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【2023大阪公立大学・理系・第3問】対数の積分と不等式評価、極限(はさみうちの原理)

2023年入試問題

【2023大阪公立大学・理系・第3問】

次の問いに答えよ.

問1 \(a\),\(b\) は実数とし,\(f(x)\) は \(a\),\(b\) が属する開区間で定義された関数とする.\(f(x)\) が連続な第 \(2\) 次導関数 \(f^{\prime\prime}(x)\) をもつとき,次の等式を証明せよ.

\(\displaystyle\int^{b}_{a}(b-x)(x-a)f^{\prime\prime}(x)dx=(b-a)\left(f(a)+f(b)\right)-2\displaystyle\int^{b}_{a}f(x)dx\)

問2 \(t\) を正の実数とする.次の不等式を証明せよ.

\(0≦\displaystyle\int^{t+1}_{t}\log{x} dx-\displaystyle\frac{1}{2}\left(\log{t}+\log{(t+1)}\right)≦\displaystyle\frac{1}{8}\left(\displaystyle\frac{1}{t}-\displaystyle\frac{1}{t+1}\right)\)

問3 次で定まる数列 \(\left\{a_{n}\right\}\) に対し,極限値 \(\displaystyle\lim_{n\rightarrow\infty}\displaystyle\frac{a_{n}}{\log{n}}\) を求めよ.

\(a_{n}=\log{(n!)}-n\log{n}+n\) ( \(n=1,2,3,\cdots\) )

解答・解説

問1 等式の証明

部分積分法

\(\displaystyle\int f(x)g^{\prime}(x)\enspace dx=f(x)g(x)-\displaystyle\int f^{\prime}(x)g(x)\enspace dx\)

部分積分すると

\(\displaystyle\int^{b}_{a}(b-x)(x-a)f^{\prime\prime}(x)dx\)

\(=\Bigl[(b-x)(x-a)f^{\prime}(x)\Bigr]^{b}_{a}-\displaystyle\int^{b}_{a}\left\{(b-x)(x-a)\right\}^{\prime}f^{\prime}(x)dx\)

\(=\displaystyle\int^{b}_{a}(2x-a-b)f^{\prime}(x)dx\)

\(=\Bigl[(2x-a-b)f(x)\Bigr]^{b}_{a}-\displaystyle\int^{b}_{a}(2x-a-b)^{\prime}f(x) dx\)

\(=(b-a)\left(f(a)+f(b)\right)-2\displaystyle\int^{b}_{a}f(x)dx\)

問2 不等式の証明

問1 において,\(a=t\),\(b=t+1\),\(f(x)=\log{x}\) とおくと

\(f^{\prime}(x)=\displaystyle\frac{1}{x}\),\(f^{\prime\prime}(x)=\displaystyle\frac{-1}{x^2}\) より

\(\displaystyle\int^{t+1}_{t}(t+1-x)(x-t)\left(-\displaystyle\frac{1}{x^2}\right)dx=\log{t}+\log{(t+1)}-2\displaystyle\int^{t+1}_{t}\log{x} dx\)

\(\displaystyle\int^{t+1}_{t}\log{x} dx-\displaystyle\frac{1}{2}\left(\log{t}+\log{(t+1)}\right)=\displaystyle\frac{1}{2}\displaystyle\int^{t+1}_{t}\displaystyle\frac{(t+1-x)(x-t)}{x^2} dx\) ・・・①

また,\(\displaystyle\frac{1}{t}-\displaystyle\frac{1}{t+1}=\Bigl[\displaystyle\frac{1}{x}\Bigr]^{t}_{t+1}\)

\(=\displaystyle\int^{t}_{t+1}\displaystyle\frac{-1}{x^2} dx\)

\(=\displaystyle\int^{t+1}_{t}\displaystyle\frac{1}{x^2} dx\) ・・・②

①,②より題意は

\(0≦\displaystyle\frac{1}{2}\displaystyle\int^{t+1}_{t}\displaystyle\frac{(t+1-x)(x-t)}{x^2} dx≦\displaystyle\frac{1}{8}\displaystyle\int^{t+1}_{t}\displaystyle\frac{1}{x^2} dx\)

つまり

\(0≦\displaystyle\int^{t+1}_{t}\displaystyle\frac{(t+1-x)(x-t)}{x^2} dx≦\displaystyle\int^{t+1}_{t}\displaystyle\frac{1}{4x^2} dx\) を示せばよい.

\(t≦x≦t+1\) より \((t+1-x)(x-t)≧0\)

\(\displaystyle\frac{1}{4x^2}-\displaystyle\frac{(t+1-x)(x-t)}{x^2}\)

\(=\displaystyle\frac{1}{4x^2}\left\{4x^2-4(2t+1)x+4t^2+4t+1\right\}\)

\(=\displaystyle\frac{1}{4x^2}\left(2x-2t-1\right)^2≧0\)

よって \(0≦\displaystyle\frac{(t+1-x)(x-t)}{x^2}≦\displaystyle\frac{1}{4x^2}\) より

\(0≦\displaystyle\int^{t+1}_{t}\displaystyle\frac{(t+1-x)(x-t)}{x^2} dx≦\displaystyle\int^{t+1}_{t}\displaystyle\frac{1}{4x^2} dx\)

したがって

\(0≦\displaystyle\int^{t+1}_{t}\log{x} dx-\displaystyle\frac{1}{2}\left(\log{t}+\log{(t+1)}\right)≦\displaystyle\frac{1}{8}\left(\displaystyle\frac{1}{t}-\displaystyle\frac{1}{t+1}\right)\)

問3 極限値 \(\displaystyle\lim_{n\rightarrow\infty}\displaystyle\frac{a_{n}}{\log{n}}\)

\(m\) を自然数とする.問2 から

\(0≦\displaystyle\int^{m+1}_{m}\log{x} dx-\displaystyle\frac{1}{2}\left(\log{m}+\log{(m+1)}\right)≦\displaystyle\frac{1}{8}\left(\displaystyle\frac{1}{m}-\displaystyle\frac{1}{m+1}\right)\) ・・・③

③に \(m=1,2,3,\cdots,n-1\) ( \(n≧2\) ) を代入して各辺加えると

\(0≦\displaystyle\int^{n}_{1}\log{x} dx-\displaystyle\frac{1}{2}\displaystyle\sum_{m=1}^{n-1}{\left(\log{m}+\log{(m+1)}\right)}≦\displaystyle\frac{1}{8}\displaystyle\sum_{m=1}^{n-1}{\left(\displaystyle\frac{1}{m}-\displaystyle\frac{1}{m+1}\right)}\)

ここで

\(\displaystyle\int^{n}_{1}\log{x} dx\)

\(=\Bigl[x\log x-x\Bigr]^{n}_{1}=n\log n-n+1\)

\(\displaystyle\sum_{m=1}^{n-1}{\left(\log{m}+\log{(m+1)}\right)}\)

\(=(\log 1+\log 2)+(\log 2+\log 3)+(\log 3+\log 4)+\cdots+(\log (n-1)+\log n)\)

\(=2(\log 1+\log 2+\log 3+\cdots+\log n)-\log 1-\log n\)

\(=2\log (n!)-\log n\)

\(\displaystyle\sum_{m=1}^{n-1}{\left(\displaystyle\frac{1}{m}-\displaystyle\frac{1}{m+1}\right)}\)

\(=\left(\displaystyle\frac{1}{1}-\displaystyle\frac{1}{2}\right)+\left(\displaystyle\frac{1}{2}-\displaystyle\frac{1}{3}\right)+\cdots+\left(\displaystyle\frac{1}{n-1}-\displaystyle\frac{1}{n}\right)\)

\(=1-\displaystyle\frac{1}{n}\) であるから

\(0≦n\log n-n+1-\displaystyle\frac{1}{2}\left\{2\log (n!)-\log n\right\}≦\displaystyle\frac{1}{8}\left(1-\displaystyle\frac{1}{n}\right)\)

\(0≦-a_{n}+\displaystyle\frac{1}{2}\log n+1≦\displaystyle\frac{1}{8}\left(1-\displaystyle\frac{1}{n}\right)\)

\(1+\displaystyle\frac{1}{2}\log n-\displaystyle\frac{1}{8}\left(1-\displaystyle\frac{1}{n}\right)≦a_{n}≦1+\displaystyle\frac{1}{2}\log n\)

\(\displaystyle\frac{1}{\log n}+\displaystyle\frac{1}{2}-\displaystyle\frac{1}{8\log n}\left(1-\displaystyle\frac{1}{n}\right)≦\displaystyle\frac{a_{n}}{\log n}≦\displaystyle\frac{1}{\log n}+\displaystyle\frac{1}{2}\)

\(\displaystyle\lim_{n\rightarrow\infty}\left\{\displaystyle\frac{1}{\log n}+\displaystyle\frac{1}{2}-\displaystyle\frac{1}{8\log n}\left(1-\displaystyle\frac{1}{n}\right)\right\}=\displaystyle\frac{1}{2}\)

\(\displaystyle\lim_{n\rightarrow\infty}\left(\displaystyle\frac{1}{\log n}+\displaystyle\frac{1}{2}\right)=\displaystyle\frac{1}{2}\)

したがって,はさみうちの原理から \(\displaystyle\lim_{n\rightarrow\infty}\displaystyle\frac{a_{n}}{\log{n}}=\displaystyle\frac{1}{2}\)

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