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?
 A: Note that $2<a_n<3$ for $n\ge5$. Since, for $n\ge5$,
\begin{eqnarray} 
&&|a_{n+3}-a_{n+2}|\\
&=&\bigg|\frac{1}{a_{n+2}^2}-\frac{1}{a_{n}^2}\bigg|=\frac{(a_{n+2}+a_{n})|a_{n+2}-a_{n}|}{a_{n+2}^2a_{n}^2}\\
&\le&\frac{6}{2^2\cdot 2^2}|a_{n+2}-a_{n}|\le\frac12|a_{n+2}-a_{n}|\\
&\le&\cdots\le\frac{1}{2^{n-2}}|a_5-a_3|
\end{eqnarray}
we conclude that $\{a_n\}$ is Cauchy and hence $\lim_{n\to\infty}a_n=L$ exists.
A: Another approach: Let $L \approx 2.3593$ be the unique solution of $L = 2 + 2/L^2$ in the interval $[2, 2.5]$, and $b_n = a_n - L$. We want to show that $b_n \to 0$.
The recursion formula becomes
$$
 b_{n+2} = \frac{1}{(L+b_{n+1})^2} + \frac{1}{(L+b_{n})^2} - \frac{2}{L^2} \, .
$$
We estimate
$$
 \left| \frac{1}{(L+b_{n})^2} - \frac{1}{L^2}\right| = \frac{|b_n|(2L+b_n)}{L^2(L+b_n)^2} \le \frac{5}{16} |b_n|
$$
for $n \ge 3$, so that
$$
|b_{n+2} | \le \frac{5}{16} (|b_{n+1}| + |b_n|) \, . 
$$
Then
$$
 2 |b_{n+2} | + |b_{n+1} | \le \frac{13}{8}|b_{n+1} | + \frac{5}{8}|b_{n}|
\le \frac{13}{16} \left(2 |b_{n+1} | + |b_{n} | \right) \, .
$$
This shows that $2 |b_{n+1} | + |b_{n} |$ decreases geometrically to zero. It follows that $b_n \to 0$, as desired.
A: 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.
A: Let's use fixed point iterations.
Observe:
Substituting $x_n > 2$ gives us a upper bound on $x_{n}$ ($n \geq 3$). And substituting this upper bound back, gives us a 'closer' lower bound. And so on. I hope to show that these lower and upper bounds converge, which effectively then says that $x_n$ must converge to that point eventually.
Define $g(x) := 2(1 + \frac{1}{x^2})$. Then the sequence we consider is $t_{n+1} = g(t_n)$. Using the fixed point iteration theorem, we have
(1) $t \in [2,3] \Rightarrow g(t) \in [2,3]$, and $g$ is continuous.
(2) $|g'(t)| = \frac{4}{t^3} < 1$
Hence $\exists \alpha \in (2,3)$ s.t. $\forall t_0, \; \in (2,3)$ $ t_n =  g^{(n)}(t_0) \rightarrow \alpha$.
Claim 1: Let $N_0, t > 0$.

*

*$(\forall n \geq N_0, \; x_n > t) \Rightarrow (\forall n \geq N_0 + 2, \; x_n < g(t))$.


*$(\forall n \geq N_0, \; x_n < t) \Rightarrow (\forall n \geq N_0 + 2, \; x_n > g(t))$
Proof. Trivial. $\square$
Claim 2: $\forall n \geq 5, \; x_n \in (2,3)$.
Proof. $(\forall n > 0, \; x_n > 0) \Rightarrow (\forall n > 0, \; x_{n+2} > 2) \Rightarrow (\forall n > 0, \; x_{n+4} < \frac{5}{2} < 3)$ $\square$
Claim 3. Let $S = \lim \sup x_n$ and $I = \lim \inf x_n$. Then
$S \leq \alpha \land I \geq \alpha$.
Proof. Fix $t \in \mathbb{N}$. As $g^{(n)}(x_5) \rightarrow \alpha$, $\exists N'$ s.t. $n \geq \max(N', t) \Rightarrow g^{(n)}(x_5) \in (\alpha - \frac{1}{t}, \alpha + \frac{1}{t})$.
Then by Claim 1 and 2, $\sup_{m \geq (5 + 2\max(t,N')) } x_m \le \alpha + \frac{1}{t} \Rightarrow \lim_{t \rightarrow \infty} \sup_{m \geq (5 + 2\max(t,N')) } x_m \le \lim_{t \rightarrow \infty} (\alpha + \frac{1}{t}) \Rightarrow S \le \alpha$.
Similarly, show $I \ge \alpha$. $\square$
Corollary. Thus $\lim_n x_n = S = I = \alpha$.
Proof. Trivial. $\square$
Edit 2. Thanks to @Martin R, to help me formalize the proof.
Edit 1. As @NN2's answer notes, this is actually an attracting stable fixed point, which I've proved using fixed point iterations. (Alternatively, you could just use the theorem (which I didn't know) - check NN2's answer).
