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?

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.
