数列极限收敛发散

若 $a_{n-1}+2a_n \rightarrow 0$,则 $a_n$ 收敛还是发散?

Colliot

设 $\{a_n\}$ 满足 $$\lim_{n\rightarrow\infty} a_{n-1}+2a_n = 0$$,试讨论 $\{a_n\}$ 的敛散性。


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$$\begin{align} &根据题意\forall\varepsilon>0,\exists N_0\in\mathbb{N},s.t.n>N_0时有-2\varepsilon<a_{n-1}+2a_n<2\varepsilon\\ &于是我们可以得到a_n\in(-\varepsilon-\frac{a_{n-1}}{2},\varepsilon-\frac{a_{n-1}}{2})\\ &从而a_{n+1}<\varepsilon-\frac{a_n}{2}<\varepsilon-(-\varepsilon-\frac{a_{n-1}}{2})/2=\frac{3\varepsilon}{2}+\frac{a_{n-1}}{4}\\ &同理有a_{n+1}>-\frac{3\varepsilon}{2}+\frac{a_{n-1}}{4}\\ &以此类推,我们可以得到a_{n+m}\in(-(2-\frac{1}{2^m})\varepsilon-(-\frac{1}{2})^m\frac{a_{n-1}}{2},(2-\frac{1}{2^m})\varepsilon-(-\frac{1}{2})^m\frac{a_{n-1}}{2})\\ &\because a_{n-1}为有限值\\ &\therefore\exists M_0\in\mathbb{N},s.t.m>M_0时有|(-\frac{1}{2})^m\frac{a_{n-1}}{2}|<\frac{\varepsilon}{2^m}\\ &从而当n>N_0+M_0时,a_n\in(-2\varepsilon,2\varepsilon),\lim_{n\to\infty}a_n=0,即\{a_n\}收敛于0\\ \end{align}$$

Created at 10/30/2018

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