,x}\\,n^{-3x}\\bigl(1+O(n^{-2})\\bigr),\n\\qquad\n|b_{n}|=\\frac{\\kappa^{\\,x}\\,(1+O(n^{-2}))}{n^{p}(\\log n)^{z}}\n         =\\bigl(1+O(n^{-2})\\bigr)\\,f(n),\n\\]\nwhere  \n\\[\nf(t):=\\frac{\\kappa^{\\,x}}{t^{p}(\\log t)^{z}},\\quad t\\ge3.\n\\]\n\nStep 3.  Absolute convergence.\n\nThe integral test for the positive series \\(\\sum n^{-p}(\\log n)^{-z}\\) gives  \n\n• if \\(p>1\\) the series converges for every \\(z\\in\\mathbb R\\);  \n\n• if \\(p=1\\) it converges precisely when \\(z>1\\);  \n\n• if \\(p<1\\) it diverges for every \\(z\\).\n\nBecause \\(|b_{n}|/f(n)\\to1\\), the same criteria apply to \\(\\sum |b_{n}|\\).\n\nHence  \nABSOLUTE CONVERGENCE ⇔  \n\n (i) \\(p>1\\);  or (ii) \\(p=1\\) and \\(z>1\\).\n\nStep 4.  Necessary vanishing of the terms.\n\nSince \\(|b_{n}|\\asymp n^{-p}(\\log n)^{-z}\\),  \n\\[\n|b_{n}|\\xrightarrow[n\\to\\infty]{}0\n\\Longleftrightarrow\n\\Bigl(p>0\\Bigr)\\;\\text{or}\\;\\Bigl(p=0\\text{ and }z>0\\Bigr).\n\\tag{4.1}\n\\]\nOutside (4.1) the series diverges.\n\nStep 5.  Rigorous monotonicity of \\(|b_{n}|\\).\n\nFor the alternating–series (Leibniz) test we must show that  \n\\(|b_{n+1}|\\le |b_{n}|\\) for all large \\(n\\) under the hypotheses of (4.1).\n\n5.1  Factorisation.  \nWrite  \n\\[\n|b_{n}| = C_{n}\\,f(n),\\qquad\nC_{n}:=\\bigl(1+\\rho n^{-2}+O(n^{-4})\\bigr)^{x}.\n\\]\nA Taylor expansion gives  \n\\[\nC_{n}=1+\\frac{\\rho x}{n^{2}}+O(n^{-4}),\\qquad \nC_{n+1}-C_{n}=O(n^{-3}). \\tag{5.1}\n\\]\n\n5.2  Discrete derivative of \\(f\\).  \nA direct computation yields  \n\\[\nf(t+1)-f(t)= -t^{-p-1}(\\log t)^{-z-1}\\bigl[p\\log t+z+O(1/t)\\bigr]\n           =-c_{0}\\,t^{-p-1}(\\log t)^{-z-1}\\bigl(1+o(1)\\bigr),\n\\]\nwith some constant \\(c_{0}>0\\) (because \\(p\\ge0\\) and \\(p,z\\) are fixed).\n\nThus  \n\\[\nf(n+1)-f(n)= -c_{1}\\,n^{-p-1}(\\log n)^{-z-1}\\bigl(1+o(1)\\bigr).\n\\tag{5.2}\n\\]\n\n5.3  Comparison of the two contributions.  \nFor large \\(n\\),\n\\[\n\\begin{aligned}\n|b_{n+1}|-|b_{n}|\n&=C_{n+1}f(n+1)-C_{n}f(n)  \\\\\n&=C_{n+1}\\bigl[f(n+1)-f(n)\\bigr]+(C_{n+1}-C_{n})f(n).\n\\end{aligned}\n\\]\nUsing (5.1)–(5.2) and \\(f(n)\\asymp n^{-p}(\\log n)^{-z}\\), we get  \n\\[\n\\bigl|(C_{n+1}-C_{n})f(n)\\bigr|\n      =O\\!\\bigl(n^{-3}\\bigr)\\cdot n^{-p}(\\log n)^{-z}\n      =O\\!\\bigl(n^{-p-3}(\\log n)^{-z}\\bigr),\n\\]\nwhereas  \n\\[\n|f(n+1)-f(n)|\n      \\asymp n^{-p-1}(\\log n)^{-z-1}.\n\\]\nBecause \\(p\\le1\\) in the region where absolute convergence fails\n(see Step 3), we have \\(-p-1> -p-3\\); hence  \n\\[\nn^{-p-1}(\\log n)^{-z-1}\\gg n^{-p-3}(\\log n)^{-z}\\qquad(n\\to\\infty).\n\\]\nTherefore the negative term \\(C_{n+1}[f(n+1)-f(n)]\\) dominates and\n\\(|b_{n+1}|\\le |b_{n}|\\) for all sufficiently large \\(n\\).\n\nConsequently \\(|b_{n}|\\) is eventually monotone decreasing whenever (4.1) is satisfied.\n\nStep 6.  Conditional convergence.\n\nAssume (4.1).  Then \\(|b_{n}|\\to0\\) and is eventually decreasing, so the alternating–series test applies and \\(S(x,y,z)\\) converges.\n\nCombining with Step 3, conditional (i.e. non-absolute) convergence occurs precisely when the alternating series converges but the series of absolute values diverges, namely  \n\n(a) \\(0<p<1\\)  (all \\(z\\));  \n\n(b) \\(p=1\\) and \\(z\\le1\\);  \n\n(c) \\(p=0\\) and \\(z>0\\).\n\nStep 7.  Final classification.\n\nRecall \\(p=3x+y\\).\n\n1.  Absolute convergence (part (a)):  \n • \\(3x+y>1\\) (any \\(z\\));  \n • \\(3x+y=1\\) with \\(z>1\\).\n\n2.  Conditional but not absolute convergence (part (b)):  \n • \\(0<3x+y<1\\) (any \\(z\\));  \n • \\(3x+y=1\\) with \\(z\\le1\\);  \n • \\(3x+y=0\\) with \\(z>0\\).\n\n3.  Divergence:  \n • \\(3x+y<0\\);  \n • \\(3x+y=0\\) with \\(z\\le0\\).\n\nThis completes the proof.\n\n————————————————————————————————————",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.703411",
        "was_fixed": false,
        "difficulty_analysis": "• Higher Dimension:  the parameter space is three-dimensional\n\\((x,y,z)\\) instead of one or two.  \n• Multiple Small-Angle Expansions:  both a secant and a cosecant term,\neach requiring different series expansions, must be combined while\ncarefully cancelling the dominant \\(n^{3}\\) singularity of the cosecant.  \n• Mixed Growth Rates:  the bracketed expression behaves like\n\\(n^{-3}\\), demanding a non-trivial asymptotic synthesis before any\ncomparison test can be applied.  \n• Absolute vs Conditional Convergence:  the problem asks for a full\nclassification of both regimes, forcing the solver to invoke and\ncoordinate the integral test, limit comparison, Leibniz, and Dirichlet\ncriteria.  \n• Edge-Case Analysis:  delicate logarithmic factors at the critical\nborder \\(3x+y=1\\) and \\(3x+y=0\\) must be handled separately.  \n\nThese added layers render the enhanced variant markedly more intricate\nthan either the original problem or the current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Determine, with proof, the complete set of real triples \n\\[\n(x,y,z)\\in\\mathbb R^{3}\n\\]\nfor which the alternating series  \n\\[\n\\boxed{\\,S(x,y,z)\\;=\\;\n\\sum_{n=3}^{\\infty}(-1)^{n}\\,\n\\frac{\\Bigl[\n\\,3\\sec\\!\\bigl(\\tfrac{3}{n^{5/2}}\\bigr)-3\n\\;+\\;\n2\\bigl(\\csc\\!\\bigl(\\tfrac{2}{n^{3}}\\bigr)-\\tfrac{n^{3}}{2}\\bigr)\n\\Bigr]^{\\,x}}\n{\\,n^{\\,y}\\,(\\log n)^{\\,z}}}\\qquad(\\log=\\text{natural logarithm})\n\\]\n\n(a) converges absolutely;  \n(b) converges conditionally, i.e. converges but not absolutely. \n\n(Real powers are understood through the principal branch; the bracket is strictly positive for every \\(n\\ge 3\\), so all terms are real for every real \\(x\\).)\n\n————————————————————————————————————",
      "solution": "Notation.  Put  \n\\[\na_{n}:=\n3\\sec\\!\\left(\\frac{3}{n^{5/2}}\\right)-3+\n2\\left(\\csc\\!\\left(\\frac{2}{n^{3}}\\right)-\\frac{n^{3}}{2}\\right),\n\\qquad n\\ge 3 ,\n\\]\n\\[\nb_{n}:=\\frac{a_{n}^{\\,x}}{n^{y}(\\log n)^{z}}, \n\\qquad \nS(x,y,z)=\\sum_{n=3}^{\\infty}(-1)^{n} b_{n},\n\\]\nand  \n\\[\np:=3x+y\\quad(\\text{the effective power of }n).\n\\]\n\nStep 1.  Precise asymptotics of \\(a_{n}\\).\n\nUsing the standard small–angle expansions  \n\\[\n\\sec t=1+\\frac{t^{2}}{2}+O(t^{4}),\\qquad \n\\csc t=t^{-1}+\\frac{t}{6}+O(t^{3})\\quad(t\\to0),\n\\]\none finds  \n\\[\n\\begin{aligned}\n3\\sec\\!\\bigl(\\tfrac{3}{n^{5/2}}\\bigr)-3\n   &=\\frac{27}{2n^{5}}+O(n^{-10}),\\\\\n2\\Bigl(\\csc\\!\\bigl(\\tfrac{2}{n^{3}}\\bigr)-\\tfrac{n^{3}}{2}\\Bigr)\n   &=\\frac{2}{3n^{3}}+O(n^{-9}),\n\\end{aligned}\n\\]\nhence  \n\\[\na_{n}= \\frac{2}{3n^{3}}+\\frac{27}{2n^{5}}+O(n^{-7})\n     =\\kappa n^{-3}\\!\\Bigl(1+\\rho n^{-2}+O(n^{-4})\\Bigr),\n\\qquad \n\\kappa:=\\frac{2}{3},\\;\\; \\rho:=\\frac{81}{4}.\n\\]\nIn particular \\(a_{n}>0\\) for every \\(n\\ge3\\), so \\(a_{n}^{x}\\) is well defined for all real \\(x\\).\n\nStep 2.  Leading form of \\(b_{n}\\).\n\nBecause \\(a_{n}\\sim\\kappa n^{-3}\\) with \\(\\kappa>0\\),\n\\[\na_{n}^{\\,x}= \\kappa^{\\,x}\\,n^{-3x}\\bigl(1+O(n^{-2})\\bigr),\n\\qquad\n|b_{n}|=\\frac{\\kappa^{\\,x}\\,(1+O(n^{-2}))}{n^{p}(\\log n)^{z}}\n         =\\bigl(1+O(n^{-2})\\bigr)\\,f(n),\n\\]\nwhere  \n\\[\nf(t):=\\frac{\\kappa^{\\,x}}{t^{p}(\\log t)^{z}},\\quad t\\ge3.\n\\]\n\nStep 3.  Absolute convergence.\n\nThe integral test for the positive series \\(\\sum n^{-p}(\\log n)^{-z}\\) gives  \n\n• if \\(p>1\\) the series converges for every \\(z\\in\\mathbb R\\);  \n\n• if \\(p=1\\) it converges precisely when \\(z>1\\);  \n\n• if \\(p<1\\) it diverges for every \\(z\\).\n\nBecause \\(|b_{n}|/f(n)\\to1\\), the same criteria apply to \\(\\sum |b_{n}|\\).\n\nHence  \nABSOLUTE CONVERGENCE ⇔  \n\n (i) \\(p>1\\);  or (ii) \\(p=1\\) and \\(z>1\\).\n\nStep 4.  Necessary vanishing of the terms.\n\nSince \\(|b_{n}|\\asymp n^{-p}(\\log n)^{-z}\\),  \n\\[\n|b_{n}|\\xrightarrow[n\\to\\infty]{}0\n\\Longleftrightarrow\n\\Bigl(p>0\\Bigr)\\;\\text{or}\\;\\Bigl(p=0\\text{ and }z>0\\Bigr).\n\\tag{4.1}\n\\]\nOutside (4.1) the series diverges.\n\nStep 5.  Rigorous monotonicity of \\(|b_{n}|\\).\n\nFor the alternating–series (Leibniz) test we must show that  \n\\(|b_{n+1}|\\le |b_{n}|\\) for all large \\(n\\) under the hypotheses of (4.1).\n\n5.1  Factorisation.  \nWrite  \n\\[\n|b_{n}| = C_{n}\\,f(n),\\qquad\nC_{n}:=\\bigl(1+\\rho n^{-2}+O(n^{-4})\\bigr)^{x}.\n\\]\nA Taylor expansion gives  \n\\[\nC_{n}=1+\\frac{\\rho x}{n^{2}}+O(n^{-4}),\\qquad \nC_{n+1}-C_{n}=O(n^{-3}). \\tag{5.1}\n\\]\n\n5.2  Discrete derivative of \\(f\\).  \nA direct computation yields  \n\\[\nf(t+1)-f(t)= -t^{-p-1}(\\log t)^{-z-1}\\bigl[p\\log t+z+O(1/t)\\bigr]\n           =-c_{0}\\,t^{-p-1}(\\log t)^{-z-1}\\bigl(1+o(1)\\bigr),\n\\]\nwith some constant \\(c_{0}>0\\) (because \\(p\\ge0\\) and \\(p,z\\) are fixed).\n\nThus  \n\\[\nf(n+1)-f(n)= -c_{1}\\,n^{-p-1}(\\log n)^{-z-1}\\bigl(1+o(1)\\bigr).\n\\tag{5.2}\n\\]\n\n5.3  Comparison of the two contributions.  \nFor large \\(n\\),\n\\[\n\\begin{aligned}\n|b_{n+1}|-|b_{n}|\n&=C_{n+1}f(n+1)-C_{n}f(n)  \\\\\n&=C_{n+1}\\bigl[f(n+1)-f(n)\\bigr]+(C_{n+1}-C_{n})f(n).\n\\end{aligned}\n\\]\nUsing (5.1)–(5.2) and \\(f(n)\\asymp n^{-p}(\\log n)^{-z}\\), we get  \n\\[\n\\bigl|(C_{n+1}-C_{n})f(n)\\bigr|\n      =O\\!\\bigl(n^{-3}\\bigr)\\cdot n^{-p}(\\log n)^{-z}\n      =O\\!\\bigl(n^{-p-3}(\\log n)^{-z}\\bigr),\n\\]\nwhereas  \n\\[\n|f(n+1)-f(n)|\n      \\asymp n^{-p-1}(\\log n)^{-z-1}.\n\\]\nBecause \\(p\\le1\\) in the region where absolute convergence fails\n(see Step 3), we have \\(-p-1> -p-3\\); hence  \n\\[\nn^{-p-1}(\\log n)^{-z-1}\\gg n^{-p-3}(\\log n)^{-z}\\qquad(n\\to\\infty).\n\\]\nTherefore the negative term \\(C_{n+1}[f(n+1)-f(n)]\\) dominates and\n\\(|b_{n+1}|\\le |b_{n}|\\) for all sufficiently large \\(n\\).\n\nConsequently \\(|b_{n}|\\) is eventually monotone decreasing whenever (4.1) is satisfied.\n\nStep 6.  Conditional convergence.\n\nAssume (4.1).  Then \\(|b_{n}|\\to0\\) and is eventually decreasing, so the alternating–series test applies and \\(S(x,y,z)\\) converges.\n\nCombining with Step 3, conditional (i.e. non-absolute) convergence occurs precisely when the alternating series converges but the series of absolute values diverges, namely  \n\n(a) \\(0<p<1\\)  (all \\(z\\));  \n\n(b) \\(p=1\\) and \\(z\\le1\\);  \n\n(c) \\(p=0\\) and \\(z>0\\).\n\nStep 7.  Final classification.\n\nRecall \\(p=3x+y\\).\n\n1.  Absolute convergence (part (a)):  \n • \\(3x+y>1\\) (any \\(z\\));  \n • \\(3x+y=1\\) with \\(z>1\\).\n\n2.  Conditional but not absolute convergence (part (b)):  \n • \\(0<3x+y<1\\) (any \\(z\\));  \n • \\(3x+y=1\\) with \\(z\\le1\\);  \n • \\(3x+y=0\\) with \\(z>0\\).\n\n3.  Divergence:  \n • \\(3x+y<0\\);  \n • \\(3x+y=0\\) with \\(z\\le0\\).\n\nThis completes the proof.\n\n————————————————————————————————————",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.549188",
        "was_fixed": false,
        "difficulty_analysis": "• Higher Dimension:  the parameter space is three-dimensional\n\\((x,y,z)\\) instead of one or two.  \n• Multiple Small-Angle Expansions:  both a secant and a cosecant term,\neach requiring different series expansions, must be combined while\ncarefully cancelling the dominant \\(n^{3}\\) singularity of the cosecant.  \n• Mixed Growth Rates:  the bracketed expression behaves like\n\\(n^{-3}\\), demanding a non-trivial asymptotic synthesis before any\ncomparison test can be applied.  \n• Absolute vs Conditional Convergence:  the problem asks for a full\nclassification of both regimes, forcing the solver to invoke and\ncoordinate the integral test, limit comparison, Leibniz, and Dirichlet\ncriteria.  \n• Edge-Case Analysis:  delicate logarithmic factors at the critical\nborder \\(3x+y=1\\) and \\(3x+y=0\\) must be handled separately.  \n\nThese added layers render the enhanced variant markedly more intricate\nthan either the original problem or the current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}