数列 \(\left\{ a _ n \right\} , \left\{ b _ n \right\}\) を \[\begin{align} a _ n & = \displaystyle\int _ {-\frac{\pi}{6}}^{\frac{\pi}{6}} e^{n \sin \theta} \, d \theta , \\ b _ n & = \displaystyle\int _ {-\frac{\pi}{6}}^{\frac{\pi}{6}} e^{n \sin \theta} \cos \theta \, d \theta \quad ( n = 1, 2, 3, \cdots ) \ . \end{align}\] で定める. ただし, \(e\) は自然対数の底とする.
(1) 一般項 \(\left\{ b _ n \right\}\) を求めよ.
(2) すべての \(n\) について, \(b _ n \leqq a _ n \leqq \dfrac{2}{\sqrt{3}} b _ n\) が成り立つことを示せ.
(3) \(\displaystyle\lim _ {n \rightarrow \infty} \dfrac{1}{n} \log \left( na _ n \right)\) を求めよ. ただし, 対数は自然対数とする.
【 解 答 】
(1)
\[ e^{n \sin \theta} \cos \theta = \dfrac{1}{n} \left( e^{n \sin \theta} \right)' \ . \] なので \[\begin{align} b _ n & = \dfrac{1}{n} \left[ e^{n \sin \theta} \right] _ {-\frac{\pi}{6}}^{\frac{\pi}{6}} \\ & = \underline{\dfrac{1}{n} \left( e^{\frac{n}{2}} -e^{-\frac{n}{2}} \right)} \ . \end{align}\]
(2)
\(-\dfrac{\pi}{6} \leqq \theta \leqq \dfrac{\pi}{6}\) において \[ \dfrac{\sqrt{3}}{2} \leqq \cos \theta \leqq 1 , \ e^{n \sin \theta} \gt 0 \ . \] なので \[ \dfrac{\sqrt{3}}{2} e^{n \sin \theta} \leqq e^{n \sin \theta} \cos \theta \leqq e^{n \sin \theta} \ . \] この区間で辺々を積分すれば \[\begin{gather} \dfrac{\sqrt{3}}{2} a _ n \leqq b _ n \leqq a _ n \\ \text{∴} \quad b _ n \leqq a _ n \leqq \dfrac{2}{\sqrt{3}} a _ n \ . \end{gather}\]
(3)
(2) の結果を用いれば \[ \dfrac{1}{n} \log \left( nb _ n \right) \leqq \dfrac{1}{n} \log \left( na _ n \right) \leqq \dfrac{1}{n} \log \left( \dfrac{2}{\sqrt{3}}nb _ n \right) \quad ... [1] \ . \] ここで (1) の結果を用いれば, [1] について \[\begin{align} ( \text{左辺} ) & = \dfrac{1}{n} \log e^{\frac{n}{2}} \left( 1 -e^{-n} \right) \\ & = \dfrac{1}{2} +\dfrac{1}{n} \log \left( 1 -e^{-n} \right) \\ & \rightarrow \dfrac{1}{2} \quad ( \ n \rightarrow \infty \text{のとき} ) \ , \\ ( \text{右辺} ) & = \dfrac{1}{n} \log \dfrac{2}{\sqrt{3}} e^{\frac{n}{2}} \left( 1 -e^{-n} \right) \\ & = \dfrac{1}{2} +\dfrac{1}{n} \log \left( 1 -e^{-n} \right) +\dfrac{1}{n} \log \dfrac{2}{\sqrt{3}} \\ & \rightarrow \dfrac{1}{2} \quad ( \ n \rightarrow \infty \text{のとき} ) \ . \end{align}\] よって, はさみうちの原理より \[ \displaystyle\lim _ {n \rightarrow \infty} \dfrac{1}{n} \log \left( na _ n \right) = \underline{\dfrac{1}{2}} \ . \]