{
  "index": "1950-A-2",
  "type": "ANA",
  "tag": [
    "ANA",
    "NT"
  ],
  "difficulty": "",
  "question": "2. Answer both (i) and (ii). Test for convergence the series\n(i) \\( \\frac{1}{\\log (2!)}+\\frac{1}{\\log (3!)}+\\frac{1}{\\log (4!)}+\\cdots+\\frac{1}{\\log (n!)}+\\cdots \\)\n(ii) \\( \\frac{1}{3}+\\frac{1}{3 \\sqrt{3}}+\\frac{1}{3 \\sqrt{3} \\sqrt[3]{3}}+\\cdots+\\frac{1}{3 \\sqrt{3} \\sqrt[3]{3} \\cdots \\sqrt[n]{3}}+\\cdots \\).",
  "solution": "Solution. For \\( n \\geq 2 \\), we have \\( n^{n}>n! \\), hence \\( n \\log n>\\log (n!) \\) and\n\\[\n\\frac{1}{\\log (n!)}>\\frac{1}{n \\log n}\n\\]\n\nSeries (i) therefore dominates the series\n\\[\n\\sum_{n=2}^{\\infty} \\frac{1}{n \\log n}\n\\]\n\nSince\n\\[\n\\int_{2}^{x} \\frac{d t}{t \\log t}=\\log \\log x-\\log \\log 2\n\\]\nthe improper integral \\( \\int_{2}^{\\infty} d t /(t \\log t) \\) diverges, and hence, by the integral test, so does (1). Therefore series (i) is divergent.\n\nThe denominator of the \\( n \\)th term of series (ii) is \\( 3^{1+(1 / 2)+\\cdots+(1 / n)} \\), and \\( 1+\\frac{1}{2} \\) \\( +\\cdots+(1 / n) \\sim \\log n \\). Hence the \\( n \\)th term of (ii) is about\n\\[\n\\frac{1}{3^{\\log n}}=\\frac{1}{n^{\\log 3}}\n\\]\n\nNow \\( \\Sigma n^{-p} \\) converges if \\( p>1 \\) and \\( \\log 3>1 \\), so series (ii) converges.\nWe shall give the details of this argument. Since\n\\[\n\\sum_{k=1}^{n} \\frac{1}{k}>\\sum_{k=1}^{n} \\int_{k}^{k+1} \\frac{d t}{t}=\\int_{1}^{\\dot{n}+1} \\frac{d t}{t}=\\log (n+1)>\\log n\n\\]\nwe have\n\\[\n3^{1+(1 / 2)+\\cdots+(1 / n)}>3^{\\log n}=n^{\\log 3}\n\\]\n\nHence series (ii) is dominated by \\( \\Sigma n^{-\\log 3} \\). Since the latter converges, so does (ii).",
  "vars": [
    "n",
    "t",
    "x",
    "k"
  ],
  "params": [
    "p"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "n": "indexvar",
        "t": "integrand",
        "x": "upperlimit",
        "k": "summand",
        "p": "exponent"
      },
      "question": "2. Answer both (i) and (ii). Test for convergence the series\n(i) \\( \\frac{1}{\\log (2!)}+\\frac{1}{\\log (3!)}+\\frac{1}{\\log (4!)}+\\cdots+\\frac{1}{\\log (indexvar!)}+\\cdots \\)\n(ii) \\( \\frac{1}{3}+\\frac{1}{3 \\sqrt{3}}+\\frac{1}{3 \\sqrt{3} \\sqrt[3]{3}}+\\cdots+\\frac{1}{3 \\sqrt{3} \\sqrt[3]{3} \\cdots \\sqrt[indexvar]{3}}+\\cdots \\).",
      "solution": "Solution. For \\( indexvar \\geq 2 \\), we have \\( indexvar^{indexvar}>indexvar! \\), hence \\( indexvar \\log indexvar>\\log (indexvar!) \\) and\n\\[\n\\frac{1}{\\log (indexvar!)}>\\frac{1}{indexvar \\log indexvar}\n\\]\n\nSeries (i) therefore dominates the series\n\\[\n\\sum_{indexvar=2}^{\\infty} \\frac{1}{indexvar \\log indexvar}\n\\]\n\nSince\n\\[\n\\int_{2}^{upperlimit} \\frac{d integrand}{integrand \\log integrand}=\\log \\log upperlimit-\\log \\log 2\n\\]\nthe improper integral \\( \\int_{2}^{\\infty} d integrand /(integrand \\log integrand) \\) diverges, and hence, by the integral test, so does (1). Therefore series (i) is divergent.\n\nThe denominator of the \\( indexvar \\)th term of series (ii) is \\( 3^{1+(1 / 2)+\\cdots+(1 / indexvar)} \\), and \\( 1+\\frac{1}{2} \\) \\( +\\cdots+(1 / indexvar) \\sim \\log indexvar \\). Hence the \\( indexvar \\)th term of (ii) is about\n\\[\n\\frac{1}{3^{\\log indexvar}}=\\frac{1}{indexvar^{\\log 3}}\n\\]\n\nNow \\( \\Sigma indexvar^{-exponent} \\) converges if \\( exponent>1 \\) and \\( \\log 3>1 \\), so series (ii) converges.\nWe shall give the details of this argument. Since\n\\[\n\\sum_{summand=1}^{indexvar} \\frac{1}{summand}>\\sum_{summand=1}^{indexvar} \\int_{summand}^{summand+1} \\frac{d integrand}{integrand}=\\int_{1}^{\\dot{indexvar}+1} \\frac{d integrand}{integrand}=\\log (indexvar+1)>\\log indexvar\n\\]\nwe have\n\\[\n3^{1+(1 / 2)+\\cdots+(1 / indexvar)}>3^{\\log indexvar}=indexvar^{\\log 3}\n\\]\n\nHence series (ii) is dominated by \\( \\Sigma indexvar^{-\\log 3} \\). Since the latter converges, so does (ii)."
    },
    "descriptive_long_confusing": {
      "map": {
        "n": "pebblejar",
        "t": "lanterns",
        "x": "mooncraft",
        "k": "windshall",
        "p": "melonseed"
      },
      "question": "2. Answer both (i) and (ii). Test for convergence the series\n(i) \\( \\frac{1}{\\log (2!)}+\\frac{1}{\\log (3!)}+\\frac{1}{\\log (4!)}+\\cdots+\\frac{1}{\\log (pebblejar!)}+\\cdots \\)\n(ii) \\( \\frac{1}{3}+\\frac{1}{3 \\sqrt{3}}+\\frac{1}{3 \\sqrt{3} \\sqrt[3]{3}}+\\cdots+\\frac{1}{3 \\sqrt{3} \\sqrt[3]{3} \\cdots \\sqrt[pebblejar]{3}}+\\cdots \\).",
      "solution": "Solution. For \\( pebblejar \\geq 2 \\), we have \\( pebblejar^{pebblejar}>pebblejar! \\), hence \\( pebblejar \\log pebblejar>\\log (pebblejar!) \\) and\n\\[\n\\frac{1}{\\log (pebblejar!)}>\\frac{1}{pebblejar \\log pebblejar}\n\\]\n\nSeries (i) therefore dominates the series\n\\[\n\\sum_{pebblejar=2}^{\\infty} \\frac{1}{pebblejar \\log pebblejar}\n\\]\n\nSince\n\\[\n\\int_{2}^{mooncraft} \\frac{d lanterns}{lanterns \\log lanterns}=\\log \\log mooncraft-\\log \\log 2\n\\]\nthe improper integral \\( \\int_{2}^{\\infty} d lanterns /(lanterns \\log lanterns) \\) diverges, and hence, by the integral test, so does (1). Therefore series (i) is divergent.\n\nThe denominator of the \\( pebblejar \\)th term of series (ii) is \\( 3^{1+(1 / 2)+\\cdots+(1 / pebblejar)} \\), and \\( 1+\\frac{1}{2}+\\cdots+(1 / pebblejar) \\sim \\log pebblejar \\). Hence the \\( pebblejar \\)th term of (ii) is about\n\\[\n\\frac{1}{3^{\\log pebblejar}}=\\frac{1}{pebblejar^{\\log 3}}\n\\]\n\nNow \\( \\Sigma pebblejar^{-\\melonseed} \\) converges if \\( melonseed>1 \\) and \\( \\log 3>1 \\), so series (ii) converges.\nWe shall give the details of this argument. Since\n\\[\n\\sum_{windshall=1}^{pebblejar} \\frac{1}{windshall}>\\sum_{windshall=1}^{pebblejar} \\int_{windshall}^{windshall+1} \\frac{d lanterns}{lanterns}=\\int_{1}^{\\dot{pebblejar}+1} \\frac{d lanterns}{lanterns}=\\log (pebblejar+1)>\\log pebblejar\n\\]\nwe have\n\\[\n3^{1+(1 / 2)+\\cdots+(1 / pebblejar)}>3^{\\log pebblejar}=pebblejar^{\\log 3}\n\\]\n\nHence series (ii) is dominated by \\( \\Sigma pebblejar^{-\\log 3} \\). Since the latter converges, so does (ii)."
    },
    "descriptive_long_misleading": {
      "map": {
        "n": "irrational",
        "t": "timeless",
        "x": "constant",
        "k": "unindexed",
        "p": "logarithm"
      },
      "question": "2. Answer both (i) and (ii). Test for convergence the series\n(i) \\( \\frac{1}{\\log (2!)}+\\frac{1}{\\log (3!)}+\\frac{1}{\\log (4!)}+\\cdots+\\frac{1}{\\log (irrational!)}+\\cdots \\)\n(ii) \\( \\frac{1}{3}+\\frac{1}{3 \\sqrt{3}}+\\frac{1}{3 \\sqrt{3} \\sqrt[3]{3}}+\\cdots+\\frac{1}{3 \\sqrt{3} \\sqrt[3]{3} \\cdots \\sqrt[irrational]{3}}+\\cdots \\).",
      "solution": "Solution. For \\( irrational \\geq 2 \\), we have \\( irrational^{irrational}>irrational! \\), hence \\( irrational \\log irrational>\\log (irrational!) \\) and\n\\[\n\\frac{1}{\\log (irrational!)}>\\frac{1}{irrational \\log irrational}\n\\]\n\nSeries (i) therefore dominates the series\n\\[\n\\sum_{irrational=2}^{\\infty} \\frac{1}{irrational \\log irrational}\n\\]\n\nSince\n\\[\n\\int_{2}^{constant} \\frac{d timeless}{timeless \\log timeless}=\\log \\log constant-\\log \\log 2\n\\]\nthe improper integral \\( \\int_{2}^{\\infty} d timeless /(timeless \\log timeless) \\) diverges, and hence, by the integral test, so does (1). Therefore series (i) is divergent.\n\nThe denominator of the \\( irrational \\)th term of series (ii) is \\( 3^{1+(1 / 2)+\\cdots+(1 / irrational)} \\), and \\( 1+\\frac{1}{2} +\\cdots+(1 / irrational) \\sim \\log irrational \\). Hence the \\( irrational \\)th term of (ii) is about\n\\[\n\\frac{1}{3^{\\log irrational}}=\\frac{1}{irrational^{\\log 3}}\n\\]\n\nNow \\( \\Sigma irrational^{-\\logarithm} \\) converges if \\( logarithm>1 \\) and \\( \\log 3>1 \\), so series (ii) converges.\nWe shall give the details of this argument. Since\n\\[\n\\sum_{unindexed=1}^{irrational} \\frac{1}{unindexed}>\\sum_{unindexed=1}^{irrational} \\int_{unindexed}^{unindexed+1} \\frac{d timeless}{timeless}=\\int_{1}^{\\dot{irrational}+1} \\frac{d timeless}{timeless}=\\log (irrational+1)>\\log irrational\n\\]\nwe have\n\\[\n3^{1+(1 / 2)+\\cdots+(1 / irrational)}>3^{\\log irrational}=irrational^{\\log 3}\n\\]\n\nHence series (ii) is dominated by \\( \\Sigma irrational^{-\\log 3} \\). Since the latter converges, so does (ii)."
    },
    "garbled_string": {
      "map": {
        "n": "qzxwvtnp",
        "t": "hjgrksla",
        "x": "mncbvier",
        "k": "dwexmisk",
        "p": "rufbolte"
      },
      "question": "2. Answer both (i) and (ii). Test for convergence the series\n(i) \\( \\frac{1}{\\log (2!)}+\\frac{1}{\\log (3!)}+\\frac{1}{\\log (4!)}+\\cdots+\\frac{1}{\\log (qzxwvtnp!)}+\\cdots \\)\n(ii) \\( \\frac{1}{3}+\\frac{1}{3 \\sqrt{3}}+\\frac{1}{3 \\sqrt{3} \\sqrt[3]{3}}+\\cdots+\\frac{1}{3 \\sqrt{3} \\sqrt[3]{3} \\cdots \\sqrt[\\qzxwvtnp]{3}}+\\cdots \\).",
      "solution": "Solution. For \\( qzxwvtnp \\geq 2 \\), we have \\( qzxwvtnp^{qzxwvtnp}>qzxwvtnp! \\), hence \\( qzxwvtnp \\log qzxwvtnp>\\log (qzxwvtnp!) \\) and\n\\[\n\\frac{1}{\\log (qzxwvtnp!)}>\\frac{1}{qzxwvtnp \\log qzxwvtnp}\n\\]\n\nSeries (i) therefore dominates the series\n\\[\n\\sum_{qzxwvtnp=2}^{\\infty} \\frac{1}{qzxwvtnp \\log qzxwvtnp}\n\\]\n\nSince\n\\[\n\\int_{2}^{mncbvier} \\frac{d hjgrksla}{hjgrksla \\log hjgrksla}=\\log \\log mncbvier-\\log \\log 2\n\\]\nthe improper integral \\( \\int_{2}^{\\infty} d hjgrksla /(hjgrksla \\log hjgrksla) \\) diverges, and hence, by the integral test, so does (1). Therefore series (i) is divergent.\n\nThe denominator of the \\( qzxwvtnp \\)th term of series (ii) is \\( 3^{1+(1 / 2)+\\cdots+(1 / qzxwvtnp)} \\), and \\( 1+\\frac{1}{2}+\\cdots+(1 / qzxwvtnp) \\sim \\log qzxwvtnp \\). Hence the \\( qzxwvtnp \\)th term of (ii) is about\n\\[\n\\frac{1}{3^{\\log qzxwvtnp}}=\\frac{1}{qzxwvtnp^{\\log 3}}\n\\]\n\nNow \\( \\Sigma qzxwvtnp^{-rufbolte} \\) converges if \\( rufbolte>1 \\) and \\( \\log 3>1 \\), so series (ii) converges.\nWe shall give the details of this argument. Since\n\\[\n\\sum_{dwexmisk=1}^{qzxwvtnp} \\frac{1}{dwexmisk}>\\sum_{dwexmisk=1}^{qzxwvtnp} \\int_{dwexmisk}^{dwexmisk+1} \\frac{d hjgrksla}{hjgrksla}=\\int_{1}^{\\dot{qzxwvtnp}+1} \\frac{d hjgrksla}{hjgrksla}=\\log (qzxwvtnp+1)>\\log qzxwvtnp\n\\]\nwe have\n\\[\n3^{1+(1 / 2)+\\cdots+(1 / qzxwvtnp)}>3^{\\log qzxwvtnp}=qzxwvtnp^{\\log 3}\n\\]\n\nHence series (ii) is dominated by \\( \\Sigma qzxwvtnp^{-\\log 3} \\). Since the latter converges, so does (ii)."
    },
    "kernel_variant": {
      "question": "Answer both (i) and (ii).\n\n(i)  For real parameters α,β consider  \n\n          S(α,β) =  Σ_{n=3}^{∞}  (−1)^{n}\\,\n                              n^{\\,α}\\,\\Bigl[\\log\\bigl(n!\\bigr)\\Bigr]^{−β}.\n     Determine, with proof, exactly for which pairs (α,β) the series  \n\n          • converges absolutely,  \n          • converges conditionally (i.e. converges but not absolutely),  \n          • diverges.  \n\n(ii)  Fix a real constant c>1 and, for s>0, set  \n\n          H_{n}^{(s)} = Σ_{k=1}^{n} k^{−s}.  \n\n     Investigate the series  \n\n          T_s(c) = Σ_{n=1}^{∞}  c^{−\\,H_{n}^{(s)}}                      (★)\n\n     and specify precisely for which values of the exponent s it converges\n     or diverges.",
      "solution": "(i)  Let  \n\n          a_n := (−1)^n n^{α}\\,[\\log(n!)]^{−β}, n≥3.\n\n     1.  Size of |a_n|.  \n         Stirling’s formula in logarithmic form gives  \n\n              log(n!) = n log n − n + ½ log(2π n) + O(1/n) ,\n\n         hence, as n→∞,  \n\n              [log(n!)]^{β} = (n log n)^{β}\\bigl(1+o(1)\\bigr).\n\n         Therefore  \n\n              |a_n| = n^{α−β}(log n)^{−β}\\bigl(1+o(1)\\bigr).  (1)\n\n         Introduce  \n\n              p := β − α.                                            (2)\n\n         Then |a_n| behaves like n^{−p}(log n)^{−β}.\n\n     2.  Absolute convergence.  \n         The comparison series  \n\n              Σ n^{−p}(log n)^{−β}\n\n         is a classical p-series with logarithmic correction; one obtains by\n         Cauchy condensation (or the integral test)\n\n              Σ n^{−p}(log n)^{−β}  converges\n                 ⇔  p>1  or  (p=1 and β>1).                         (3)\n\n         Via (2) this translates to\n\n              Absolute convergence\n              ⇔ β − α > 1  or  (β − α = 1 and β>1).                  (4)\n\n     3.  Conditional convergence.  \n         Whenever (4) fails, the series cannot be absolutely convergent, but\n         it may converge by alternating cancellation.  The Leibniz test\n         requires\n\n              (i) |a_n| → 0, (ii) |a_n| is eventually monotone.\n\n         From (1) we have |a_n|→0  exactly when  \n\n              α−β < 0, or α = β and β>0.                           (5)\n\n         To check monotonicity write  \n\n              g(x)=x^{α−β}(log x)^{−β}, x≥3,\n\n              g′(x)/g(x)= (α−β)/x − β/(x log x).\n\n         For large x the dominant term is (α−β)/x; thus\n\n              g′(x)<0 eventually if α−β<0,                         (6a)\n              g′(x)<0 eventually if α−β=0 and β>0.                 (6b)\n\n         Hence both Leibniz conditions are satisfied exactly under (5).\n         Intersecting this set with the complement of (4) gives\n\n              Conditional convergence\n              ⇔\n              (a) 0 < β−α < 1,  or\n              (b) β−α = 1 with β≤1,  or\n              (c) α = β > 0.                                       (7)\n\n     4.  Divergence.  \n         The only remaining cases are\n\n              • α > β  (|a_n| does not tend to 0);  \n              • α = β ≤ 0 (|a_n| does not tend to 0).\n\n         In both situations the necessary condition for convergence fails, so\n         the series diverges.\n\n     Final classification.  \n\n       Absolute convergence  ⇔  β−α>1  or  (β−α=1 & β>1);\n\n       Conditional convergence ⇔ \n            (i) 0<β−α<1, or  \n            (ii) β−α=1 with β≤1, or  \n            (iii) α=β>0;\n\n       Divergence             ⇔  α>β,  or  α=β≤0.   ■\n\n\n\n(ii)  Write b_n := c^{−H_{n}^{(s)}}, c>1.\n\n     A.  0 < s < 1.  \n         Euler–Maclaurin gives  \n\n              H_{n}^{(s)} = n^{\\,1−s}/(1−s) + ζ(s) + O(n^{−s}), n→∞,\n\n         so there exist positive constants C,κ with  \n\n              b_n = C·exp(−κ n^{1−s}).\n\n         Exponential decay implies absolute convergence of T_s(c).\n\n     B.  s = 1.  \n         H_{n}^{(1)} = log n + γ + o(1), γ Euler’s constant, hence  \n\n              b_n = c^{−log n + O(1)} = K·n^{−log c}(1+o(1)).\n\n         Comparison with the p-series Σ n^{−p} shows\n\n              T_1(c) converges  iff  log c > 1  (⇔ c>e).\n\n     C.  s > 1.  \n         Here H_{n}^{(s)} ↓ ζ(s) > 1, so b_n → c^{−ζ(s)} > 0; therefore the\n         terms do not even tend to zero and the series diverges.\n\n     Summary.  \n\n            0 < s < 1  → T_s(c) convergent (any c>1);  \n            s = 1    → convergent  iff  c>e;  \n            s > 1    → divergent  (any c>1).   ■",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.428088",
        "was_fixed": false,
        "difficulty_analysis": "• Parameter space:  Part (i) asks for a full (α,β) phase diagram – far more\n  intricate than the single numerical case in the original.  \n• Advanced asymptotics:  Stirling’s formula and logarithmic refinements are\n  essential; a simple comparison with 1/(n log n) no longer suffices.  \n• Mixed convergence notions:  One must distinguish absolute from conditional\n  convergence, invoking Cauchy condensation, integral tests, and Leibniz’s\n  criterion in concert.  \n• Interaction with special functions:  Part (ii) links the series to the\n  generalized harmonic numbers and the Riemann zeta function, and requires\n  an exponential/​polynomial dichotomy analysis unavailable in the original.  \n• Multiple regimes & thresholds:  Convergence hinges on inequalities such as\n  β−α > 1, log c > 1, or s<1—each derived from different analytic tools.  \nOverall, the enhanced variant demands deeper theoretical insight, more\ndelicate estimates, and a wider repertoire of convergence tests than either\nthe original problem or the first kernel modification."
      }
    },
    "original_kernel_variant": {
      "question": "Answer both (i) and (ii).\n\n(i)  For real parameters α,β consider  \n\n          S(α,β) =  Σ_{n=3}^{∞}  (−1)^{n}\\,\n                              n^{\\,α}\\,\\Bigl[\\log\\bigl(n!\\bigr)\\Bigr]^{−β}.\n     Determine, with proof, exactly for which pairs (α,β) the series  \n\n          • converges absolutely,  \n          • converges conditionally (i.e. converges but not absolutely),  \n          • diverges.  \n\n(ii)  Fix a real constant c>1 and, for s>0, set  \n\n          H_{n}^{(s)} = Σ_{k=1}^{n} k^{−s}.  \n\n     Investigate the series  \n\n          T_s(c) = Σ_{n=1}^{∞}  c^{−\\,H_{n}^{(s)}}                      (★)\n\n     and specify precisely for which values of the exponent s it converges\n     or diverges.",
      "solution": "(i)  Let  \n\n          a_n := (−1)^n n^{α}\\,[\\log(n!)]^{−β}, n≥3.\n\n     1.  Size of |a_n|.  \n         Stirling’s formula in logarithmic form gives  \n\n              log(n!) = n log n − n + ½ log(2π n) + O(1/n) ,\n\n         hence, as n→∞,  \n\n              [log(n!)]^{β} = (n log n)^{β}\\bigl(1+o(1)\\bigr).\n\n         Therefore  \n\n              |a_n| = n^{α−β}(log n)^{−β}\\bigl(1+o(1)\\bigr).  (1)\n\n         Introduce  \n\n              p := β − α.                                            (2)\n\n         Then |a_n| behaves like n^{−p}(log n)^{−β}.\n\n     2.  Absolute convergence.  \n         The comparison series  \n\n              Σ n^{−p}(log n)^{−β}\n\n         is a classical p-series with logarithmic correction; one obtains by\n         Cauchy condensation (or the integral test)\n\n              Σ n^{−p}(log n)^{−β}  converges\n                 ⇔  p>1  or  (p=1 and β>1).                         (3)\n\n         Via (2) this translates to\n\n              Absolute convergence\n              ⇔ β − α > 1  or  (β − α = 1 and β>1).                  (4)\n\n     3.  Conditional convergence.  \n         Whenever (4) fails, the series cannot be absolutely convergent, but\n         it may converge by alternating cancellation.  The Leibniz test\n         requires\n\n              (i) |a_n| → 0, (ii) |a_n| is eventually monotone.\n\n         From (1) we have |a_n|→0  exactly when  \n\n              α−β < 0, or α = β and β>0.                           (5)\n\n         To check monotonicity write  \n\n              g(x)=x^{α−β}(log x)^{−β}, x≥3,\n\n              g′(x)/g(x)= (α−β)/x − β/(x log x).\n\n         For large x the dominant term is (α−β)/x; thus\n\n              g′(x)<0 eventually if α−β<0,                         (6a)\n              g′(x)<0 eventually if α−β=0 and β>0.                 (6b)\n\n         Hence both Leibniz conditions are satisfied exactly under (5).\n         Intersecting this set with the complement of (4) gives\n\n              Conditional convergence\n              ⇔\n              (a) 0 < β−α < 1,  or\n              (b) β−α = 1 with β≤1,  or\n              (c) α = β > 0.                                       (7)\n\n     4.  Divergence.  \n         The only remaining cases are\n\n              • α > β  (|a_n| does not tend to 0);  \n              • α = β ≤ 0 (|a_n| does not tend to 0).\n\n         In both situations the necessary condition for convergence fails, so\n         the series diverges.\n\n     Final classification.  \n\n       Absolute convergence  ⇔  β−α>1  or  (β−α=1 & β>1);\n\n       Conditional convergence ⇔ \n            (i) 0<β−α<1, or  \n            (ii) β−α=1 with β≤1, or  \n            (iii) α=β>0;\n\n       Divergence             ⇔  α>β,  or  α=β≤0.   ■\n\n\n\n(ii)  Write b_n := c^{−H_{n}^{(s)}}, c>1.\n\n     A.  0 < s < 1.  \n         Euler–Maclaurin gives  \n\n              H_{n}^{(s)} = n^{\\,1−s}/(1−s) + ζ(s) + O(n^{−s}), n→∞,\n\n         so there exist positive constants C,κ with  \n\n              b_n = C·exp(−κ n^{1−s}).\n\n         Exponential decay implies absolute convergence of T_s(c).\n\n     B.  s = 1.  \n         H_{n}^{(1)} = log n + γ + o(1), γ Euler’s constant, hence  \n\n              b_n = c^{−log n + O(1)} = K·n^{−log c}(1+o(1)).\n\n         Comparison with the p-series Σ n^{−p} shows\n\n              T_1(c) converges  iff  log c > 1  (⇔ c>e).\n\n     C.  s > 1.  \n         Here H_{n}^{(s)} ↓ ζ(s) > 1, so b_n → c^{−ζ(s)} > 0; therefore the\n         terms do not even tend to zero and the series diverges.\n\n     Summary.  \n\n            0 < s < 1  → T_s(c) convergent (any c>1);  \n            s = 1    → convergent  iff  c>e;  \n            s > 1    → divergent  (any c>1).   ■",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.371577",
        "was_fixed": false,
        "difficulty_analysis": "• Parameter space:  Part (i) asks for a full (α,β) phase diagram – far more\n  intricate than the single numerical case in the original.  \n• Advanced asymptotics:  Stirling’s formula and logarithmic refinements are\n  essential; a simple comparison with 1/(n log n) no longer suffices.  \n• Mixed convergence notions:  One must distinguish absolute from conditional\n  convergence, invoking Cauchy condensation, integral tests, and Leibniz’s\n  criterion in concert.  \n• Interaction with special functions:  Part (ii) links the series to the\n  generalized harmonic numbers and the Riemann zeta function, and requires\n  an exponential/​polynomial dichotomy analysis unavailable in the original.  \n• Multiple regimes & thresholds:  Convergence hinges on inequalities such as\n  β−α > 1, log c > 1, or s<1—each derived from different analytic tools.  \nOverall, the enhanced variant demands deeper theoretical insight, more\ndelicate estimates, and a wider repertoire of convergence tests than either\nthe original problem or the first kernel modification."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}