8-26 Given the sequence $\left\{a_{n}\right\}$ satisfies
$$a_{1}=\frac{1}{2}, a_{1}+a_{2}+\cdots+a_{n}=n^{2} a_{n}, n \geqslant 1 .$$

Find the general term formula for $a_{n}$.
[Solution] When $n \geqslant 2$, since
$$\begin{array}{l}
a_{1}+a_{2}+\cdots+a_{n}=n^{2} a_{n} \\
a_{1}+a_{2}+\cdots+a_{n-1}=(n-1)^{2} a_{n-1}
\end{array}$$

Therefore,
we can get
$$a_{n}=n^{2} a_{n}-(n-1)^{2} a_{n-1} .$$

Thus, we obtain $\quad a_{n}=\frac{1}{n(n+1)}, n=1,2,3, \cdots$