How to prove this recursive sequence converges to $\sqrt 2$?

Let $a_0,a_1>0$ be given. Consider the recursive sequence $$a_{n+2}=\frac{1}{a_{n+1}}+\frac{1}{a_n}$$ Prove that $a_n\to\sqrt2$.

I attempted to find a bound for $a_n$ but so far I have only got $a_n>0$. Somebody hints that I might want to use limsup/liminf, but I want to at least put a bound on them first.

Aside from limsup/liminf (which I don't know how to make use of here yet), is there any other method? That, if possible, will be preferable for me since I'm not very used to limsup/liminf.

Or, can we prove that this second-order recursion is stable in some way?

Thanks in advance!

Solution 1:

For those who are interested, one can solve this using $\liminf$, $\limsup$ (So this does not quite meet the OP's requirement). The idea is described in another answer linked in the comment.

First of all one can show that $a_n$ is bounded. Let $$\ell = \liminf_{n\to \infty} a_n, \ \ L = \limsup_{n\to \infty} a_n.$$

By the definition of $\ell$, one can find a subsequence $\{a_{n_k}\}$ so that $a_{n_k} \to \ell$. By picking a subsequence, we also assume that

$$a_{n_k - 1} \to \ell_1,\ \ a_{n_k -2} \to \ell_2\ \ \ \text{as }k\to \infty.$$

Then taking $k\to \infty$ of

$$a_{n_k} = \frac{1}{a_{n_k-1}} + \frac{1}{a_{n_k-2}},$$

we have

$$\ell = \frac{1}{\ell_1 } + \frac{1}{\ell_2} \ge \frac{1}{L} + \frac{1}{L} = \frac{2}{L} \Rightarrow \ell L \ge 2$$

Similarly by taking a subsequence going to $L$, we have $\ell L \le 2$. Thus $\ell L = 2$.

Now we show that $\ell =L =\sqrt 2$. Then this implies $a_n \to \sqrt 2$ as $n\to \infty$.

Similarly, let $a_{n_k}$ converges to $L$ and $a_{n_k-1} \to \ell_1$, $a_{n_k -2} \to \ell_2$ and $a_{n_k - 3} \to \ell_3$. Then we have

$$\frac {2}{\ell} = L = \frac{1}{\ell_1} + \frac{1}{\ell_2}$$

and

$$\ell_1 = \frac{1}{\ell_2} + \frac{1}{\ell_3}.$$

The first equality actually forces $\ell_1 = \ell_2 = \ell$. Put this into the second equality gives

$$\frac{2}{L} = \ell = \frac{1}{\ell } + \frac{1}{\ell_3}.$$

This again forces $\ell = \ell_3 = L$. Thus $\ell = L = \sqrt 2$.

Solution 2:

Hint:

From experimental data, one sees that the difference with $\sqrt2$ decreases exponentially with a common ratio $0.70\cdots$, so that one is tempted to try $a_n=\sqrt2+ar^n$.

Then for large $n$,

$$\sqrt2+ar^{n+2}=\frac1{\sqrt2+ar^{n+1}}+\frac1{\sqrt2+ar^n}\approx \frac1{\sqrt2}-\frac {ar^{n+1}}2+\frac1{\sqrt2}-\frac {ar^{n}}2.$$

This is a linear recurrence and the characteristic equation yields

$$r=\frac{-1\pm i\sqrt7}{4}=\frac1{\sqrt2}e^{\pm i\arctan(\sqrt7)}.$$

It should be possible to show that the absolute error is bounded by $$\frac a{\sqrt2^n}$$ for some $a$ and $N\le n$ (function of the initial conditions).