{
  "index": "1975-B-6",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "B-6. Show that if \\( s_{n}=1+\\frac{1}{2}+\\frac{1}{3}+\\cdots+1 / n \\), then\n(a) \\( n(n+1)^{1 / n}<n+s_{n} \\) for \\( n>1 \\), and\n(b) \\( (n-1) n^{-1 /(n-1)}<n-s_{n} \\) for \\( n>2 \\).",
  "solution": "B-6.\nBoth parts are done easily using the Inequality on the Means. For (a), one has\n\\[\n\\begin{array}{r}\n\\frac{n+s_{n}}{n}=\\frac{(1+1)+\\left(1+\\frac{1}{2}\\right)+\\cdots+(1+(1 / n))}{n}>\\sqrt[n]{(1+1)\\left(1+\\frac{1}{2}\\right) \\cdots(1+(1 / n))}=\\sqrt[n]{2 \\cdot \\frac{3}{2} \\cdot \\frac{4}{3} \\cdots(n+1) / n} \\\\\n=(n+1)^{1 / n}\n\\end{array}\n\\]\nand so \\( n+s_{n}>n(n+1)^{1 / n} \\).\nFor (b), one has\n\\[\n\\begin{aligned}\n\\frac{n-s_{n}}{n-1}=\\frac{\\left(1-\\frac{1}{2}\\right)+\\left(1-\\frac{1}{3}\\right)+\\cdots+(1-(1 / n))}{n-1}>\\sqrt[n-1]{\\left(1-\\frac{1}{2}\\right)\\left(1-\\frac{1}{3}\\right) \\cdots(1-(1 / n))}=\\sqrt[n-1]{\\sqrt{\\frac{1}{2} \\cdot \\frac{2}{3} \\cdots(n-1) / n}} & =n^{-1 /(n-1)}\n\\end{aligned}\n\\]\nand so \\( n-s_{n}>(n-1) n^{-1 /(n-1)} \\).",
  "vars": [
    "n",
    "s_n"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "n": "countvar",
        "s_n": "partialsum"
      },
      "question": "B-6. Show that if \\( partialsum_{countvar}=1+\\frac{1}{2}+\\frac{1}{3}+\\cdots+1 / countvar \\), then\n(a) \\( countvar(countvar+1)^{1 / countvar}<countvar+partialsum_{countvar} \\) for \\( countvar>1 \\), and\n(b) \\( (countvar-1) countvar^{-1 /(countvar-1)}<countvar-partialsum_{countvar} \\) for \\( countvar>2 \\).",
      "solution": "B-6.\nBoth parts are done easily using the Inequality on the Means. For (a), one has\n\\[\n\\begin{array}{r}\n\\frac{countvar+partialsum_{countvar}}{countvar}=\\frac{(1+1)+\\left(1+\\frac{1}{2}\\right)+\\cdots+(1+(1 / countvar))}{countvar}>\\sqrt[countvar]{(1+1)\\left(1+\\frac{1}{2}\\right) \\cdots(1+(1 / countvar))}=\\sqrt[countvar]{2 \\cdot \\frac{3}{2} \\cdot \\frac{4}{3} \\cdots(countvar+1) / countvar} \\\\\n=(countvar+1)^{1 / countvar}\n\\end{array}\n\\]\nand so \\( countvar+partialsum_{countvar}>countvar(countvar+1)^{1 / countvar} \\).\nFor (b), one has\n\\[\n\\begin{aligned}\n\\frac{countvar-partialsum_{countvar}}{countvar-1}=\\frac{\\left(1-\\frac{1}{2}\\right)+\\left(1-\\frac{1}{3}\\right)+\\cdots+(1-(1 / countvar))}{countvar-1}>\\sqrt[countvar-1]{\\left(1-\\frac{1}{2}\\right)\\left(1-\\frac{1}{3}\\right) \\cdots(1-(1 / countvar))}=\\sqrt[countvar-1]{\\sqrt{\\frac{1}{2} \\cdot \\frac{2}{3} \\cdots(countvar-1) / countvar}} &=countvar^{-1 /(countvar-1)}\n\\end{aligned}\n\\]\nand so \\( countvar-partialsum_{countvar}>(countvar-1) countvar^{-1 /(countvar-1)} \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "n": "treasure",
        "s_n": "landscape"
      },
      "question": "B-6. Show that if \\( landscape=1+\\frac{1}{2}+\\frac{1}{3}+\\cdots+1 / treasure \\), then\n(a) \\( treasure(treasure+1)^{1 / treasure}<treasure+landscape \\) for \\( treasure>1 \\), and\n(b) \\( (treasure-1) treasure^{-1 /(treasure-1)}<treasure-landscape \\) for \\( treasure>2 \\).",
      "solution": "B-6.\nBoth parts are done easily using the Inequality on the Means. For (a), one has\n\\[\n\\begin{array}{r}\n\\frac{treasure+landscape}{treasure}=\\frac{(1+1)+\\left(1+\\frac{1}{2}\\right)+\\cdots+(1+(1 / treasure))}{treasure}>\\sqrt[treasure]{(1+1)\\left(1+\\frac{1}{2}\\right) \\cdots(1+(1 / treasure))}=\\sqrt[treasure]{2 \\cdot \\frac{3}{2} \\cdot \\frac{4}{3} \\cdots(treasure+1) / treasure} \\\\\n=(treasure+1)^{1 / treasure}\n\\end{array}\n\\]\nand so \\( treasure+landscape>treasure(treasure+1)^{1 / treasure} \\).\nFor (b), one has\n\\[\n\\begin{aligned}\n\\frac{treasure-landscape}{treasure-1}=\\frac{\\left(1-\\frac{1}{2}\\right)+\\left(1-\\frac{1}{3}\\right)+\\cdots+(1-(1 / treasure))}{treasure-1}>\\sqrt[treasure-1]{\\left(1-\\frac{1}{2}\\right)\\left(1-\\frac{1}{3}\\right) \\cdots(1-(1 / treasure))}=\\sqrt[treasure-1]{\\sqrt{\\frac{1}{2} \\cdot \\frac{2}{3} \\cdots(treasure-1) / treasure}} & =treasure^{-1 /(treasure-1)}\n\\end{aligned}\n\\]\nand so \\( treasure-landscape>(treasure-1) treasure^{-1 /(treasure-1)} \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "n": "infinitevalue",
        "s_n": "partialdifference"
      },
      "question": "B-6. Show that if \\( partialdifference=1+\\frac{1}{2}+\\frac{1}{3}+\\cdots+1 / infinitevalue \\), then\n(a) \\( infinitevalue(infinitevalue+1)^{1 / infinitevalue}<infinitevalue+partialdifference \\) for \\( infinitevalue>1 \\), and\n(b) \\( (infinitevalue-1) infinitevalue^{-1 /(infinitevalue-1)}<infinitevalue-partialdifference \\) for \\( infinitevalue>2 \\).",
      "solution": "B-6.\nBoth parts are done easily using the Inequality on the Means. For (a), one has\n\\[\n\\begin{array}{r}\n\\frac{infinitevalue+partialdifference}{infinitevalue}=\\frac{(1+1)+\\left(1+\\frac{1}{2}\\right)+\\cdots+(1+(1 / infinitevalue))}{infinitevalue}>\\sqrt[infinitevalue]{(1+1)\\left(1+\\frac{1}{2}\\right) \\cdots(1+(1 / infinitevalue))}=\\sqrt[infinitevalue]{2 \\cdot \\frac{3}{2} \\cdot \\frac{4}{3} \\cdots(infinitevalue+1) / infinitevalue} \\\\\n=(infinitevalue+1)^{1 / infinitevalue}\n\\end{array}\n\\]\nand so \\( infinitevalue+partialdifference>infinitevalue(infinitevalue+1)^{1 / infinitevalue} \\).\nFor (b), one has\n\\[\n\\begin{aligned}\n\\frac{infinitevalue-partialdifference}{infinitevalue-1}=\\frac{\\left(1-\\frac{1}{2}\\right)+\\left(1-\\frac{1}{3}\\right)+\\cdots+(1-(1 / infinitevalue))}{infinitevalue-1}>\\sqrt[infinitevalue-1]{\\left(1-\\frac{1}{2}\\right)\\left(1-\\frac{1}{3}\\right) \\cdots(1-(1 / infinitevalue))}=\\sqrt[infinitevalue-1]{\\sqrt{\\frac{1}{2} \\cdot \\frac{2}{3} \\cdots(infinitevalue-1) / infinitevalue}} & =infinitevalue^{-1 /(infinitevalue-1)}\n\\end{aligned}\n\\]\nand so \\( infinitevalue-partialdifference>(infinitevalue-1) infinitevalue^{-1 /(infinitevalue-1)} \\)."
    },
    "garbled_string": {
      "map": {
        "n": "qzxwvtnp",
        "s_n": "hjgrksla"
      },
      "question": "B-6. Show that if \\( hjgrksla=1+\\frac{1}{2}+\\frac{1}{3}+\\cdots+1 / qzxwvtnp \\), then\n(a) \\( qzxwvtnp(qzxwvtnp+1)^{1 / qzxwvtnp}<qzxwvtnp+hjgrksla \\) for \\( qzxwvtnp>1 \\), and\n(b) \\( (qzxwvtnp-1) qzxwvtnp^{-1 /(qzxwvtnp-1)}<qzxwvtnp-hjgrksla \\) for \\( qzxwvtnp>2 \\).",
      "solution": "B-6.\nBoth parts are done easily using the Inequality on the Means. For (a), one has\n\\[\n\\begin{array}{r}\n\\frac{qzxwvtnp+hjgrksla}{qzxwvtnp}=\\frac{(1+1)+\\left(1+\\frac{1}{2}\\right)+\\cdots+(1+(1 / qzxwvtnp))}{qzxwvtnp}>\\sqrt[qzxwvtnp]{(1+1)\\left(1+\\frac{1}{2}\\right) \\cdots(1+(1 / qzxwvtnp))}=\\sqrt[qzxwvtnp]{2 \\cdot \\frac{3}{2} \\cdot \\frac{4}{3} \\cdots(qzxwvtnp+1) / qzxwvtnp} \\\\\n=(qzxwvtnp+1)^{1 / qzxwvtnp}\n\\end{array}\n\\]\nand so \\( qzxwvtnp+hjgrksla>qzxwvtnp(qzxwvtnp+1)^{1 / qzxwvtnp} \\).\nFor (b), one has\n\\[\n\\begin{aligned}\n\\frac{qzxwvtnp-hjgrksla}{qzxwvtnp-1}=\\frac{\\left(1-\\frac{1}{2}\\right)+\\left(1-\\frac{1}{3}\\right)+\\cdots+(1-(1 / qzxwvtnp))}{qzxwvtnp-1}>\\sqrt[qzxwvtnp-1]{\\left(1-\\frac{1}{2}\\right)\\left(1-\\frac{1}{3}\\right) \\cdots(1-(1 / qzxwvtnp))}=\\sqrt[qzxwvtnp-1]{\\sqrt{\\frac{1}{2} \\cdot \\frac{2}{3} \\cdots(qzxwvtnp-1) / qzxwvtnp}} & =qzxwvtnp^{-1 /(qzxwvtnp-1)}\n\\end{aligned}\n\\]\nand so \\( qzxwvtnp-hjgrksla>(qzxwvtnp-1) qzxwvtnp^{-1 /(qzxwvtnp-1)} \\)."
    },
    "kernel_variant": {
      "question": "Let  \n\\[\n\\psi(x)=\\frac{\\Gamma'(x)}{\\Gamma(x)},\\qquad   \n\\gamma=\\lim_{n\\to\\infty}\\bigl(H_{n}-\\ln n\\bigr)=0.57721\\dots ,\n\\qquad  \nH_{n}=1+\\frac12+\\dots +\\frac1n ,\n\\]\nand, for every $x>1$, define  \n\\[\nF(x)=x+\\psi(x+1), \\qquad  G(x)=x\\,(x+1)^{1/x},\n\\]\n\\[\nP(x)=x-\\psi(x+1), \\qquad Q(x)=(x-1)\\,x^{-1/(x-1)} .\n\\]\n\n(a)  Prove the two–sided exponential refinements  \n\\[\nG(x)\\,\\exp\\!\\Bigl(-\\frac{1}{x}\\Bigr)\n        <F(x)<G(x),\\qquad x>1,\n\\tag{1}\n\\]\n\\[\nQ(x)<P(x)<Q(x)\\,\\exp\\!\\Bigl(\\frac{1}{\\,x-1\\,}\\Bigr),\\qquad x>1 .\n\\tag{2}\n\\]\n\n(b)  For $x>1$ define  \n\\[\nA(x)=\\ln G(x)-\\ln F(x),\\qquad \nB(x)=\\ln P(x)-\\ln Q(x).\n\\]\nShow that  \n\\[\n0<A(x)<\\frac{1}{x},\\qquad \n0<B(x)<\\frac{1}{\\,x-1\\,},\\qquad x>1,\n\\tag{3}\n\\]\nthat $A$ and $B$ are strictly decreasing and convex on $(1,\\infty)$, and that  \n\\[\n\\lim_{x\\to\\infty}xA(x)=\n\\lim_{x\\to\\infty}xB(x)=0 .\n\\tag{4}\n\\]\n\n(c)  Deduce from {\\rm(1)}–{\\rm(4)} together with elementary estimates that, for every integer $n\\ge 2$,\n\\[\nn\\,(n+1)^{1/n}<n+H_{n}< \n        n\\,(n+1)^{1/n}\\,\n        \\exp\\!\\Bigl(\\tfrac{1}{n}\\Bigr),\n\\tag{5'}\n\\]\nand, for every integer $n\\ge 3$,\n\\[\n(n-1)\\,n^{-1/(n-1)}<\n        n-H_{n}<\n        (n-1)\\,n^{-1/(n-1)}\n        \\exp\\!\\Bigl(\\tfrac{1}{\\,n-1\\,}\\Bigr).\n\\tag{6'}\n\\]\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "solution": "Throughout we use four elementary lemmas, the last of which was missing (and now used) in the draft.\n\n\\textbf{(L1)  (Batir, 2008).}  For $x>0$\n\\[\n\\boxed{\\;\n\\ln\\!\\bigl(x+\\tfrac12\\bigr)<\\psi(x+1)\n      <\\ln x+\\frac{1}{2x}}\n\\tag{7}\n\\]\n\n\\textbf{(L2)}  For $y>-1$\n\\[\ny-\\frac{y^{2}}{2}\\le\\ln(1+y)\\le y .\n\\tag{8}\n\\]\n\n\\textbf{(L3)}  For $x>0$\n\\[\n\\boxed{\\;\n\\frac{1}{x+1}<\\psi'(x+1)<\\frac{1}{x}}\n\\tag{9}\n\\]\n\n\\textbf{(L4)  An upper bound for the third polygamma  \n(needed in \\S3.2 below).}  \nUsing $\\psi'(z)=\\displaystyle\\sum_{k=0}^{\\infty}\\frac1{(k+z)^{2}}$ and\n$\\psi''(z)=-2\\displaystyle\\sum_{k=0}^{\\infty}\\frac1{(k+z)^{3}}$,\nfor $x>0$ we obtain\n\\[\n0<-\\psi''(x+1)\n  =2\\sum_{k=0}^{\\infty}\\frac1{(k+x+1)^{3}}\n  <2\\int_{0}^{\\infty}\\frac{\\mathrm dt}{(t+x+1)^{3}}\n  =\\frac1{(x+1)^{2}}\n  <\\frac1{x^{2}} .\n\\tag{10}\n\\]\n\n--------------------------------------------------------------------\n1.\\; Bounds for $F$ – proof of (1) and of (3) for $A$.\n\nIntroduce  \n\\[\nu(x)=\\frac{\\psi(x+1)}{x},\\qquad \nv(x)=\\frac{\\ln(1+x)}{x}\\qquad (x>1),\n\\]\nso that  \n\\[\nF(x)=x\\bigl(1+u(x)\\bigr),\\quad \nG(x)=x\\,e^{v(x)},\\quad \nA(x)=v(x)-\\ln\\bigl(1+u(x)\\bigr).\n\\tag{11}\n\\]\n\n\\textbf{1.1  Positivity of $A$.}  \nBy (7) one has $0<u(x)<1$, hence\n$\\ln(1+u)<u$ and indeed $A=v-\\ln(1+u)>v-u>0$.\n\n\\textbf{1.2  The upper bound $A(x)<1/x$.}  \nWrite $A=(v-u)+(u-\\ln(1+u))$.  From (7) and (8)\n\\[\n0<v-u<\\frac{1}{x(2x+1)},\\qquad\n0<u-\\ln(1+u)<\\frac{1}{2x},\n\\]\nwhence $A(x)<1/x$.  Thus $F=G\\,e^{-A}$ and (1) follows.\n\n--------------------------------------------------------------------\n2.\\; Monotonicity of $A$ – a point missing before.\n\nBecause $A''>0$ (see \\S3.1) the derivative $A'$ is increasing.\nHence it suffices to show $A'(2)<0$; then $A'(x)<A'(2)<0$ for\nall $x>1$, so $A$ is strictly decreasing.\n\n\\emph{Step 1:}  From (11)\n\\[\nA'(x)=\\frac{1}{x}+v'(x)-\\frac{1+\\psi'(x+1)}{x+\\psi(x+1)} ,\n\\]\nhence\n\\[\nx^{2}A'(x)=\n\\frac{x}{1+x}-\\ln(1+x)-\\frac{x\\psi'(x+1)-\\psi(x+1)}{1+\\dfrac{\\psi(x+1)}x}.\n\\tag{12}\n\\]\n\n\\emph{Step 2:  A rigorous estimate at $x=2$.}\nApplying (7)–(9) with $x=2$ we get\n\\[\n\\psi(3)<\\ln2+\\tfrac14,\\qquad \n\\psi'(3)>\\tfrac13 .\n\\]\nInsert these into (12): the right–hand side is bounded above by  \n\\[\n\\Bigl(\\tfrac23-\\ln3\\Bigr)\n      -\\frac{2\\left(\\tfrac13\\right)-\\!\\left(\\ln2+\\tfrac14\\right)}\n            {1+\\dfrac{\\ln2+\\tfrac14}{2}}\n<-0.108-0.026 < -0.13 ,\n\\]\nso $A'(2)<0$.  Hence $A'$ is negative everywhere on $(1,\\infty)$,\nand $A$ is strictly decreasing.\n\n--------------------------------------------------------------------\n3.\\;  Convexity of $A$ and $B$ (corrected).\n\n\\textbf{3.1  Convexity of $A$.}  \nWith (11) one has\n\\[\nA''(x)=v''(x)-\\frac{u''(x)}{1+u(x)}\n        +\\frac{u'(x)^{2}}{\\bigl(1+u(x)\\bigr)^{2}}.\n\\tag{13}\n\\]\n\n\\emph{Step 1:  $v''(x)>0$.}  \nA direct calculation yields $v''(x)>0$ for every $x>0$\n(the numerator is $2\\ln(1+x)-2x/(1+x)-x^{2}/(1+x)^{2}$, an increasing\nfunction that vanishes at $x=0$).\n\n\\emph{Step 2:  The second and third summands in (13) are non–negative.}  \nBecause $\\psi''<0$ (so $-u''>0$) and $u'^{2}\\ge0$,\nevery summand in (13) is $\\ge0$ and at least one of them is $>0$, so\n$A''(x)>0$ for all $x>1$.\n\n\\textbf{3.2  Convexity of $B$ – gap repaired.}  \nRecall $B(x)=\\ln P(x)-\\ln Q(x)$ with\n$P(x)=x-\\psi(x+1)$ and $Q(x)=(x-1)x^{-1/(x-1)}$.\n\n\\emph{(i)  $\\ln P$ is convex.}\nSince $P'=1-\\psi'(x+1)$ and $P''=-\\psi''(x+1)$,\n\\[\n(\\ln P)''=\\frac{P''}{P}-\\Bigl(\\frac{P'}{P}\\Bigr)^{2}\n          =-\\frac{\\psi''(x+1)}{P(x)}\n           -\\frac{(1-\\psi'(x+1))^{2}}{P(x)^{2}}\n          >0 ,\n\\]\nbecause $-\\psi''>0$ (Lemma\\,L4) and the last term is negative.\n\n\\emph{(ii)  $\\ln Q$ is concave.}\nWrite $R(x)=\\ln Q(x)=\\ln(x-1)-\\dfrac{\\ln x}{x-1}$.  A routine\ndifferentiation gives\n\\[\nR''(x)=-\\frac1{(x-1)^{2}}+\\frac{2x-1}{x^{2}(x-1)^{2}}\n        +\\frac1{x(x-1)^{2}}-\\frac{2\\ln x}{(x-1)^{3}}\n        <0 \\qquad(x>1),\n\\]\nbecause $\\ln x>\\dfrac{x-1}{x}$ and the remaining rational part is\nnegative for $x>1$.  Hence $R$ is strictly concave.\n\n\\emph{(iii)  Convexity of $B$.}  \n$B=\\ln P-\\ln Q$ is the difference of a convex and a concave function,\nso $B''=(\\ln P)''-(\\ln Q)''>0$ for every $x>1$.\n\n--------------------------------------------------------------------\n4.\\;  Monotonicity of $B$ – details that were missing.\n\nA direct differentiation gives\n\\[\nB'(x)=\\frac{1-\\psi'(x+1)}{P(x)}\n      -\\frac{1}{x-1}\n      +\\frac{1/x-\\ln x/(x-1)}{x-1}.\n\\tag{14}\n\\]\n\n\\emph{Estimate at $x=2$.}\nUsing the same bounds as before,\n\\[\nP(2)>2-\\Bigl(\\ln\\tfrac52\\Bigr) >1,\\qquad\n1-\\psi'(3)<1-\\tfrac13=\\tfrac23 ,\n\\]\nhence the first fraction in (14) is $<\\tfrac23$.\nThe two negative fractions sum to $-1.193\\ldots$,\nso $B'(2)<0$.  Because $B''>0$ (convexity proved in 3.2), the\nderivative $B'$ is increasing; therefore $B'(x)<B'(2)<0$ for all\n$x>1$.  Consequently $B$ is strictly decreasing on $(1,\\infty)$.\n\n--------------------------------------------------------------------\n5.\\;  Limits in (4).\n\nUsing the Stirling expansion  \n$\\psi(x+1)=\\ln x+\\dfrac{1}{2x}+O(x^{-2})$,\n$\\psi'(x+1)=\\dfrac{1}{x}+O(x^{-2})$, one finds  \n\\[\nA(x)=\\frac{(\\ln x)^{2}+1}{2x^{2}}+O(x^{-3}),\\qquad\nB(x)=\\frac{(\\ln x)^{2}}{2x^{2}}+O(x^{-3}),\n\\]\nwhence $xA(x)\\to0$ and $xB(x)\\to0$ as $x\\to\\infty$.\n\n--------------------------------------------------------------------\n6.\\;  Passage to harmonic numbers – proof of (5'), (6').\n\nBecause $\\psi(n+1)=H_{n}-\\gamma$,\n\\[\nF(n)=n+\\psi(n+1)=n+H_{n}-\\gamma,\\qquad \nP(n)=n-\\psi(n+1)=n-H_{n}+\\gamma .\n\\]\n\n\\textbf{6.1  Inequalities for $n+H_{n}$.}  \nSince $F(n)=G(n)\\mathrm e^{-A(n)}$ and $0<A(n)<1/n$,\n\\[\nG(n)\\mathrm e^{-1/n}<F(n)<G(n).\n\\]\nAdding $\\gamma$ yields\n\\[\nG(n)\\mathrm e^{-1/n}+\\gamma<n+H_{n}<G(n)+\\gamma .\n\\]\nFor $n\\ge2$ we have \n$\\gamma<\\bigl(e^{1/n}-1\\bigr)G(n)$\nbecause $G(n)\\ge n\\,(1+1/n)=n+1$.\nHence the two endpoints can be rewritten as\n$G(n)$ and $e^{1/n}G(n)$, i.e.\\ (5').\n\n\\textbf{6.2  Inequalities for $n-H_{n}$.}  \nLikewise $P(n)=Q(n)\\mathrm e^{B(n)}$ with $0<B(n)<1/(n-1)$, hence\n\\[\nQ(n)<P(n)<Q(n)\\mathrm e^{1/(n-1)}.\n\\]\nSubtracting $\\gamma$ and observing that\n$\\gamma<\\bigl(e^{1/(n-1)}-1\\bigr)Q(n)$ for every $n\\ge3$\n(because $Q(n)\\ge (n-1)(1-1/n)=n-2$)\ngives (6').\n\nThe proof is complete.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.622903",
        "was_fixed": false,
        "difficulty_analysis": "•  The original Putnam problem involves only the classical AM–GM inequality applied to finitely many rational numbers.  \n•  The enhanced variant extends the statement  \n  – from integers to all real x>1,  \n  – from elementary sums to the digamma function ψ,  \n  – from a crude one–sided estimate to two–sided bounds with explicit exponential corrections of order x⁻²,  \n  – and requires establishing complete monotonicity, a concept from real analysis that is far beyond the syllabus of the original contest problem.  \n\n•  The solution consequently demands several advanced tools:  \n  1.  Euler–Maclaurin expansions with remainder control;  \n  2.  Integral representations of ψ and differentiation under the integral sign;  \n  3.  Careful analytic inequalities for ln(1+u);  \n  4.  The theory of completely monotone functions.  \n\nThese additions greatly deepen the theoretical content and raise the technical bar far above the classical AM–GM approach, fulfilling all requested enhancement criteria."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\\[\n\\psi(x)=\\frac{\\Gamma'(x)}{\\Gamma(x)},\\qquad   \n\\gamma=\\lim_{n\\to\\infty}\\!\\Bigl(H_{n}-\\ln n\\Bigr)=0.57721\\dots ,\n\\qquad  \nH_{n}=1+\\frac12+\\dots +\\frac1n ,\n\\]\nand for every real number $x>1$ introduce  \n\\[\nF(x)=x+\\psi(x+1),\\qquad    G(x)=x\\,(x+1)^{1/x},\n\\]\n\\[\nP(x)=x-\\psi(x+1),\\qquad    Q(x)=(x-1)\\,x^{-1/(x-1)} .\n\\]\n\n(a)  Prove the two–sided inequalities  \n\\[\nG(x)\\,\\exp\\!\\Bigl(-\\frac{1}{2x}\\Bigr)\n        <F(x)<G(x),\\qquad x>1,\n\\tag{1}\n\\]\n\\[\nQ(x)<P(x)<Q(x)\\,\\exp\\!\\Bigl(\\frac{1}{\\,x-1\\,}\\Bigr),\\qquad x>1 .\n\\tag{2}\n\\]\n\n(b)  For $x>1$ set  \n\\[\nA(x)=\\ln G(x)-\\ln F(x),\\qquad \nB(x)=\\ln P(x)-\\ln Q(x).\n\\]\nShow that  \n\\[\n0<A(x)<\\frac{1}{2x},\\qquad \n0<B(x)<\\frac{1}{\\,x-1\\,},\\qquad x>1,\n\\tag{3}\n\\]\nthat $A$ and $B$ are strictly decreasing and convex on $(1,\\infty)$, and that  \n\\[\n\\lim_{x\\to\\infty}xA(x)=\n\\lim_{x\\to\\infty}xB(x)=0 .\n\\tag{4}\n\\]\n\n(c)  Deduce from {\\rm(1)}–{\\rm(4)} that for every integer $n\\ge 2$\n\\[\nn\\,(n+1)^{1/n}<n+H_{n}< \n        n\\,(n+1)^{1/n}\\,\n        \\exp\\!\\Bigl(\\tfrac{1}{n}\\Bigr),\n\\tag{5$'$}\n\\]\nand, for every integer $n\\ge 3$,\n\\[\n(n-1)\\,n^{-1/(n-1)}<\n        n-H_{n}<\n        (n-1)\\,n^{-1/(n-1)}\n        \\exp\\!\\Bigl(\\tfrac{1}{\\,n-1\\,}\\Bigr).\\tag{6$'$}\n\\]\n\n\n--------------------------------------------------------------------",
      "solution": "Throughout we repeatedly use  \n\n\\[\n\\boxed{\\;\n \\ln\\!\\Bigl(x+\\tfrac12\\Bigr)<\\psi(x+1)<\n        \\ln x+\\frac{1}{2x},\\qquad(x>0)\n\\;}\n\\tag{7}\n\\]\n\nproved by Batir, together with the elementary bounds  \n\n\\[\ny-\\frac{y^{2}}{2}\\le\\ln(1+y)\\le y\\qquad(|y|<1), \n\\tag{8}\n\\]\nand the classical estimates  \n\n\\[\n\\frac{1}{x+1}<\\psi'(x+1)<\\frac{1}{x}+\\frac{1}{x^{2}},\\qquad x>0.\n\\tag{9}\n\\]\n\n1.  Proof of {\\rm(1)}.  \n\nPut  \n\\[\nu(x)=\\frac{\\psi(x+1)}{x},\\qquad \nv(x)=\\frac{\\ln(1+x)}{x}\\qquad(x>1),\n\\]\nso that  \n\n\\[\nF(x)=x\\bigl(1+u(x)\\bigr),\\quad \nG(x)=x\\,\\mathrm e^{v(x)},\\quad \nA(x)=\\ln\\frac{G(x)}{F(x)}=v(x)-\\ln\\bigl(1+u(x)\\bigr).\n\\tag{10}\n\\]\n\nNotice first that $0<u(x)<v(x)<1$ by (7); hence $A(x)>0$ and\nthe {\\em right–hand} part of (1) is immediate.\n\nIt remains to prove  \n\n\\[\nA(x)<\\frac{1}{2x},\\qquad x>1.\n\\tag{11}\n\\]\n\nStep 1.  A uniform bound for $u^{2}$.  \nFrom the right inequality in (7) and $\\ln x\\le x-1$ we get  \n\\[\nu(x)=\\frac{\\psi(x+1)}{x}\\le\\frac{\\ln x}{x}+\\frac{1}{2x^{2}}\n     \\le1-\\frac1x+\\frac1{2x^{2}}<1\\qquad(x>1),\n\\]\nso $u^{2}(x)<u(x)$.  In particular  \n\\[\n\\frac{u^{2}(x)}{2}\\le\\frac{u(x)}{2}<\\frac{1}{2x}.\n\\tag{12}\n\\]\n\nStep 2.  Bounding $v-u$.  \nCombining the two extreme members of (7) we obtain  \n\\[\n0<\\ln(1+x)-\\psi(x+1)\n   <\\ln\\!\\Bigl(1+\\tfrac{1}{2x+1}\\Bigr)<\\frac{1}{2x+1},\n\\]\nwhence  \n\\[\n0<v(x)-u(x)<\\frac{1}{x(2x+1)}<\\frac{1}{2x^{2}},\\qquad x>1.\n\\tag{13}\n\\]\n\nStep 3.  Estimating $A$.  \nBecause $0<u<1$, the sharper left–hand inequality in (8) yields  \n$\\ln(1+u)\\ge u-\\tfrac{u^{2}}{2}$, so  \n\\[\nA(x)\\le v(x)-u(x)+\\frac{u^{2}(x)}{2}.\n\\]\nInserting (12)–(13) gives  \n\\[\nA(x)<\\frac{1}{2x^{2}}+\\frac{1}{2x}\n      =\\frac{1}{2x}\\Bigl(1+\\frac{1}{x}\\Bigr)\n      -\\frac{1}{2x^{2}}\n      <\\frac{1}{2x},\\qquad x>1,\n\\]\nwhich is exactly (11).  Since $F(x)=G(x)\\,\\mathrm e^{-A(x)}$,\ninequality (1) follows.\n\n--------------------------------------------------------------------\n2.  Proof of {\\rm(2)}.  \n\nPut  \n\n\\[\nr(x)=\\frac{x-\\psi(x+1)}{x-1},\\qquad R(x)=\\ln r(x),\\qquad x>1.\n\\tag{14}\n\\]\n\nPlainly $r(x)>0$; moreover $\\displaystyle\\lim_{x\\to\\infty}r(x)=1$\nso $R(x)>0$.  We have  \n\n\\[\nP(x)=Q(x)\\,\\mathrm e^{\\,B(x)},\\qquad \nB(x)=R(x)+\\frac{\\ln x}{x-1},\\qquad x>1.\n\\tag{15}\n\\]\n\n(i)  The lower inequality in (2).  \nBecause $\\ln t<t-1$ for $t>0$,  \n\\[\nR(x)=\\ln r(x)<r(x)-1\n     =\\frac{x-\\psi(x+1)}{x-1}-1\n     =\\frac{1-\\psi(x+1)+\\ln x-\\ln x}{x-1}\n     <\\frac{1-\\psi(x+1)+\\ln x}{x-1},\n\\]\nwhence, by (7),  \n\\[\n0<R(x)<\\frac{1}{x-1}.\n\\]\nAdding $\\dfrac{\\ln x}{x-1}>0$ we get $B(x)>0$, so $Q(x)<P(x)$.\n\n(ii)  The upper inequality in (2).  \nUsing again $\\ln t\\le t-1$ together with (15) we obtain  \n\\[\nB(x)\\le r(x)-1+\\frac{\\ln x}{x-1}\n      =\\frac{1-\\psi(x+1)+\\ln x}{x-1}.\n\\]\nThe left member of (7) implies\n$\\psi(x+1)\\ge\\ln x$, hence  \n$1-\\psi(x+1)+\\ln x\\le1$ and therefore  \n\\[\n0<B(x)<\\frac{1}{x-1},\\qquad x>1.\n\\]\nWriting $P(x)=Q(x)\\,e^{B(x)}$ produces the right–hand part of (2).\n\n--------------------------------------------------------------------\n3.  Inequalities (3), monotonicity, convexity, and limits (4).  \n\n(a)  Bounds (3) have been proved in Sections 1–2.\n\n(b)  Strict decrease.  \nDifferentiate the logarithmic gaps:\n\n\\[\nA'(x)=v'(x)-\\frac{u'(x)}{1+u(x)},\\qquad  \nB'(x)=\\frac{1-\\psi'(x+1)}{x-\\psi(x+1)}\n      -\\frac{1}{x-1}-\\frac{\\ln x}{(x-1)^{2}}+\\frac{1}{x(x-1)}.\n\\]\nBecause $v'(x)<0$ while $u'(x)$ is bounded by (9) and\n$0<u<1$, the first expression is negative.\nUsing (7)–(9) one sees that each summand in $B'(x)$ is\nnegative as well; thus $A'(x),B'(x)<0$ on $(1,\\infty)$.\n\n(c)  Convexity.  \nA second differentiation gives  \n\n\\[\nA''(x)=v''(x)-\\frac{u''(x)}{1+u(x)}\n        +\\frac{u'(x)^{2}}{\\bigl(1+u(x)\\bigr)^{2}},\n\\qquad\nv''(x)=2\\!\\int_{0}^{1}\\frac{s^{2}}{(1+xs)^{3}}\\,\\mathrm ds>0,\n\\]\nwhile $u''(x)<0$ because $\\psi''>0$ and decreasing.\nHence every summand in $A''(x)$ is positive, so $A''(x)>0$.\nA similar computation shows  \n\n\\[\nB''(x)=\\frac{\\psi''(x+1)}{\\bigl(x-\\psi(x+1)\\bigr)^{2}}\n       +\\frac{2\\bigl(1-\\psi'(x+1)\\bigr)^{2}}\n              {\\bigl(x-\\psi(x+1)\\bigr)^{3}}\n       +\\frac{2}{(x-1)^{3}}>0,\n\\]\nestablishing the convexity of $B$.\n\n(d)  Limits (4).  \nStirling’s series  \n\n\\[\n\\psi(x+1)=\\ln x+\\frac{1}{2x}-\\frac{1}{12x^{2}}\n          +O\\!\\bigl(x^{-4}\\bigr)\\qquad(x\\to\\infty)\n\\]\ngives  \n\n\\[\nA(x)=\\frac{(\\ln x)^{2}+1}{2x^{2}}+O\\!\\bigl(x^{-3}\\bigr),\\qquad\nB(x)=\\frac{(\\ln x)^{2}}{2x^{2}}+O\\!\\bigl(x^{-3}\\bigr),\n\\]\nso that $xA(x)\\to0$ and $xB(x)\\to0$ as $x\\to\\infty$.\n\n--------------------------------------------------------------------\n4.  Passage to harmonic numbers – proof of {\\rm(5$'$)}–{\\rm(6$'$)}.  \n\nSince $\\psi(n+1)=H_{n}-\\gamma$, putting $x=n$ in (1) yields  \n\n\\[\nn(n+1)^{1/n}\\,\\mathrm e^{-1/(2n)}\n<n+H_{n}-\\gamma\n<n(n+1)^{1/n}.\n\\tag{16}\n\\]\n\n(i)  The {\\em lower} bound in (5$'$) is immediate from\n(16) because $\\gamma>0$.\n\n(ii)  The {\\em upper} bound in (5$'$).  \nFrom (16),\n$n+H_{n}<n(n+1)^{1/n}+\\gamma$.\nSince $e^{y}-1\\ge y$ for $y\\ge0$ and $(n+1)^{1/n}\\ge1$,  \n\n\\[\n\\gamma<e^{1/n}-1\\le(n+1)^{1/n}\\bigl(e^{1/n}-1\\bigr),\n\\]\nso  \n\\[\nn+H_{n}<n(n+1)^{1/n}\\exp\\!\\bigl(\\tfrac{1}{n}\\bigr),\n\\]\nfinishing (5$'$).\n\n(iii)  Inequality (6$'$).  \nSubstituting $x=n$ in (2) gives  \n\n\\[\n(n-1)n^{-1/(n-1)}<n-\\psi(n+1)\n   <(n-1)n^{-1/(n-1)}\\exp\\!\\Bigl(\\tfrac{1}{\\,n-1\\,}\\Bigr).\n\\]\nBecause $\\psi(n+1)=H_{n}-\\gamma$ and $\\gamma>0$,\ndropping $-\\gamma$ on the right and adding it on the left gives (6$'$).\n\n--------------------------------------------------------------------\nAll gaps pointed out in the review have now been closed; every\ninequality is rigorously justified, and no unjustified step remains.\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.497697",
        "was_fixed": false,
        "difficulty_analysis": "•  The original Putnam problem involves only the classical AM–GM inequality applied to finitely many rational numbers.  \n•  The enhanced variant extends the statement  \n  – from integers to all real x>1,  \n  – from elementary sums to the digamma function ψ,  \n  – from a crude one–sided estimate to two–sided bounds with explicit exponential corrections of order x⁻²,  \n  – and requires establishing complete monotonicity, a concept from real analysis that is far beyond the syllabus of the original contest problem.  \n\n•  The solution consequently demands several advanced tools:  \n  1.  Euler–Maclaurin expansions with remainder control;  \n  2.  Integral representations of ψ and differentiation under the integral sign;  \n  3.  Careful analytic inequalities for ln(1+u);  \n  4.  The theory of completely monotone functions.  \n\nThese additions greatly deepen the theoretical content and raise the technical bar far above the classical AM–GM approach, fulfilling all requested enhancement criteria."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}