8. Let the sequence $a_{n}=\left[(\sqrt{2}+1)^{n}+\left(\frac{1}{2}\right)^{n}\right], n \geq 0$, where $[x]$ denotes the greatest integer less than or equal to $x$. Then
$$
\sum_{n=1}^{\infty} \frac{1}{a_{n-1} a_{n+1}}=
$$
8. $\frac{1}{8}$

Analysis: Let $A_{n}=(1+\sqrt{2})^{n}+(1-\sqrt{2})^{n}$, it is easy to know that $A_{n} \in \mathbb{Z}$, so $a_{n}=\left[A_{n}+\left(\frac{1}{2}\right)^{n}-(1-\sqrt{2})^{n}\right]=A_{n}+\left[\left(\frac{1}{2}\right)^{n}-(1-\sqrt{2})^{n}\right]=A_{n}=(1+\sqrt{2})^{n}+(1-\sqrt{2})^{n}$. Therefore, we know that $a_{n+1}=2 a_{n}+a_{n-1}$, and since $a_{0}=a_{1}=2$, we know that when $n \geq 1$, $a_{n}$ is increasing. Thus, $\sum_{i=1}^{\infty} \frac{1}{a_{i-1} a_{i+1}}=\sum_{i=1}^{\infty} \frac{a_{i}}{a_{i-1} a_{i} a_{i+1}}=\frac{1}{2} \sum_{i=1}^{\infty} \frac{a_{i+1}-a_{i-1}}{a_{i-1} a_{i} a_{i+1}}=\frac{1}{2} \sum_{i=1}^{\infty}\left(\frac{1}{a_{i-1} a_{i}}-\frac{1}{a_{i} a_{i+1}}\right)=\frac{1}{2} \cdot \frac{1}{a_{0} a_{1}}=\frac{1}{8}$