6. The sequence $a_{1}, a_{2}, a_{3}, \cdots$ satisfies (1) $a_{1}=\frac{1}{2}$, (2) $a_{1}+a_{2}+\cdots+a_{n}=n^{2} a_{n}(n \geqslant 2)$, then $a_{n}=$ $\qquad$
6. $\frac{1}{n(n+1)}$.

When $n \geqslant 2$, since $a_{1}+a_{2}+\cdots+a_{n}=n^{2} a_{n}, a_{1}+a_{2}+\cdots+a_{n-1}=(n-1)^{2} a_{n-1}$, we have $a_{n}=n^{2} a_{n}-(n-1)^{2} a_{n-1}$.
From this, we get $a_{n}=\frac{n-1}{n+1} a_{n-1}, n=2,3, \cdots$
Thus, we obtain $\quad a_{n}=\frac{1}{n(n+1)}, n=1,2,3, \cdots$