2. Given the sequence of real numbers $\left\{a_{n}\right\}$ defined as
$$
a_{0}=\frac{1}{2}, a_{n+1}=\frac{1}{2}\left(a_{n}+\frac{1}{5 a_{n}}\right)(n \in \mathbf{N}) \text {. }
$$

Let $A_{n}=\frac{5}{5 a_{n}^{2}-1}$. Then $\left\{A_{n}\right\}$ contains a perfect square $\qquad$.

2. Infinite.

Let $a_{n}=\frac{p_{n}}{q_{n}}\left(p_{n}, q_{n} \in \mathbf{N}_{+},\left(p_{n}, q_{n}\right)=1\right)$. Then $a_{n+1}=\frac{p_{n+1}}{q_{n+1}}=\frac{1}{2}\left(\frac{p_{n}}{q_{n}}+\frac{q_{n}}{5 p_{n}}\right)=\frac{5 p_{n}^{2}+q_{n}^{2}}{10 p_{n} q_{n}}$
Given $a_{0}=\frac{1}{2}$, we get $a_{1}=\frac{9}{20}$.
If $5 \mid q_{n}$, then from $\frac{p_{n+1}}{q_{n+1}}=\frac{p_{n}^{2}+\frac{q_{n}^{2}}{5}}{2 p_{n} q_{n}}$, we know $5 \mid q_{n+1}$.
Therefore, when $n \geqslant 1$, $5 \mid q_{n}$.
Also, from equation (1), we know that when $n \geqslant 1$, $p_{n}$ is odd and $q_{n}$ is even. Thus,
$p_{n+1}=p_{n}^{2}+\frac{q_{n}^{2}}{5}, q_{n+1}=2 p_{n} q_{n}$.
Then $A_{n}=\frac{5}{5 a_{n}^{2}-1}=\frac{5 q_{n}^{2}}{5 p_{n}^{2}-q_{n}^{2}}=\frac{q_{n}^{2}}{p_{n}^{2}-\frac{q_{n}^{2}}{5}}$.
By induction, we know $p_{n}^{2} \frac{q_{n}^{2}}{5}=1$.
Therefore, $A_{n}=q_{n}^{2}(n \geqslant 1)$ is a perfect square.