{
  "index": "1954-B-7",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "7. Show that\n\\[\n\\lim _{n \\rightarrow \\infty} \\sum_{s=1}^{n}\\left(\\frac{a+s}{n}\\right)^{n} \\quad(a>0)\n\\]\nlies between \\( e^{a} \\) and \\( e^{a+1} \\).",
  "solution": "Solution. We shall evaluate the limit exactly using the facts:\n\\[\n\\left(1+\\frac{x}{n}\\right)^{n} \\leq e^{x}\n\\]\nif \\( n>0 \\) and \\( 1+x / n>0 \\), and\n\\[\n\\lim _{n \\rightarrow \\infty}\\left(1+\\frac{x}{n}\\right)^{n}=e^{x}\n\\]\nfor all real \\( x \\).\nLet\n\\[\nS_{n}=\\sum_{s=1}^{n}\\left(\\frac{a+s}{n}\\right)^{n}=\\sum_{r=0}^{\\prime-1}\\left(1+\\frac{a-r}{n}\\right)^{n} .\n\\]\n\nRecalling that \\( a>0 \\) and using (1), we have\n\\[\nS_{n} \\leq \\sum_{r=0}^{\\prime-1} \\exp (a-r)<\\sum_{r=0}^{\\infty} \\exp (a-r)=\\frac{e^{a+1}}{e-1} .\n\\]\n\nTherefore\n\\[\n\\limsup _{n-\\infty} S_{n} \\leq \\frac{e^{u+1}}{e-1} .\n\\]\n\nAlso, for fixed \\( k \\) and \\( n>k \\)\n\\[\nS_{n} \\geq \\sum_{r=0}^{k}\\left(1+\\frac{a-r}{n}\\right)^{\\prime \\prime} .\n\\]\n\nThe limit on the right exists as \\( n \\rightarrow \\infty \\) by (2), so\n\\[\n\\liminf _{n-\\infty} S_{n} \\geq \\sum_{r=0}^{k} \\exp (a-r) .\n\\]\n\nSince \\( k \\) is arbitrary, we get\n\\[\n\\liminf _{n-\\infty} S_{n} \\geq \\sum_{r=0}^{\\infty} \\exp (a-r)=\\frac{e^{a+1}}{e-\\frac{1}{1}} .\n\\]\n\nComparing (3) and (4), we obtain\n\\[\n\\lim S_{n}=\\frac{e^{a+1}}{e-1}\n\\]\n\nEvidently\n\\[\n1<\\frac{e}{e-1}<e, \\quad \\text { so } \\quad e^{a}<\\lim S_{n}<e^{a+1},\n\\]\nas required.\nRemarks. The inequality (1) follows immediately from the well-known inequality \\( \\log (1+y) \\leq y \\), while (2), even for complex values of \\( x \\), follows from the fact that \\( \\log (1+y) \\) has a power series representation convergent for \\( |y|<1 \\).",
  "vars": [
    "n",
    "s",
    "x",
    "r",
    "k",
    "u",
    "y",
    "S_n"
  ],
  "params": [
    "a"
  ],
  "sci_consts": [
    "e"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "n": "indexn",
        "s": "stepvar",
        "x": "inputx",
        "r": "rindex",
        "k": "kindex",
        "u": "ualias",
        "y": "yvalue",
        "S_n": "sumtotal",
        "a": "offseta"
      },
      "question": "7. Show that\n\\[\n\\lim _{indexn \\rightarrow \\infty} \\sum_{stepvar=1}^{indexn}\\left(\\frac{offseta+stepvar}{indexn}\\right)^{indexn} \\quad(offseta>0)\n\\]\nlies between \\( e^{offseta} \\) and \\( e^{offseta+1} \\).",
      "solution": "Solution. We shall evaluate the limit exactly using the facts:\n\\[\n\\left(1+\\frac{inputx}{indexn}\\right)^{indexn} \\leq e^{inputx}\n\\]\nif \\( indexn>0 \\) and \\( 1+inputx / indexn>0 \\), and\n\\[\n\\lim _{indexn \\rightarrow \\infty}\\left(1+\\frac{inputx}{indexn}\\right)^{indexn}=e^{inputx}\n\\]\nfor all real \\( inputx \\).\nLet\n\\[\nsumtotal=\\sum_{stepvar=1}^{indexn}\\left(\\frac{offseta+stepvar}{indexn}\\right)^{indexn}=\\sum_{rindex=0}^{\\prime-1}\\left(1+\\frac{offseta-rindex}{indexn}\\right)^{indexn} .\n\\]\n\nRecalling that \\( offseta>0 \\) and using (1), we have\n\\[\nsumtotal \\leq \\sum_{rindex=0}^{\\prime-1} \\exp (offseta-rindex)<\\sum_{rindex=0}^{\\infty} \\exp (offseta-rindex)=\\frac{e^{offseta+1}}{e-1} .\n\\]\n\nTherefore\n\\[\n\\limsup _{indexn-\\infty} sumtotal \\leq \\frac{e^{ualias+1}}{e-1} .\n\\]\n\nAlso, for fixed \\( kindex \\) and \\( indexn>kindex \\)\n\\[\nsumtotal \\geq \\sum_{rindex=0}^{kindex}\\left(1+\\frac{offseta-rindex}{indexn}\\right)^{\\prime \\prime} .\n\\]\n\nThe limit on the right exists as \\( indexn \\rightarrow \\infty \\) by (2), so\n\\[\n\\liminf _{indexn-\\infty} sumtotal \\geq \\sum_{rindex=0}^{kindex} \\exp (offseta-rindex) .\n\\]\n\nSince \\( kindex \\) is arbitrary, we get\n\\[\n\\liminf _{indexn-\\infty} sumtotal \\geq \\sum_{rindex=0}^{\\infty} \\exp (offseta-rindex)=\\frac{e^{offseta+1}}{e-\\frac{1}{1}} .\n\\]\n\nComparing (3) and (4), we obtain\n\\[\n\\lim sumtotal=\\frac{e^{offseta+1}}{e-1}\n\\]\n\nEvidently\n\\[\n1<\\frac{e}{e-1}<e, \\quad \\text { so } \\quad e^{offseta}<\\lim sumtotal<e^{offseta+1},\n\\]\nas required.\nRemarks. The inequality (1) follows immediately from the well-known inequality \\( \\log (1+yvalue) \\leq yvalue \\), while (2), even for complex values of \\( inputx \\), follows from the fact that \\( \\log (1+yvalue) \\) has a power series representation convergent for \\( |yvalue|<1 \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "n": "marshmallow",
        "s": "raincloud",
        "x": "butterscotch",
        "r": "goldfinch",
        "k": "seashells",
        "u": "paperback",
        "y": "blueberry",
        "S_n": "peppermint",
        "a": "sunflower"
      },
      "question": "\\[\n\\lim _{marshmallow \\rightarrow \\infty} \\sum_{raincloud=1}^{marshmallow}\\left(\\frac{sunflower+raincloud}{marshmallow}\\right)^{marshmallow} \\quad(sunflower>0)\n\\]",
      "solution": "Solution. We shall evaluate the limit exactly using the facts:\n\\[\n\\left(1+\\frac{butterscotch}{marshmallow}\\right)^{marshmallow} \\leq e^{butterscotch}\n\\]\nif \\( marshmallow>0 \\) and \\( 1+butterscotch / marshmallow>0 \\), and\n\\[\n\\lim _{marshmallow \\rightarrow \\infty}\\left(1+\\frac{butterscotch}{marshmallow}\\right)^{marshmallow}=e^{butterscotch}\n\\]\nfor all real \\( butterscotch \\).\nLet\n\\[\npeppermint=\\sum_{raincloud=1}^{marshmallow}\\left(\\frac{sunflower+raincloud}{marshmallow}\\right)^{marshmallow}=\\sum_{goldfinch=0}^{\\prime-1}\\left(1+\\frac{sunflower-goldfinch}{marshmallow}\\right)^{marshmallow} .\n\\]\n\nRecalling that \\( sunflower>0 \\) and using (1), we have\n\\[\npeppermint \\leq \\sum_{goldfinch=0}^{\\prime-1} \\exp (sunflower-goldfinch)<\\sum_{goldfinch=0}^{\\infty} \\exp (sunflower-goldfinch)=\\frac{e^{sunflower+1}}{e-1} .\n\\]\n\nTherefore\n\\[\n\\limsup _{marshmallow-\\infty} peppermint \\leq \\frac{e^{paperback+1}}{e-1} .\n\\]\n\nAlso, for fixed \\( seashells \\) and \\( marshmallow>seashells \\)\n\\[\npeppermint \\geq \\sum_{goldfinch=0}^{seashells}\\left(1+\\frac{sunflower-goldfinch}{marshmallow}\\right)^{\\prime \\prime} .\n\\]\n\nThe limit on the right exists as \\( marshmallow \\rightarrow \\infty \\) by (2), so\n\\[\n\\liminf _{marshmallow-\\infty} peppermint \\geq \\sum_{goldfinch=0}^{seashells} \\exp (sunflower-goldfinch) .\n\\]\n\nSince \\( seashells \\) is arbitrary, we get\n\\[\n\\liminf _{marshmallow-\\infty} peppermint \\geq \\sum_{goldfinch=0}^{\\infty} \\exp (sunflower-goldfinch)=\\frac{e^{sunflower+1}}{e-\\frac{1}{1}} .\n\\]\n\nComparing (3) and (4), we obtain\n\\[\n\\lim peppermint=\\frac{e^{sunflower+1}}{e-1}\n\\]\n\nEvidently\n\\[\n1<\\frac{e}{e-1}<e, \\quad \\text { so } \\quad e^{sunflower}<\\lim peppermint<e^{sunflower+1},\n\\]\nas required.\n\nRemarks. The inequality (1) follows immediately from the well-known inequality \\( \\log (1+blueberry) \\leq blueberry \\), while (2), even for complex values of \\( butterscotch \\), follows from the fact that \\( \\log (1+blueberry) \\) has a power series representation convergent for \\( |blueberry|<1 \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "n": "tinycount",
        "s": "maxindex",
        "x": "fixedvalue",
        "r": "advanceindex",
        "k": "unboundedindex",
        "u": "clarityvar",
        "y": "resultvalue",
        "S_n": "emptysum",
        "a": "negativeconstant"
      },
      "question": "7. Show that\n\\[\n\\lim _{tinycount \\rightarrow \\infty} \\sum_{maxindex=1}^{tinycount}\\left(\\frac{negativeconstant+maxindex}{tinycount}\\right)^{tinycount} \\quad(negativeconstant>0)\n\\]\nlies between \\( e^{negativeconstant} \\) and \\( e^{negativeconstant+1} \\).",
      "solution": "Solution. We shall evaluate the limit exactly using the facts:\n\\[\n\\left(1+\\frac{fixedvalue}{tinycount}\\right)^{tinycount} \\leq e^{fixedvalue}\n\\]\nif \\( tinycount>0 \\) and \\( 1+fixedvalue / tinycount>0 \\), and\n\\[\n\\lim _{tinycount \\rightarrow \\infty}\\left(1+\\frac{fixedvalue}{tinycount}\\right)^{tinycount}=e^{fixedvalue}\n\\]\nfor all real \\( fixedvalue \\).\nLet\n\\[\nemptysum=\\sum_{maxindex=1}^{tinycount}\\left(\\frac{negativeconstant+maxindex}{tinycount}\\right)^{tinycount}=\\sum_{advanceindex=0}^{\\prime-1}\\left(1+\\frac{negativeconstant-advanceindex}{tinycount}\\right)^{tinycount} .\n\\]\n\nRecalling that \\( negativeconstant>0 \\) and using (1), we have\n\\[\nemptysum \\leq \\sum_{advanceindex=0}^{\\prime-1} \\exp (negativeconstant-advanceindex)<\\sum_{advanceindex=0}^{\\infty} \\exp (negativeconstant-advanceindex)=\\frac{e^{negativeconstant+1}}{e-1} .\n\\]\n\nTherefore\n\\[\n\\limsup _{tinycount-\\infty} emptysum \\leq \\frac{e^{clarityvar+1}}{e-1} .\n\\]\n\nAlso, for fixed \\( unboundedindex \\) and \\( tinycount>unboundedindex \\)\n\\[\nemptysum \\geq \\sum_{advanceindex=0}^{unboundedindex}\\left(1+\\frac{negativeconstant-advanceindex}{tinycount}\\right)^{\\prime \\prime} .\n\\]\n\nThe limit on the right exists as \\( tinycount \\rightarrow \\infty \\) by (2), so\n\\[\n\\liminf _{tinycount-\\infty} emptysum \\geq \\sum_{advanceindex=0}^{unboundedindex} \\exp (negativeconstant-advanceindex) .\n\\]\n\nSince \\( unboundedindex \\) is arbitrary, we get\n\\[\n\\liminf _{tinycount-\\infty} emptysum \\geq \\sum_{advanceindex=0}^{\\infty} \\exp (negativeconstant-advanceindex)=\\frac{e^{negativeconstant+1}}{e-\\frac{1}{1}} .\n\\]\n\nComparing (3) and (4), we obtain\n\\[\n\\lim emptysum=\\frac{e^{negativeconstant+1}}{e-1}\n\\]\n\nEvidently\n\\[\n1<\\frac{e}{e-1}<e, \\quad \\text { so } \\quad e^{negativeconstant}<\\lim emptysum<e^{negativeconstant+1},\n\\]\nas required.\nRemarks. The inequality (1) follows immediately from the well-known inequality \\( \\log (1+resultvalue) \\leq resultvalue \\), while (2), even for complex values of \\( fixedvalue \\), follows from the fact that \\( \\log (1+resultvalue) \\) has a power series representation convergent for \\( |resultvalue|<1 \\)."
    },
    "garbled_string": {
      "map": {
        "n": "qzxwvtnp",
        "s": "hjgrksla",
        "x": "cvblremq",
        "r": "opqsdjki",
        "k": "mvnrytua",
        "u": "wqplmzne",
        "y": "asdfghjk",
        "S_n": "fxghrjkl",
        "a": "zmxncwle"
      },
      "question": "\\[\n\\lim _{qzxwvtnp \\rightarrow \\infty} \\sum_{hjgrksla=1}^{qzxwvtnp}\\left(\\frac{zmxncwle+hjgrksla}{qzxwvtnp}\\right)^{qzxwvtnp} \\quad(zmxncwle>0)\n\\]",
      "solution": "Solution. We shall evaluate the limit exactly using the facts:\n\\[\n\\left(1+\\frac{cvblremq}{qzxwvtnp}\\right)^{qzxwvtnp} \\leq e^{cvblremq}\n\\]\nif \\( qzxwvtnp>0 \\) and \\( 1+cvblremq / qzxwvtnp>0 \\), and\n\\[\n\\lim _{qzxwvtnp \\rightarrow \\infty}\\left(1+\\frac{cvblremq}{qzxwvtnp}\\right)^{qzxwvtnp}=e^{cvblremq}\n\\]\nfor all real \\( cvblremq \\).\nLet\n\\[\nfxghrjkl=\\sum_{hjgrksla=1}^{qzxwvtnp}\\left(\\frac{zmxncwle+hjgrksla}{qzxwvtnp}\\right)^{qzxwvtnp}=\\sum_{opqsdjki=0}^{\\prime-1}\\left(1+\\frac{zmxncwle-opqsdjki}{qzxwvtnp}\\right)^{qzxwvtnp} .\n\\]\n\nRecalling that \\( zmxncwle>0 \\) and using (1), we have\n\\[\nfxghrjkl \\leq \\sum_{opqsdjki=0}^{\\prime-1} \\exp (zmxncwle-opqsdjki)<\\sum_{opqsdjki=0}^{\\infty} \\exp (zmxncwle-opqsdjki)=\\frac{e^{zmxncwle+1}}{e-1} .\n\\]\n\nTherefore\n\\[\n\\limsup _{qzxwvtnp-\\infty} fxghrjkl \\leq \\frac{e^{wqplmzne+1}}{e-1} .\n\\]\n\nAlso, for fixed \\( mvnrytua \\) and \\( qzxwvtnp>mvnrytua \\)\n\\[\nfxghrjkl \\geq \\sum_{opqsdjki=0}^{mvnrytua}\\left(1+\\frac{zmxncwle-opqsdjki}{qzxwvtnp}\\right)^{\\prime \\prime} .\n\\]\n\nThe limit on the right exists as \\( qzxwvtnp \\rightarrow \\infty \\) by (2), so\n\\[\n\\liminf _{qzxwvtnp-\\infty} fxghrjkl \\geq \\sum_{opqsdjki=0}^{mvnrytua} \\exp (zmxncwle-opqsdjki) .\n\\]\n\nSince \\( mvnrytua \\) is arbitrary, we get\n\\[\n\\liminf _{qzxwvtnp-\\infty} fxghrjkl \\geq \\sum_{opqsdjki=0}^{\\infty} \\exp (zmxncwle-opqsdjki)=\\frac{e^{zmxncwle+1}}{e-\\frac{1}{1}} .\n\\]\n\nComparing (3) and (4), we obtain\n\\[\n\\lim fxghrjkl=\\frac{e^{zmxncwle+1}}{e-1}\n\\]\n\nEvidently\n\\[\n1<\\frac{e}{e-1}<e, \\quad \\text { so } \\quad e^{zmxncwle}<\\lim fxghrjkl<e^{zmxncwle+1},\n\\]\nas required.\n\nRemarks. The inequality (1) follows immediately from the well-known inequality \\( \\log (1+asdfghjk) \\leq asdfghjk \\), while (2), even for complex values of \\( cvblremq \\), follows from the fact that \\( \\log (1+asdfghjk) \\) has a power series representation convergent for \\( |asdfghjk|<1 \\)."
    },
    "kernel_variant": {
      "question": "Let\n\\[\n\\Omega:=\\bigl\\{\\,z\\in\\mathbb{C}\\;:\\;\\operatorname{Re}z>0\\,\\bigr\\}.\n\\]\nFor every positive integer $n$ put $N:=2n-1$ and define\n\\[\nS_{n}\\colon\\Omega\\longrightarrow\\mathbb{C},\n\\qquad\nS_{n}(z):=\\sum_{s=0}^{N}\\Bigl(\\frac{z+s}{N}\\Bigr)^{\\! N}. \\tag{$\\star$}\n\\]\n\n1.\\;Prove that the sequence $(S_{n})_{n\\ge 1}$ converges uniformly on every compact subset of $\\Omega$ and denote its limit by $S$.\n\n2.\\;Show directly from your description of $S$ that $S$ extends to an entire function and find an explicit closed formula for $S$.\n\n3.\\;For every compact $K\\subset\\Omega$ prove that a constant $C_{K}>0$ exists such that\n\\[\n\\lvert S_{n}(z)-S(z)\\rvert\\le\\frac{C_{K}}{n}\\qquad (n\\ge 1,\\;z\\in K). \\tag{$\\dagger$}\n\\]\n\n4.\\;Deduce that for every $z$ with $\\operatorname{Re}z>0$ one has\n\\[\ne^{\\operatorname{Re}z}\\;<\\;\\lvert S(z)\\rvert\\;<\\;e^{\\operatorname{Re}z+1}. \\tag{$\\ddagger$}\n\\]",
      "solution": "Throughout let $K\\subset\\Omega$ be a fixed non-empty compact set and put\n\\[\n\\rho:=\\min_{z\\in K}\\operatorname{Re}z>0,\\qquad\n\\zeta:=\\max_{z\\in K}\\operatorname{Re}z,\\qquad\nM:=\\max_{z\\in K}\\lvert z\\rvert .\n\\]\nWrite $N:=2n-1$ (so $n=(N+1)/2$).  For every integer $0\\le s\\le N$ set\n$r:=N-s$ and abbreviate\n\\[\nf_{n,r}(z):=\\Bigl(1+\\frac{z-r}{N}\\Bigr)^{\\! N},\\qquad\nw:=z-r. \\tag{1}\n\\]\n\n\\textbf{Step 1 – Pointwise limit of the summands.}\nFor fixed $r$ the classical limit $(1+u/N)^{N}\\!\\to e^{u}$ ($N\\to\\infty$) gives\n\\[\n\\lim_{n\\to\\infty}f_{n,r}(z)=e^{\\,z-r}\\qquad(z\\in\\Omega,\\;r\\in\\mathbb{N}). \\tag{2}\n\\]\n\n\\textbf{Step 2 – A crude growth bound (only for the tail estimate).}\nFor $z\\in K$ and $0\\le s\\le N$\n\\[\n\\Bigl\\lvert\\frac{z+s}{N}\\Bigr\\rvert\n\\le\\frac{\\lvert z\\rvert+s}{N}\n\\le\\frac{M+N}{N}=1+\\frac{M}{N}\n\\le 1+M ,\n\\]\nhence\n\\[\n\\lvert f_{n,r}(z)\\rvert\\le(1+M)^{N}.\n\\]\n\n\\textbf{Step 3 – Uniform convergence on compacta (Problem 1).}\n\nFix a number $\\beta$ with $\\tfrac34<\\beta<1$ and split\n\\[\nH_{n}(z):=\\sum_{0\\le r\\le\\beta N}f_{n,r}(z),\\qquad\nT_{n}(z):=\\sum_{\\beta N< r\\le N}f_{n,r}(z).\n\\]\n\n\\emph{3(a) The tail is exponentially small.}\nWriting $r=N-s$, $0\\le s\\le(1-\\beta)N$, one obtains exactly the same geometric\nfactor as in the kernel problem; hence\n\\[\n\\sup_{z\\in K}\\lvert T_{n}(z)\\rvert\\;\\xrightarrow[n\\to\\infty]{}\\;0. \\tag{3}\n\\]\n\n\\emph{3(b) Quantitative control of the head.}\n\n\\smallskip\n\\underline{\\textbf{Lemma A (Uniform Euler estimate).}}\nFix $\\alpha$ with $0<\\alpha<1$.  Define\n\\[\nG(u):=\\frac{\\log(1+u)-u}{u^{2}},\\qquad G(0):=-\\tfrac12 .\n\\]\n$G$ is holomorphic on the disc $\\{\\,\\lvert u\\rvert<1\\,\\}$; put\n\\[\nM(\\alpha):=\\sup_{\\lvert u\\rvert\\le\\alpha}\\lvert G(u)\\rvert,\\qquad\nm(\\alpha):=-\\inf_{\\lvert u\\rvert\\le\\alpha}\\operatorname{Re}G(u)\\;>\\;0 .\n\\]\nLet $N\\ge 1$ and $w\\in\\mathbb{C}$ with $\\lvert w\\rvert\\le\\alpha N$.\nSet $u:=w/N$ and $\\xi:=N(\\log(1+u)-u)=G(u)\\,w^{2}/N$.  Then\n\\[\n\\lvert\\xi\\rvert\\le\\frac{M(\\alpha)\\lvert w\\rvert^{2}}{N},\n\\qquad\n\\operatorname{Re}\\xi\\le-\\frac{m(\\alpha)\\lvert w\\rvert^{2}}{N}\\le 0,\n\\]\nand\n\\[\n\\Bigl\\lvert(1+\\tfrac{w}{N})^{N}-e^{w}\\Bigr\\rvert\n\\le \\frac{M(\\alpha)\\lvert w\\rvert^{2}e^{\\operatorname{Re}w}}{N}. \\tag{4}\n\\]\n\n\\smallskip\n\\underline{\\textbf{Application.}}\nBecause $r\\le\\beta N$ and $z\\in K$,\n\\[\n\\lvert w\\rvert=\\lvert z-r\\rvert\\le M+\\beta N=: \\beta' N,\n\\qquad\n\\beta':=\\beta+\\frac{M}{N}<1\\quad(N\\text{ large}).\n\\]\nChoose $\\alpha$ with $\\beta<\\alpha<1$ and $N$ so large that $\\beta'\\le\\alpha$.\nLemma~A then gives\n\\[\n\\lvert f_{n,r}(z)-e^{\\,z-r}\\rvert\n\\le \\widetilde C_{K}\\,(M+r)^{2}\\,\n\\frac{e^{\\,\\rho-r}}{N},\n\\quad 0\\le r\\le \\beta N, \\tag{5}\n\\]\nwhere\n\\[\n\\widetilde C_{K}:=\\frac{M(\\alpha)}{1-\\alpha}\\,e^{\\,\\zeta-\\rho}.\n\\]\nNote that $\\widetilde C_{K}$ depends only on the fixed compact $K$ (through\n$\\rho,\\zeta,M$) and on the chosen $\\alpha$.\n\nSince\n\\[\n\\sum_{r=0}^{\\infty}(M+r)^{2}\\,e^{\\,\\rho-r}<\\infty,\n\\]\nthe Weierstrass $M$-test shows that the series\n\\[\nH(z):=\\sum_{r=0}^{\\infty}e^{\\,z-r} \\tag{6}\n\\]\nis the uniform limit of $H_{n}$ on $K$, and\n\\[\n\\sup_{z\\in K}\\lvert H_{n}(z)-H(z)\\rvert\n\\le\\frac{C_{1,K}}{N}\n=\\frac{2C_{1,K}}{N+1}\n\\le\\frac{C_{1,K}}{n}, \\tag{7}\n\\]\nfor a suitable constant $C_{1,K}>0$.\n\n\\emph{3(c) Conclusion of Step 3.}\nCombining (3) and (7) we obtain\n\\[\n\\sup_{z\\in K}\\lvert S_{n}(z)-H(z)\\rvert\n\\le\\frac{C_{2,K}}{n}\\qquad(n\\ge 1). \\tag{8}\n\\]\nHence $(S_{n})$ converges uniformly on $K$; we set\n\\[\nS(z):=H(z)\\qquad(z\\in\\Omega).\n\\]\nBecause $K$ was arbitrary, uniform convergence on every compact subset of\n$\\Omega$ is proved.\n\n\\textbf{Step 4 – Entire continuation and closed formula (Problem 2).}\n\nSince $\\lvert e^{\\,z-r}\\rvert=e^{\\operatorname{Re}z-r}$ and\n$\\sum_{r\\ge 0}e^{-r}$ converges, the series (6) converges for every\n$z\\in\\mathbb{C}$; thus $S$ extends holomorphically to $\\mathbb{C}$ and\n\\[\nS(z)=e^{\\,z}\\sum_{r=0}^{\\infty}e^{-r}\n     =e^{\\,z}\\cdot\\frac{1}{1-e^{-1}}\n     =\\frac{e^{\\,z+1}}{e-1},\\qquad z\\in\\mathbb{C}. \\tag{9}\n\\]\n\n\\textbf{Step 5 – Quantitative $O(1/n)$ estimate ($\\dagger$).}\n\nInequality (8) is exactly the desired bound:\n\\[\n\\lvert S_{n}(z)-S(z)\\rvert\\le\\frac{C_{2,K}}{n}\n\\qquad(n\\ge 1,\\;z\\in K). \\tag{10}\n\\]\n\n\\textbf{Step 6 – Two-sided bounds ($\\ddagger$).}\n\nFrom (9)\n\\[\n\\lvert S(z)\\rvert=\\frac{e^{\\,\\operatorname{Re}z+1}}{e-1}.\n\\]\nBecause $1<\\dfrac{e}{e-1}<e$, multiplying by $e^{\\,\\operatorname{Re}z}$\ngives\n\\[\ne^{\\,\\operatorname{Re}z}\\;<\\;\\lvert S(z)\\rvert\\;<\\;e^{\\,\\operatorname{Re}z+1},\n\\qquad\\operatorname{Re}z>0.\\;\\;\\square\n\\]",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.472947",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension/variables: the problem is lifted from a single real parameter to a complex variable, requiring control of both modulus and argument.  \n• Additional structure: we now study an entire function defined by an infinite series limit, not merely a numerical limit.  \n• Deeper theory: uniform convergence on compacts (M–test), holomorphic limits (Weierstrass theorem), and explicit analytic continuation are essential.  \n• Technical estimates: an explicit exponential rate of convergence (†) must be proved, necessitating binomial–exponential comparison inequalities.  \n• Interacting concepts: asymptotics, complex analysis, and infinite series all interplay.  \nTogether these elements raise the problem well beyond the original’s elementary bounding, demanding a fuller analytical toolkit and multiple layers of argument."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n  Ω := { z ∈ ℂ : Re z > 0 }.  \nFor every positive integer n put N := 2n − 1 and define  \n  Sₙ : Ω → ℂ,   Sₙ(z) := ∑_{s=0}^{N}\\Bigl(\\dfrac{z+s}{N}\\Bigr)^{\\!N}.   (★)\n\n1. Prove that the sequence (Sₙ)ₙ converges uniformly on every compact subset of Ω and denote its limit by S.  \n2. Show directly from your description of S that S extends to an entire function and find an explicit closed formula for S.  \n3. For every compact K ⊂ Ω prove that a constant C_K > 0 exists such that  \n  |Sₙ(z) − S(z)| ≤ C_K / n   (n ≥ 1, z ∈ K).   (†)  \n4. Deduce that for every z with Re z > 0 one has  \n  e^{Re z} < |S(z)| < e^{Re z + 1}.   (‡)",
      "solution": "Throughout let K ⊂ Ω be a fixed non–empty compact set and put  \n ρ := min_{z∈K} Re z > 0, M := max_{z∈K}|z|.  \nWrite N := 2n − 1 (hence n = (N+1)/2).  For every integer 0 ≤ s ≤ N set r := N−s and abbreviate  \n f_{n,r}(z) := (1 + (z−r)/N)^{N},  w:=z−r.  (1)\n\nStep 1 – Point-wise limit of the summands.  \nFor fixed r the classical limit (1+u/N)^{N} → e^{u} (N→∞) implies  \n lim_{n→∞} f_{n,r}(z) = e^{z−r}  (z ∈ Ω, r ∈ ℕ).  (2)\n\nStep 2 – A crude growth bound (needed only for the tail estimate).  \nFor z ∈ K and 0 ≤ s ≤ N\n |(z+s)/N| ≤ (|z|+s)/N ≤ (M+N)/N = 1+M/N ≤ 1+M,  \nhence  \n |f_{n,r}(z)| ≤ (1+M)^{N}.  \n(The bound depends on N, so the family (Sₙ) is not locally bounded in the usual sense; we shall never use such a property.)\n\nStep 3 – Uniform convergence on compacta (Problem 1).\n\nChoose β with ¾ < β < 1 and split  \n H_{n}(z) := ∑_{0≤r≤βN} f_{n,r}(z),  T_{n}(z) := ∑_{βN<r≤N} f_{n,r}(z).\n\n3 (a) The tail is exponentially small.  \nWriting r = N−s, 0 ≤ s ≤ (1−β)N and using (2) as in the original solution we obtain a factor q^{N} with q < 1 and therefore  \n sup_{z∈K}|T_{n}(z)| → 0 (n→∞).                                                (3)\n\n3 (b) Quantitative control of the head.\n\nRepaired Lemma A (Uniform Euler estimate).  \nFix α with 0 < α < 1.  Define the holomorphic function  \n G(u) := ( log(1+u) − u )/u^{2}, G(0):=−½.  \nBecause G is continuous on the closed disc Ȧ := { |u|≤α }, the numbers  \n M(α):=sup_{Ȧ}|G|, m(α):=−inf_{Ȧ}Re G (>0)  \nare finite.  \n\nLet N ≥ 1 and w ∈ ℂ with |w| ≤ αN.  Put u:=w/N and  \n ξ := N( log(1+u) − u )=G(u)·w^{2}/N.\n\nThen  \n |ξ| ≤ M(α)|w|^{2}/N,  Re ξ ≤−m(α)|w|^{2}/N ≤ 0.\n\nSince Re ξ ≤ 0 we have |e^{ξ}−1| ≤ |ξ| (indeed e^{ξ}−1 = ξ∫_{0}^{1}e^{tξ}dt).  Therefore\n\n |(1+w/N)^{N}−e^{w}|  \n  = |e^{w}|·|e^{ξ}−1|  \n  ≤ M(α)\\,|w|^{2}e^{Re w}/N.                                        (4)\n\nThe constants M(α), m(α) depend only on α and are independent of N.\n\nApplication to the present situation.  \nBecause r ≤ βN and z ∈ K,  \n |w|=|z−r| ≤ M+βN=:β' N with β' := β+M/N < 1 for all large N.  \nFix α with β < α < 1 and take n so large that β' ≤ α.  Lemma A with this α gives\n\n |f_{n,r}(z)−e^{z−r}| ≤ C_{K}\\,(M+r)^{2}e^{ρ−r}/N  (0 ≤ r ≤ βN),    (5)\n\nwhere C_{K}:=M(α)/(1−α) depends only on K (through ρ,M) and α.\n\nBecause ∑_{r=0}^{∞}(M+r)^{2}e^{ρ−r}<∞, the M–test shows that the series  \n H(z):=∑_{r=0}^{∞}e^{z−r}                                             (6)  \nis the uniform limit of H_{n} on K, and\n\n sup_{z∈K}|H_{n}(z)−H(z)| ≤ C_{1,K}/N = 2C_{1,K}/(N+1) ≤ C_{1,K}/n.     (7)\n\n3 (c) Conclusion of Step 3.  \nFrom (3) and (7)\n\n sup_{z∈K}|S_{n}(z) − H(z)| ≤ C_{2,K}/n  (n≥1).                     (8)\n\nHence (Sₙ) converges uniformly on K; we set  \n S(z) := H(z) (z ∈ Ω).  \nUniform convergence on every compact subset of Ω is proved.\n\nStep 4 – Entire continuation and closed formula (Problem 2).\n\nBecause |e^{z−r}| = e^{Re z}e^{−r} and ∑_{r≥0}e^{−r} converges, series (6) converges for every z ∈ ℂ, so S extends holomorphically to ℂ and\n\n S(z) = e^{z}\\sum_{r=0}^{∞}e^{−r}\n    = e^{z}\\cdot\\frac{1}{1−e^{−1}}\n    = \\frac{e^{z+1}}{e−1},  z ∈ ℂ.                                   (9)\n\nThus S is an entire function.\n\nStep 5 – Quantitative O(1/n) estimate (†).\n\nInequality (8) gives exactly the requested bound:\n\n |Sₙ(z)−S(z)| ≤ C_{2,K}/n  (n ≥ 1, z ∈ K).                         (10)\n\nStep 6 – Two-sided bounds (‡).\n\nFrom (9)\n\n |S(z)| = e^{Re z+1}/(e−1).\n\nBecause 1 < e/(e−1) < e, multiplying by e^{Re z} yields\n\n e^{Re z} < |S(z)| < e^{Re z+1},  (Re z > 0).∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.397628",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension/variables: the problem is lifted from a single real parameter to a complex variable, requiring control of both modulus and argument.  \n• Additional structure: we now study an entire function defined by an infinite series limit, not merely a numerical limit.  \n• Deeper theory: uniform convergence on compacts (M–test), holomorphic limits (Weierstrass theorem), and explicit analytic continuation are essential.  \n• Technical estimates: an explicit exponential rate of convergence (†) must be proved, necessitating binomial–exponential comparison inequalities.  \n• Interacting concepts: asymptotics, complex analysis, and infinite series all interplay.  \nTogether these elements raise the problem well beyond the original’s elementary bounding, demanding a fuller analytical toolkit and multiple layers of argument."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}