Determining convergence of sequence $a_n=a_{n-1}^{-1}+a_{n-2}^{-1}$

Question

I'm trying to prove convergence for the sequence $$a_n=\frac{1}{a_{n-1}}+\frac{1}{a_{n-2}}$$ with $a_0=a_1=1$. If it does converge then it converges to $\sqrt{2}$, which agrees with numerical tests. $$$$I've managed to prove two facts about the sequence, although I'm not sure they're relevant to the question or not. Firstly, for every set of three consecutive terms in the sequence, at least one is bigger than $\sqrt{2}$ and at least one is smaller than $\sqrt{2}$. Secondly, if two consecutive terms are in the interval $(\alpha,\,2\alpha^{-1})$ for some $\alpha\in[1,\,\sqrt{2})$, then all terms that follow are in this interval as well. $$$$Does anyone have an idea for how to prove convergence, with or without these facts?

Answer 1

By induction, we can easily show that $1 \leq a_n \leq 2$ for all $n$. Next, note that the recurrence relation can be recast as

$$ a_n - \sqrt{2} = -\frac{a_{n-1} - \sqrt{2}}{\sqrt{2}a_{n-1}} - \frac{a_{n-2} - \sqrt{2}}{\sqrt{2}a_{n-2}}. $$

Applying this relation to the factor $a_{n-1} - \sqrt{2}$ on the right-hand side again,

\begin{align*} a_n - \sqrt{2} &= -\frac{1}{\sqrt{2}a_{n-1}}\left( -\frac{a_{n-2} - \sqrt{2}}{\sqrt{2}a_{n-2}} - \frac{a_{n-3} - \sqrt{2}}{\sqrt{2}a_{n-3}} \right) - \frac{a_{n-2} - \sqrt{2}}{\sqrt{2}a_{n-2}} \\ &= - \frac{\sqrt{2} - a_{n-1}^{-1}}{2a_{n-2}} (a_{n-2} - \sqrt{2}) + \frac{1}{2a_{n-1}a_{n-3}}(a_{n-3} - \sqrt{2}). \end{align*}

From this and the bounds for $(a_n)$ altogether, we get

$$ \left|a_n - \sqrt{2}\right| \leq \frac{\sqrt{2}-\frac{1}{2}}{2}\left|a_{n-2}-\sqrt{2}\right| + \frac{1}{2}\left|a_{n-3} - \sqrt{2}\right|. $$

From this, we find that $\varepsilon := \limsup_{n\to\infty} |a_n - \sqrt{2}|$ satisfies

$$ \varepsilon \leq \frac{\sqrt{2}-\frac{1}{2}}{2} \varepsilon + \frac{1}{2} \varepsilon = \frac{2\sqrt{2}+1}{4} \varepsilon. $$

Since $\frac{2\sqrt{2}+1}{4} < 1$, this inequality is satisfied only when $\varepsilon = 0$, and therefore the claim follows.