1. (1)　すべての自然数 \(k\) に対して, 次の不等式を示せ. \[ \dfrac{1}{2(k+1)} \lt \int _ 0^1 \dfrac{1-x}{k+x} dx \lt \dfrac{1}{2k} \]
2. (2)　\(m \gt n\) であるようなすべての自然数 \(m , n\) に対して, 次の不等式を示せ. \[ \dfrac{m-n}{2(m+1)(n+1)} \lt \log \dfrac{m}{n} -\textstyle\sum\limits _ {k=n+1}^m \dfrac{1}{k} \lt \dfrac{m-n}{2mn} \]

【 解 答 】

(1)

\(0 \lt x \lt 1\) において, \(k \lt k+x \lt k+1\) なので \[\begin{align} \dfrac{1}{k+1} & \lt \dfrac{1}{k+x} \lt \dfrac{1}{k} \\ \dfrac{1-x}{k+1} & \lt \dfrac{1-x}{k+x} \lt \dfrac{1-x}{k} \quad ( \ \text{∵} \ 1-x \gt 0 \ ) \\ \text{∴} \quad \displaystyle\int _ 0^1 \dfrac{1-x}{k+1} \, dx & \lt \displaystyle\int _ 0^1 \dfrac{1-x}{k+x} \, dx \lt \int _ 0^1 \dfrac{1-x}{k} \, dx \end{align}\] \(\displaystyle\int _ 0^1 (1-x) \, dx = \dfrac{1}{2} \cdot 1 \cdot 1 = \dfrac{1}{2}\) なので \[ \dfrac{1}{2(k+1)} \lt \int _ 0^1 \dfrac{1-x}{k+x} dx \lt \dfrac{1}{2k} \]

(2)

\[\begin{align} \displaystyle\int _ 0^1 \dfrac{1-x}{k+x} \, dx & = \int _ 0^1 \left( \dfrac{k+1}{k+x} -1 \right) \, dx \\ & = \left[ (k+1) \log (k+x) -x \right] _ 0^1 \\ & = (k+1) \left\{ \log (k+1) -\log k \right\} -1 \end{align}\]

(1) の結果にこれを用いて, 辺々を \(k+1\) で割ると \[ \dfrac{1}{2(k+1)^2} \lt \left\{ \log (k+1) -\log k \right\} -\dfrac{1}{k+1} \lt \dfrac{1}{2k(k+1)} \] ここで \[\begin{align} \dfrac{1}{2(k+1)^2} & \gt \dfrac{1}{2(k+1)(k+2)} = \dfrac{1}{2} \left( \dfrac{1}{k+1} - \dfrac{1}{k+2} \right) , \\ \dfrac{1}{2k(k+1)} & = \dfrac{1}{2} \left( \dfrac{1}{k} - \dfrac{1}{k+1} \right) \end{align}\] なので \[ \dfrac{1}{2} \left( \dfrac{1}{k+1} - \dfrac{1}{k+2} \right) \lt \left\{ \log (k+1) -\log k \right\} -\dfrac{1}{k+1} \lt \dfrac{1}{2} \left( \dfrac{1}{k} - \dfrac{1}{k+1} \right) \] この式に \(k = n , \cdots , m-1\) を代入した式を辺々加えると, \[\begin{align} \textstyle\sum\limits _ {k=n}^{m-1} \left( \dfrac{1}{k+1} - \dfrac{1}{k+2} \right) & = \left( \dfrac{1}{n+1} - \dfrac{1}{n+2} \right) + \cdots +\left( \dfrac{1}{m} - \dfrac{1}{m+1} \right) \\ & = \dfrac{1}{n+1} - \dfrac{1}{m+1} = \dfrac{m-n}{(n+1)(m+1)} , \\ \textstyle\sum\limits _ {k=n}^{m-1} \left( \dfrac{1}{k} - \dfrac{1}{k+1} \right) & = \dfrac{1}{n} - \dfrac{1}{m} = \dfrac{m-n}{nm} , \\ \textstyle\sum\limits _ {k=n}^{m-1} \left\{ \log (k+1) -\log k \right\} & = \log m -\log n = \log \dfrac{m}{n} , \\ \textstyle\sum\limits _ {k=n}^{m-1} \dfrac{1}{k+1} & = \textstyle\sum\limits _ {k=n+1}^m \dfrac{1}{k} \end{align}\] 以上より \[ \dfrac{m-n}{2(m+1)(n+1)} \lt \log \dfrac{m}{n} - \textstyle\sum\limits _ {k=n+1}^m \dfrac{1}{k} \lt \dfrac{m-n}{2mn} \]