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?

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.
