次の問いに答えよ.
(1) 関数 \(f(x) = \dfrac{\log (1-x)}{x}\) は \(0 \lt x \lt 1\) の範囲で減少することを示せ.
(2) 極限値 \[ \displaystyle\lim _ {n \rightarrow \infty} \dfrac{1}{n} \textstyle\sum\limits _ {k=1}^{n} \dfrac{1}{\tan \left( \dfrac{(n+k) \pi}{6n} \right)} \] を求めよ.
【 解 答 】
(1)
\[\begin{align} f'(x) & = -\dfrac{1}{1-x} \cdot \dfrac{1}{x} +\log ( 1-x ) \left( -\dfrac{1}{x^2} \right) \\ & = -\dfrac{x +(1-x) \log ( 1-x )}{x^2 ( 1-x )} \end{align}\] 分子を \(g(x)\) とおくと \[\begin{align} g'(x) & = 1 -\log ( 1-x ) +( 1-x ) \left( -\dfrac{1}{1-x} \right) \\ & = -\log ( 1-x ) \gt 0 \quad ( \ \text{∵} \ 1-x \lt 1 \ ) \end{align}\] ゆえに, \(g(x)\) は単調増加で \[ g(x) \gt g(0) = 0 \] したがって, \(1-x \gt 0\) にも注意すれば \[ f'(x) \lt 0 \] よって, \(0 \lt x \lt 1\) において, \(f(x)\) は単調減少する.
(2)
求める値を \(S\) とおけば \[\begin{align} S & = \displaystyle\lim _ {n \rightarrow \infty} \dfrac{1}{n} \textstyle\sum\limits _ {k=1}^{n} \dfrac{1}{\tan \dfrac{\pi}{6} \left( 1 +\dfrac{k}{n} \right)} \\ & = \displaystyle\int _ {0}^{1} \dfrac{1}{\tan \dfrac{\pi}{6} ( 1+x )} \, dx \\ & = \dfrac{6}{\pi} \displaystyle\int _ {0}^{1} \dfrac{\left\{ \sin \dfrac{\pi}{6} ( 1+x ) \right\}'}{\sin \dfrac{\pi}{6} ( 1+x )} \, dx \\ & = \dfrac{6}{\pi} \left[ \log \left| \sin \dfrac{\pi}{6} ( 1+x ) \right| \right] _ {0}^{1} \\ & = \dfrac{6}{\pi} \left( \log \sin \dfrac{\pi}{3} -\log \sin \dfrac {\pi}{6} \right) \\ & = \dfrac{6}{\pi} \log \dfrac{\dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}} \\ & = \underline{\dfrac{3}{\pi} \log 3} \end{align}\]