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計算證明題4.
設數列\(\langle\;a_n\rangle\;\)滿足\(\displaystyle a_n=\int_0^1 (1-x^2)^{\frac{n}{2}}dx\),\(n=0,1,2,3,\ldots\)。
(1)證明:\(\displaystyle a_n=\frac{n}{n+1}a_{n-2},n\ge 2\)。
(2)試求\(\displaystyle \lim_{n\to \infty}\frac{a_{n+1}}{a_n}\)的值。
[解答]
計算4(1)
\(a_{n}=\int^{1}_{0}(1-x^{2})^{n/2}dx\)
\( \ \ \ =\int^{\pi/2}_{0}\cos^{n}{\theta}d{\sin{\theta}}(=\int^{\pi/2}_{0}\cos^{n+1}{\theta}d{\theta}\))
\( \ \ \ =\cos^{n}{\theta}\sin{\theta}|^{\pi/2}_{0}+\int^{\pi/2}_{0}n\cos^{n-1}{\theta}\sin^{2}{\theta}d{\theta}\)
\( \ \ \ =n\int^{\pi/2}_{0}\cos^{n-1}{\theta}(1-\cos^{2}{\theta})d{\theta}\)
\( \ \ \ =n\int^{\pi/2}_{0}\cos^{n-1}{\theta}d{\theta}-n\int^{\pi/2}_{0}\cos^{n+1}{\theta}d{\theta}\)
\( \ \ \ =na_{n-2}-na_{n}\)
整理得
\(\displaystyle a_{n}=\frac{n}{n+1}a_{n-2}\)
計算4(2)
考慮 \(0\leq \theta \leq \pi/2\Rightarrow0\leq \cos{\theta}\leq 1\Rightarrow\cos^{n}{\theta}\) 遞減\(\Rightarrow a_{n}\) 遞減
即 \(\displaystyle \frac{a_{n+2}}{a_{n}}\leq \frac{a_{n+1}}{a_{n}}\leq \frac{a_{n+1}}{a_{n+1}}=1\)
又因 \(\displaystyle \lim\limits_{n\rightarrow \infty}\frac{a_{n+2}}{a_{n}}=1\)
由夾擠定理得 \(\displaystyle \lim\limits_{n\rightarrow \infty}\frac{a_{n+1}}{a_{n}}=1\)