大阪大学 後期理系 2015年度 問2

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解答作成者: 森 宏征

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入試情報

大学名 大阪大学
学科・方式 後期理系
年度 2015年度
問No 問2
学部 理学部 ・ 医学部 ・ 歯学部 ・ 薬学部 ・ 工学部 ・ 基礎工学部
カテゴリ 関数と極限 ・ 積分法の応用
状態 解答 解説なし ウォッチリスト

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\documentclass[a4paper,12pt,fleqn,dvipdfmx]{jreport} \usepackage{amsmath} \usepackage{amssymb} \usepackage{ascmac} \usepackage{vector3} \setlength{\topmargin}{-25mm} \setlength{\oddsidemargin}{2.5mm} \setlength{\textwidth}{420pt} \setlength{\textheight}{700pt} \usepackage{color} \ExecuteOptions{usename} \def\Op{{\mathrm{O}}} \def\LLL{{\lll}} \def\RRR{{\rrr}} \def\LL{{\ll}} \def\RR{{\rr}} \usepackage{graphicx} \usepackage{pifont} \usepackage{fancybox} \usepackage{custom_mori} \begin{document} \setlength{\abovedisplayskip}{0.5zw} \setlength{\belowdisplayskip}{0.5zw} \begin{FRAME}  数列 $\{a_n\}$ を \smallskip% $a_n = \dfrac{n!}{\sqrt{\vphantom{b} n}\,n^n e^{-n}}$ で定める. このとき $\lim\limits_{n \to \infty} a_n = \sqrt{\vphantom{b} 2\pi}$ であることを, 以下の手順で示せ. \begin{enumerate} \item[(1)]  数列 $\{b_n\}$ を % $b_n = \dfrac{2^{2n}(n!)^2}{\sqrt{\vphantom{b} n}\,(2n)!}$ で 定める. $0 < x < \dfrac{\pi}{2}$ のとき \[ \sin^{2n+1}x < \sin^{2n}x < \sin^{2n-1}x \quad (n = 1,\ 2,\ 3,\ \cdots) \] であることを用いて, $\lim\limits_{n \to \infty} b_n = \sqrt{\vphantom{b} \pi}$ で あることを示せ. %\medskip \item[(2)]  すべての自然数 $n$ に対して \begin{align*} 0 < \log\frac{a_n}{a_{n+1}} < \frac{100}{n(n+1)} \end{align*} が成り立つことを示せ. %\medskip \item[(3)]  $\lim\limits_{n \to \infty} \dfrac{a_n}{a_{2n}} = 1$ で あることを示せ. %\medskip \item[(4)]  $\lim\limits_{n \to \infty} a_n = \sqrt{\vphantom{b} 2\pi}$ で あることを示せ. \end{enumerate} \end{FRAME} \vskip 2mm \noindent{\color[named]{BurntOrange}\bfseries \Ovalbox{解答}} \begin{enumerate} \renewcommand{\labelenumi}{(\arabic{enumi})} \item  $\forall n \in \mathbb{N} \cup \{0\}$ に 対して $S_n = \displaystyle\int_0^\frac{\pi}{2} \sin^n x\,dx$ とおく. \begin{align*} S_{n+2} &= \int_0^\frac{\pi}{2} \sin^{n+1}x(-\cos x)'\,dx \\[1mm] &= -\LL \sin^{n+1}x\cos x \RR_0^\frac{\pi}{2} + (n + 1)\int_0^\frac{\pi}{2} \sin^n x(1 - \sin^2 x)\,dx \\[1mm] &= (n + 1)S_n - (n + 1)S_{n+2} \\ \therefore \enskip S_{n+2} &= \frac{n + 1}{n+2}S_n \end{align*} $S_0 = \displaystyle\int_0^\frac{\pi}{2} 1\,dx = \dfrac{\pi}{2},\enskip S_1 = \displaystyle\int_0^\frac{\pi}{2} \sin x\,dx = 1$ より, \begin{align*} S_{2n} &= \frac{2n-1}{2n} \cdot \frac{2n-3}{2n-2} \cdots \frac{1}{2}S_0 = \frac{(2n)!}{(2^n \cdot n!)^2} \cdot \frac{\pi}{2} \\[1mm] &= \frac{\pi}{2} \cdot \frac{1}{\sqrt{\vphantom{b} n}\,b_n} \tag*{$\cdott\MARU{1}$} \\[1mm] S_{2n+1} &= \frac{2n}{2n+1} \cdot \frac{2n-2}{2n -1} \cdots \frac{2}{3}S_1 = \frac{(2^n \cdot n!)^2}{(2n + 1)(2n)!} \\[1mm] &= \frac{\sqrt{\vphantom{b} n}\,b_n}{2n + 1} \tag*{$\cdott\MARU{2}$} \displaybreak[0]\\[1mm] S_{2n-1} &= \frac{2n-2}{2n-1} \cdot \frac{2n-4}{2n-3} \cdots \frac{2}{3}S_1 = \frac{(2^n \cdot n!)^2}{2n \cdot (2n)!} \\[1mm] &= \frac{\sqrt{\vphantom{b} n}\,b_n}{2n} \tag*{$\cdott\MARU{3}$} \end{align*} $0 \leqq x \leqq \dfrac{\pi}{2}$ のとき $\sin^{2n+1}x \leqq \sin^{2n}x \leqq \sin^{2n-1}x$ だから, \begin{gather*} S_{2n+1} \leqq S_{2n} \leqq S_{2n-1} \end{gather*} \MARU{1},\enskip\MARU{2},\enskip\MARU{3}より \begin{gather*} \frac{\sqrt{\vphantom{b} n}\,b_n}{2n + 1} \leqq \frac{\pi}{2} \cdot \frac{1}{\sqrt{\vphantom{b} n}\,b_n} \leqq \frac{\sqrt{\vphantom{b} n}\,b_n}{2n} \end{gather*} $b_n > 0$ より \begin{gather*} \pi \leqq {b_n}^2 \leqq \frac{\pi}{2} \cdot \frac{2n+1}{n} \end{gather*} $\lim\limits_{n \to \infty} \dfrac{\pi}{2} \cdot \dfrac{2n+1}{n} = \pi$ より,はさみうちの原理によって \begin{align*} \lim_{n \to \infty} {b_n}^2 = \pi \qquad \therefore \enskip \lim_{n \to \infty} b_n = \sqrt{\vphantom{b} \pi} \tag*{■} \end{align*} \item  $a_n$ の定義より \begin{gather*} \frac{a_n}{a_{n+1}} = \frac{n! \cdot e^n}{\sqrt{\vphantom{b} n}\,n^n} \cdot \frac{\sqrt{\vphantom{b} n+1}\,(n+1)^{n+1}} {(n + 1)! \cdot e^{n+1}} = \bigg(\frac{n + 1}{n}\bigg)^{\!\!n+\frac{1}{2}} \cdot \frac{1}{e} \\[1mm] \therefore \enskip \log\frac{a_n}{a_{n+1}} = \bigg(n + \frac{1}{2}\bigg) \{\log(n+1) - \log n\} - 1 \tag*{$\cdott\MARU{4}$} \end{gather*} 図のように点A,\enskip B,\enskip C,\enskip D,\enskip E,\enskip% F,\enskip Tをとる.\smallskip ただし, C,\enskip DはTにおける $y = \dfrac{1}{x}$ の接線 $l$ と 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\special{pa 3586 1728}% \special{pa 3590 1728}% \special{pa 3596 1728}% \special{pa 3600 1730}% \special{pa 3606 1730}% \special{pa 3610 1730}% \special{pa 3616 1730}% \special{pa 3620 1730}% \special{pa 3626 1732}% \special{pa 3630 1732}% \special{pa 3636 1732}% \special{pa 3640 1732}% \special{pa 3646 1732}% \special{pa 3650 1734}% \special{pa 3656 1734}% \special{pa 3660 1734}% \special{pa 3666 1734}% \special{pa 3670 1736}% \special{pa 3676 1736}% \special{pa 3680 1736}% \special{pa 3686 1736}% \special{sp}% % LINE 2 0 3 0 % 2 1345 2200 1345 1120 % \special{pn 8}% \special{pa 1346 2200}% \special{pa 1346 1120}% \special{fp}% % LINE 2 0 3 0 % 2 3145 2200 3145 1687 % \special{pn 8}% \special{pa 3146 2200}% \special{pa 3146 1688}% \special{fp}% % LINE 2 0 3 0 % 2 2245 2200 2245 1534 % \special{pn 8}% \special{pa 2246 2200}% \special{pa 2246 1534}% \special{fp}% % LINE 2 0 3 0 % 2 3145 1687 1345 1120 % \special{pn 8}% \special{pa 3146 1688}% \special{pa 1346 1120}% \special{fp}% % LINE 2 0 3 0 % 2 985 1219 3685 1885 % \special{pn 8}% \special{pa 986 1220}% \special{pa 3686 1886}% \special{fp}% % STR 2 0 3 0 % 3 3600 2250 3600 2340 2 0 % $x$ \put(36.0000,-23.4000){\makebox(0,0)[lb]{$x$}}% % STR 2 0 3 0 % 3 850 535 850 625 2 0 % \resizebox{!}{1zw}{$y=\dfrac{1}{x}$} \put(8.5000,-6.2500){\makebox(0,0)[lb]{\resizebox{!}{1zw}{$y=\dfrac{1}{x}$}}}% % STR 2 0 3 0 % 3 1310 2240 1310 2330 2 0 % \resizebox{!}{0.4zw}{$n$} \put(13.1000,-23.3000){\makebox(0,0)[lb]{\resizebox{!}{0.4zw}{$n$}}}% % STR 2 0 3 0 % 3 3000 2260 3000 2350 2 0 % \resizebox{!}{0.5zw}{$n+1$} \put(30.0000,-23.5000){\makebox(0,0)[lb]{\resizebox{!}{0.5zw}{$n+1$}}}% % STR 2 0 3 0 % 3 2110 2370 2110 2460 2 0 % \resizebox{!}{0.9zw}{$n+\dfrac{1}{2}$} \put(21.1000,-24.6000){\makebox(0,0)[lb]{\resizebox{!}{0.9zw}{$n+\dfrac{1}{2}$}}}% % STR 2 0 3 0 % 3 3720 1870 3720 1960 2 0 % \resizebox{!}{0.6zw}{$l$} \put(37.2000,-19.6000){\makebox(0,0)[lb]{\resizebox{!}{0.6zw}{$l$}}}% % STR 2 0 3 0 % 3 1340 980 1340 1070 2 0 % {\footnotesize E} \put(13.4000,-10.7000){\makebox(0,0)[lb]{{\footnotesize E}}}% % STR 2 0 3 0 % 3 3110 1540 3110 1630 2 0 % {\footnotesize F} \put(31.1000,-16.3000){\makebox(0,0)[lb]{{\footnotesize F}}}% % STR 2 0 3 0 % 3 1200 1370 1200 1460 2 0 % {\footnotesize C} \put(12.0000,-14.6000){\makebox(0,0)[lb]{{\footnotesize C}}}% % STR 2 0 3 0 % 3 3010 1820 3010 1910 2 0 % {\footnotesize D} \put(30.1000,-19.1000){\makebox(0,0)[lb]{{\footnotesize D}}}% % STR 2 0 3 0 % 3 1390 2050 1390 2140 2 0 % {\footnotesize A} \put(13.9000,-21.4000){\makebox(0,0)[lb]{{\footnotesize A}}}% % STR 2 0 3 0 % 3 3190 2050 3190 2140 2 0 % {\footnotesize B} \put(31.9000,-21.4000){\makebox(0,0)[lb]{{\footnotesize B}}}% % STR 2 0 3 0 % 3 2100 1570 2100 1660 2 0 % {\footnotesize T} \put(21.0000,-16.6000){\makebox(0,0)[lb]{{\footnotesize T}}}% \end{picture}% %\input{osaka2015s2c_zu_1} \end{center} 台形ABDCの面積を $S_1$, 台形ABFEの面積を $S_2$ とすれば, \begin{gather*} S_1 < \int_n^{n+1} \frac{1}{x}\,dx < S_2 \end{gather*} 点Eの$y$座標を $y(\E)$ のように記す. \begin{gather*} S_2 = \frac{1}{2}\{y(\E) + y(\F)\} \cdot \A\B = \frac{1}{2}\!\left(\frac{1}{n} + \frac{1}{n + 1} \right) \\[1mm] S_1 = y(\T) \cdot \A\B = \frac{1}{\>n + \dfrac{\mathstrut 1}{2}\>} \end{gather*} より \begin{align*} \frac{1}{\>n + \dfrac{\mathstrut 1}{2}\>} < \log(n + 1) - \log n < \frac{1}{2}\bigg(\frac{1}{n} + \frac{1}{n + 1}\bigg) = \frac{2n+1}{2n(n+1)} \end{align*} 辺々に $n + \dfrac{1}{2} > 0$ を掛けて整理すれば, \begin{gather*} 1 < \bigg(n + \frac{1}{2}\bigg) \{\log(n + 1) - \log n\} < \frac{(2n + 1)^2}{4n(n + 1)} \\[1mm] \therefore \enskip 0 < \bigg(n + \frac{1}{2}\bigg) \{\log(n + 1) - \log n\} - 1 < \frac{(2n + 1)^2}{4n(n + 1)} - 1 \end{gather*} \MARU{4}および $\displaystyle \frac{(2n + 1)^2}{4n(n + 1)} - 1 = \frac{1}{4n(n + 1)} < \frac{100}{n(n+1)}$ から, \begin{align*} 0 < \log\frac{a_n}{a_{n+1}} < \frac{100}{n(n+1)} \tag*{■} \end{align*} \item  $\displaystyle\frac{a_n}{a_{2n}} = \frac{a_n}{a_{n+1}} \cdot \frac{a_{n+1}}{a_{n+2}} \cdots \frac{a_{2n-1}}{a_{2n}}$ より(2)の不等式を用いると, \begin{gather*} \log\frac{a_n}{a_{2n}} = \sum_{k=n}^{2n-1}\log\frac{a_k}{a_{k+1}} > 0 \displaybreak[0] \\ \log\frac{a_n}{a_{2n}} = \sum_{k=n}^{2n-1} \log\frac{a_k}{a_{k+1}} < 100\sum_{k=n}^{2n-1} \bigg(\frac{1}{k} - \frac{1}{k+1}\bigg) = 100\bigg(\frac{1}{n} - \frac{1}{2n}\bigg) \\[1mm] \therefore \enskip 0 < \log\frac{a_n}{a_{2n}} < 100\bigg(\frac{1}{n} - \frac{1}{2n}\bigg) \end{gather*} ゆえにはさみうちの原理より \begin{align*} \lim_{n \to \infty} \log\frac{a_n}{a_{2n}} = 0 \qquad \therefore \enskip \lim_{n \to \infty} \frac{a_n}{a_{2n}} = 1 \tag*{■} \end{align*} \item  $a_n,\ a_{2n},\ b_n$ の定義より \begin{align*} \frac{{a_n}^2}{a_{2n}} = \frac{(n!)^2 e^{2n}}{n(n^n)^2} \cdot \frac{\sqrt{\vphantom{b} 2n}\,(2n)^{2n}} {(2n)!\,e^{2n}} = \sqrt{\vphantom{b} 2}\,b_n \end{align*} (1)の結果より \begin{align*} \lim_{n \to \infty} a_n &= \lim_{n \to \infty} \bigg(\frac{{a_n}^2}{a_{2n}} \cdot \frac{a_{2n}}{a_n}\bigg) = \sqrt{\vphantom{b} 2} \left(\lim_{n \to \infty} b_n \right) \cdot \left(\lim_{n \to \infty} \frac{a_{2n}}{a_n} \right) \\[1mm] &= \sqrt{\vphantom{b} 2} \cdot \sqrt{\vphantom{b} \pi} \cdot 1 = \sqrt{\vphantom{b} 2\pi} \tag*{■} \end{align*} \end{enumerate} \end{document}