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发表于 2024-5-5 05:57
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归纳地定义一有理数序列如次: \(a_1=2,\,a_{n+1}={\large\frac{a_n^2+2}{2a_n}}\;(n\in\mathbb{N}^+).\)
于是\(\{a_n\}\)为正项有理数序列由其通项的归纳定义保证,
\(\because{\small a_1^2-2>0,\;a_{n+1}^2-2=}\big(\frac{a_n^2+2}{2a_n}\big)^2{\small-2=}\big(\frac{a_n^2-2}{2a_n}\big)^2\small>0,\;\therefore a_n^2>2\_(\forall n)\)
\(\therefore a_{n+1}-a_n=-{\large\frac{a_n^2-2}{2a_n}}< 0,\;\{a_n\}\)单调减有下界. 对\(a_{n+1}=\large\frac{a_n^2+2}{2a_n}\)
两边取极限得 \(a=\displaystyle\lim_{n\to\infty}a_n,\;a=\small\frac{a^2+2}{2a},\;a=\sqrt{2}.\) |
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