Prove $\{a_n\}$ converges. 
Suppose $a_1,a_2>0$ and
$a_{n+2}=2+\dfrac{1}{a_{n+1}^2}+\dfrac{1}{a_n^2}(n\ge 1)$. Prove
$\{a_n\}$ converges.

First, we may show $\{a_n\}$ is bounded for $n\ge 3$, since $$2 \le a_{n+2}\le 2+\frac{1}{2^2}+\frac{1}{2^2}=\frac{5}{2},~~~~~~ \forall n \ge 1.$$
But how to go on?

I'm not so sure about this one, but here's my attempt-
\begin{align}
\lVert a_{n+1}-a_n  \rVert &= \left\lVert \frac{1}{a_n^2}-\frac{1}{a_{n-2}^2}  \right\rVert\\
&\leq \frac{ \lVert {a_{n-2}^2}-a_n^2   \rVert }{16} \\
&= \frac{ \lVert {a_{n-2}+a_{n}} \rVert \lVert {a_{n-2}-a_{n}}   \rVert }{16} \\
&\leq \frac{6}{16} \lVert {a_{n-2}-a_{n}} \rVert \\
&=\frac{6}{16} \left\lVert \frac{1}{a_{n-3}^2} + \frac{1}{a_{n-4}^2} - \frac{1}{a_{n-1}^2} - \frac{1}{a_{n-2}^2} \right\rVert \\
&=\frac{6}{16} \left\lVert \left(\frac{1}{a_{n-3}^2} - \frac{1}{a_{n-1}^2}\right)  + \left(\frac{1}{a_{n-4}^2} - \frac{1}{a_{n-2}^2}\right) \right\rVert \\
&\leq \frac{6}{16} \cdot \frac{6}{16} \lVert (a_{n-1}-a_{n-3})+(a_{n-2}-a_{n-4})   \rVert\\
&\ \ \vdots \\
&\leq \left(\frac{6}{16}\right)^n max(|| a_4-a_2 ||,\lVert a_3-a_1||)2^n  \\
&=\left(\frac{12}{16}\right)^n max(|| a_4-a_2 ||,\lVert a_3-a_1||)
\end{align}
(Original image: https://i.stack.imgur.com/uBBqe.jpg )
Does this seem correct? I tried using Cauchy's convergence test and the fact that the tail lies between 2 and 3. If not, I hope it provides at least some clue to the correct path. The last step is a result of the observation that each term of $a_k - a_{k-2}$ should produce two similar terms(as seen while expanding), whose index decreases linearly.
Edit: I have added the term $max(|| a_4-a_2 ||,\lVert a_3-a_1||)$ instead of the original single term to take care of the parity issue.
