Example 13 (2000 National College Entrance Examination Question) Let $\left\{a_{n}\right\}$ be a sequence of positive terms with the first term being 1, and $(n+1) a_{n+1}^{2}-$ $n a_{n}^{2}+a_{n} a_{n+1}=0(n=1,2,3, \cdots)$, then its general term $a_{n}=$ $\qquad$ .
Solve: Fill in $\frac{1}{n}$. Reason: The known recurrence relation can be transformed into $\left[(n+1) a_{n+1}-a_{n}\right]\left(a_{n+1}+a_{n}\right)=0$, but $a_{n+1}+a_{n}>0$, so $(n+1) a_{n+1}=n a_{n}$, which means $\frac{a_{n+1}}{a_{n}}=\frac{n}{n+1}$, thus
$$
a_{n}=\frac{a_{n}}{a_{n-1}} \cdot \frac{a_{n-1}}{a_{n-2}} \cdot \cdots \cdot \frac{a_{2}}{a_{1}} \cdot a_{1}=\frac{n-1}{n} \cdot \frac{n-2}{n-1} \cdot \cdots \cdot \frac{1}{2} \cdot 1=\frac{1}{n} .
$$