8. In the sequence $\left\{a_{n}\right\}$, if $a_{1}=1, a_{n} \neq 0\left(n \in \mathbf{N}_{+}\right)$, and $a_{n}+2 a_{n} a_{n-1}-a_{n-1}=$ $0\left(n \geqslant 2, n \in \mathbf{N}_{+}\right)$, then $a_{n}=$ $\qquad$
8. $a_{n}=\frac{1}{2 n-1}$.

Dividing both sides by $a_{n} a_{n-1}$ yields $\frac{1}{a_{n}}-\frac{1}{a_{n-1}}=2$, hence $\frac{1}{a_{n}}=\frac{1}{a_{1}}+(n-1) 2=2 n-1$. Therefore, $a_{n}=\frac{1}{2 n-1}$.