若\(a_1=3,b_1=2,\;a_{n+1}=a_n+2b_n,\,b_{n+1}=a_n+b_n,\)则\(a_n/b_n\to\sqrt{2}\)

elim

题：若\(a_1=3,b_1=2,\;a_{n+1}=a_n+2b_n,\,b_{n+1}=a_n+b_n,\) 则\(\displaystyle\lim_{n\to\infty}\frac{a_n}{b_n}=\sqrt{2}.\)

wilsony

这是大学生数学竞赛题吗？

luyuanhong

elim

谢谢陆老师的解答。学习了。

elim

wilsony 发表于 2024-5-5 00:19
这是大学生数学竞赛题吗？

不是的。我想构造一个有理数序列，其极限是无理数。于是想到利用\(\sqrt{2}\)的连分数的渐近分数序列:
\(x^2=2\iff (x+1)(x-1)=1\iff x-1=\large\frac{1}{2+(x-1)}\)
\(={\large\frac{1}{2+\frac{1}{2+(x-1)}}}=\cdots{=\scriptsize\dfrac{1}{2+\scriptsize\dfrac{1}{2+\scriptsize\dfrac{1}{2+\scriptsize\dfrac{1}{\ddots}}}}},\;\therefore\;\sqrt{2}=[1;2,2,2,\ldots]\)
\(\therefore\;\sqrt{2}\)的连分数渐近分数序列是\(\frac{3}{2},\frac{7}{5},\frac{17}{12},\ldots,\frac{a_n}{b_n},\frac{a_{n+1}}{b_{n+1}},\ldots\)
\(\therefore\;\;\small\dfrac{a_{n+1}}{b_{n+1}}=1+\dfrac{1}{1+\dfrac{a_n}{b_n}}=\dfrac{a_n+2b_n}{a_n+b_n}\)

elim

令\(r_n=\large\frac{a_n}{b_n}\), \(r_{n+1}\)则 \(r_{n+1}^2-2={\large\frac{2-r_n^2}{(1+r_n)^2}},\; r_{2n-1}^2> 2,\, r_{2n}^2< 2\).
\(r_{n+2}-r_n={\large\frac{2(2-r_n^2)}{(1+r_n)^2}},\{r_{2n-1}\}\)单调减,\(\{r_{2n}\}\)单调增. 设其极限依次为
\(\alpha,\beta,\)则 \(\alpha=\frac{2+\beta}{1+\beta},\;\beta=\frac{2+\alpha}{1+\alpha}\), 解得 \(\alpha=\beta=\sqrt{2}\)

若欲得到 \(r_n\)的通项公式，估计必须用陆老师的方法。

王守恩

a(n)=1, 3, 7, 17, 41, 99, 239, 577,1393, 3363, 8119, 19601, 47321,114243, 275807, 665857, 1607521,
b(n)=1, 2, 5, 12, 29, 70, 169, 408, 985,  2378, 5741, 13860, 33461, 80782, 195025, 470832, 1136689,

a(n)=\(\big[\frac{(1 + \sqrt{2})^n}{2}\big]\),   b(n)=\(\big[\frac{(1 + \sqrt{2})^n}{2\sqrt{2}}\big]\)

luyuanhong

楼上 elim 的帖子很好！已收藏。