【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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