一道递推数列极限题

2020-02-01 本文已影响0人洛玖言

设 $a_1=3,\,a_n=2a_{n-1}^2-1\;(n\geqslant2)$ ，
求 $\displaystyle\lim_{n\to\infty}\dfrac{a_n}{2^na_1a_2\cdots a_{n-1}}$

Sol:
由 $a_1=3>1,\;a_n=2a_{n-1}^2-1$ 及数学归纳法得 $a_{n}>1.$

$\begin{aligned} a_{n}^2-1=&(a_{n}+1)(a_{n}-1)\\ =&2^2a_{n-1}^2(a_{n-1}^2-1)\\ =&2^4a_{n-1}^2a_{n-2}^2(a_{n-2}^2-1)\\ =&\cdots\\ =&2^{2n-2}a_{n-1}^2a_{n-2}^2\cdots a_{1}^2(a_1^2-1)\\ =&2^{2n+1}a_{n-1}^2a_{n-2}^2\cdots a_{1}^2\\ =&2(2^na_{n-1}a_{n-2}\cdots a_{1})^2 \end{aligned}$

$\therefore\displaystyle\dfrac{a_{n}^2-1}{(2^na_{n-1}a_{n-2}\cdots a_{1})^2}=2$
又 $\because a_{n}>1$
$\therefore\displaystyle\lim_{n\to\infty}\dfrac{1}{2^na_{n-1}a_{n-2}\cdots a_{1}}=0$

$\Rightarrow\displaystyle\lim_{n\to\infty}\dfrac{1}{(2^na_{n-1}a_{n-2}\cdots a_{1})^2}=0$
$\therefore\begin{aligned} &\lim_{n\to\infty}\dfrac{a_n^2}{(2^na_{n-1}a_{n-2}\cdots a_{1})^2}\\ =&\lim_{n\to\infty}\dfrac{a_n^2-1}{(2^na_{n-1}a_{n-2}\cdots a_{1})^2}+\lim_{n\to\infty}\dfrac{1}{(2^na_{n-1}a_{n-2}\cdots a_{1})^2}\\ =&2 \end{aligned}$

$\therefore\displaystyle\lim_{n\to\infty}\dfrac{a_n}{2^na_{n-1}a_{n-2}\cdots a_{1}}=\sqrt{2}$

一道递推数列极限题

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