\(a \gt 0\) を実数とする. \(n = 1, 2, 3, \cdots\) に対し, 座標平面の \(3\) 点 \[ ( 2n \pi , 0 ) , \ \left( \left( 2n +\dfrac{1}{2} \right) \pi , \dfrac{1}{\left\{ \left( 2n +\frac{1}{2} \right) \pi \right\}^a} \right) , \ \left( (2n+1) \pi , 0 \right) \] を頂点とする三角形の面積を \(A _ n\) とし, \[ B _ n = \displaystyle\int _ {2n \pi}^{(2n+1) \pi} \dfrac{\sin x}{x^a} \, dx , \ C _ n = \displaystyle\int _ {2n \pi}^{(2n+1) \pi} \dfrac{\sin^2 x}{x^a} \, dx \] とおく.
(1) \(n = 1, 2, 3, \cdots\) に対し, 次の不等式が成り立つことを示せ. \[ \dfrac{2}{\left\{ (2n+1) \pi \right\}^a} \leqq B _ n \leqq \dfrac{2}{\left( 2n \pi \right)^a} \]
(2) 極限値 \(\displaystyle\lim _ {n \rightarrow \infty} \dfrac{A _ n}{B _ n}\) を求めよ.
(3) 極限値 \(\displaystyle\lim _ {n \rightarrow \infty} \dfrac{A _ n}{C _ n}\) を求めよ.
【 解 答 】
(1)
\(2n \pi \leqq x \leqq (2n+1) \pi\) において, \(\dfrac{1}{x^a}\) は単調減少であり, また \(\sin x \geqq 0\) なので \[ \dfrac{\sin x}{\{ (2n+1) \pi \}^a} \leqq \dfrac{\sin x}{x^a} \leqq \dfrac{\sin x}{( 2n \pi )^a} \quad ... [1] \ . \] ここで \[ \displaystyle\int _ {2n \pi}^{(2n+1) \pi} \sin x \, dx = [ -\cos x ] _ {2n \pi}^{(2n+1) \pi} = 2 \ . \] なので, [1] の辺々を積分して \[ \dfrac{2}{\left\{ (2n+1) \pi \right\}^a} \leqq B _ n \leqq \dfrac{2}{\left( 2n \pi \right)^a} \ . \]
(2)
\[ A _ n = \dfrac{1}{2} \cdot \pi \cdot \dfrac{1}{\left\{ \left( 2n +\frac{1}{2} \right) \pi \right\}^a} = \dfrac{\pi}{2} \cdot \dfrac{1}{\left\{ \left( 2n +\frac{1}{2} \right) \pi \right\}^a} \quad ... [2] \ . \] これと, (1) の結果を用いれば \[\begin{align} \dfrac{\pi}{4} \cdot \dfrac{( 2n \pi )^a}{\left\{ \left( 2n +\frac{1}{2} \right) \pi \right\}^a} & \leqq \dfrac{A _ n}{B _ n} \leqq \dfrac{\pi}{4} \cdot \dfrac{\{ (2n+1) \pi \}^a}{\left\{ \left( 2n +\frac{1}{2} \right) \pi \right\}^a} \\ \text{∴} \quad \dfrac{\pi}{4} \cdot \left( 1 -\dfrac{2}{4n+1} \right)^a & \leqq \dfrac{A _ n}{B _ n} \leqq \dfrac{\pi}{4} \cdot \left( 1 +\dfrac{2}{4n+1} \right)^a \quad ... [3] \ . \end{align}\] ここで \[ \displaystyle\lim _ {n \rightarrow \infty} \left( 1 \pm \dfrac{2}{4n+1} \right) = 1 \ . \] なので, \(n \rightarrow \infty\) のとき \[ ( [3] \text{の第1辺} ) \rightarrow \dfrac{\pi}{4} , \ ( [3] \text{の第3辺} ) \rightarrow \dfrac{\pi}{4} \ . \] よって, はさみうちの原理より \[ \displaystyle\lim _ {n \rightarrow \infty} \dfrac{A _ n}{B _ n} = \underline{\dfrac{\pi}{4}} \ . \]
(3)
\(\sin^2 x \geqq 0\) なので, (1) と同様に考えれば \[\begin{align} \displaystyle\int _ {2n \pi}^{(2n+1) \pi} \sin^2 x \, dx & = \dfrac{1}{2} \displaystyle\int _ {2n \pi}^{(2n+1) \pi} ( 1 -\cos 2x ) \, dx \\ & = \dfrac{1}{2} \left[ x -\dfrac{1}{2} \sin 2x \right] _ {2n \pi}^{(2n+1) \pi} = \dfrac{\pi}{2} \ . \end{align}\] なので \[ \dfrac{\pi}{2} \cdot \dfrac{1}{\left\{ (2n+1) \pi \right\}^a} \leqq C _ n \leqq \dfrac{\pi}{2} \cdot \dfrac{2}{\left( 2n \pi \right)^a} \ . \] これと [2] より \[\begin{align} \dfrac{( 2n \pi )^a}{\left\{ \left( 2n +\frac{1}{2} \right) \pi \right\}^a} & \leqq \dfrac{A _ n}{C _ n} \leqq \dfrac{\{ (2n+1) \pi \}^a}{\left\{ \left( 2n +\frac{1}{2} \right) \pi \right\}^a} \\ \text{∴} \quad \left( 1 -\dfrac{2}{4n+1} \right)^a & \leqq \dfrac{A _ n}{C _ n} \leqq \left( 1 +\dfrac{2}{4n+1} \right)^a \ . \end{align}\]
(2) と同様に, はさみうちの原理より \[ \displaystyle\lim _ {n \rightarrow \infty} \dfrac{A _ n}{C _ n} = \underline{1} \ . \]