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\documentclass[fleqn,12pt]{jsarticle} %\usepackage[dvips,dviout]{graphicx,color} \usepackage{ascmac,array,framed,wrapfig} \usepackage{enumerate,amssymb,amsmath} %\usepackage{picins} %\usepackage[noreplace]{otf} %\usepackage{bm} \newcommand{\mb}[1]{\mbox{\boldmath $ #1 $}} % math-italic の bold 体が使える． % 指定は \mb． 例）\mb{y} : y の bold 体 \newcommand{\MARU}[1]{{\ooalign{\hfil#1\/\hfil\crcr\raise.167ex\hbox{\mathhexbox20D}}}} \def\Noteq{\mathrel{% \setbox0\hbox{=}\hbox{=}\llap{\hbox to\wd0{\hss$\backslash$\hss}}}} \newcommand{\ssqrt}[1]{\sqrt{\smash[b]{\mathstrut #1}}} \newcommand{\Not}[1]{\ooalign{\hfil$\backslash$\hfil\crcr$#1$}} \newcommand{\Dfrac}[2]{\dfrac{\,#1\,}{\,#2\,}} \newcommand{\VeC}[1]{\overrightarrow{\mathstrut {\,#1\,}}} \newcommand{\VEC}[1]{\overrightarrow{\mathstrut {\,\mathrm{#1}\,}}} \newcommand{\Frac}[2]{\frac{\,#1\,}{\,#2\,}} \newcommand{\comb}[2]{{}_{#1}\mathrm{C}\,{}_{#2}} \newcommand{\parm}[2]{{}_{#1}\mathrm{P}\,{}_{#2}} \linespread{1.2} \def\labelenumi{(\theenumi)} \def\labelenumii{(\theenumii)} \def\labelenumiii{(\theenumiii)} \def\theenumi{\arabic{enumi}} \def\theenumii{\roman{enumii}} \def\theenumiii{\alph{enumiii}} \pagestyle{empty} \begin{document} \begin{framed} $n$ を自然数とする．$n+1$ 項の等差数列 $x_{0},\,x_{1},\,\cdots \,x_{n}$ と等比数列 $y_{0},\,y_{1},\,\cdots \,y_{n}$ が \[ 1= x_{0} < x_{1} < x_{2} < \cdots < x_{n}=2 \] \[ 1= y_{0} < y_{1} < y_{2} < \cdots < y_{n}=2 \] を満たすとし， $P(n),~Q(n),~R(n),~S(n)$ を次で定める． \[ P(n) = \Dfrac{x_{1} + x_{2} + \cdots + x_{n}}{n}~,~Q(n) = \sqrt[n]{x_{1}x_{2}\cdots x_{n}} \] \[ R(n) = \Dfrac{y_{1} + y_{2} + \cdots + y_{n}}{n}~,~S(n) = \sqrt[n]{y_{1}y_{2}\cdots y_{n}} \] このとき極限値 $\displaystyle \lim_{n \to \infty} P(n),~\lim_{n \to \infty} Q(n),~\lim_{n \to \infty} R(n),~\lim_{n \to \infty} S(n)$ をそれぞれ求めよ． \end{framed} \vspace{2mm} 等差数列 $\{x_{k}\}$ の公差 $d$ は，$x_{0} = 1,~x_{n} = x_{0} +dn = 2$ より~$d=\Dfrac{1}{n}$ よって，$x_{k} = x_{0} +\Dfrac{k}{n} = 1+\Dfrac{k}{n}~~(k=0,1,2,\cdots)$ これは，十分~$1= x_{0} < x_{1} < x_{2} < \cdots < x_{n}=2$ を満たす． \begin{align*} \lim_{n \to \infty} P(n) &= \lim_{n \to \infty} \Dfrac{x_{1} + x_{2} + \cdots + x_{n}}{n} = \lim_{n \to \infty} \Dfrac{1}{n} \sum_{k=1}^{n} x_{k}\\ &= \lim_{n \to \infty} \Dfrac{1}{n} \sum_{k=1}^{n} \left( 1+\Dfrac{k}{n} \right) = \int_{0}^{1} (1+x)dx\\ &= \Big[ x+\Dfrac{1}{2}x^{2}\Big]_{0}^{1} = \mb{\Dfrac{3}{2}} \end{align*} \vspace{1mm} $Q(n) >0$ より \begin{align*} \lim_{n \to \infty} \log Q(n) &= \lim_{n \to \infty} \log \sqrt[n]{x_{1}x_{2}\cdots x_{n}} = \lim_{n \to \infty} \Dfrac{1}{n} \sum_{k=1}^{n} \log x_{k}\\ &= \lim_{n \to \infty} \Dfrac{1}{n} \sum_{k=1}^{n} \log \left( 1+\Dfrac{k}{n} \right) = \int_{0}^{1} \log (1+x)dx\\ &= \Big[ (1+x)\log (1+x)\Big]_{0}^{1} - \int_{0}^{1} (1+x) \cdot \Dfrac{1}{1+x} dx\\ &= 2\log 2 -\Big[x]_{0}^{1} = 2\log 2-1 = \log \Dfrac{4}{e} \end{align*} （指数関数の連続性により）$\displaystyle \lim_{n \to \infty} Q(n) = \mb{\Dfrac{4}{e}}$ \vspace{2mm} $ 1= y_{0} < y_{1} < y_{2} < \cdots < y_{n}=2$ より， 等比数列 $\{y_{k}\}$ の公比 $r$ は正の実数で，$y_{0} = 1,~y_{n} = y_{0}\cdot r^{n} = 2$ より~$r=2^{\Frac{1}{n}}$ よって，$y_{k} = y_{0} \cdot 2^{\Frac{k}{n}} = 2^{\Frac{k}{n}}~~(k=0,1,2,\cdots)$ \begin{align*} \lim_{n \to \infty} R(n) &= \lim_{n \to \infty} \Dfrac{y_{1} + y_{2} + \cdots + y_{n}}{n} = \lim_{n \to \infty} \Dfrac{1}{n} \sum_{k=1}^{n} y_{k}\\ &= \lim_{n \to \infty} \Dfrac{1}{n} \sum_{k=1}^{n} 2^{\Frac{k}{n}} = \int_{0}^{1} 2^{x}dx\\ &= \Big[ \Dfrac{2^{x}}{\log 2}\Big]_{0}^{1} = \mb{\Dfrac{1}{\log 2}} \end{align*} \vspace{1mm} $S(n) >0$ より \begin{align*} \lim_{n \to \infty} \log S(n) &= \lim_{n \to \infty} \log \sqrt[n]{y_{1}y_{2}\cdots y_{n}} = \lim_{n \to \infty} \Dfrac{1}{n} \sum_{k=1}^{n} \log y_{k}\\ &= \lim_{n \to \infty} \Dfrac{1}{n} \sum_{k=1}^{n} \Dfrac{k}{n}\log 2 = \log 2 \int_{0}^{1} xdx\\ &= \log2 \Big[ \Dfrac{1}{2}x^{2}\Big]_{0}^{1} = \Dfrac{1}{2}\log 2 = \log 2^{\Frac{1}{2}} \end{align*} （指数関数の連続性により）$\displaystyle \lim_{n \to \infty} Q(n) = 2^{\Frac{1}{2}} = \mb{\ssqrt{2}}$ \end{document}