13. 已知数列 $\left\{a_{n}\right\}$ 满足 $a_{1}=-1, a_{n+1}=a_{n}+\frac{1}{n(n+1)}, n \in \mathbf{N}^{*}$, 则通项公式 $a_{n}=$
$\because a_{n+1}-a_{n}=\frac{1}{n(n+1)}$,

$\therefore a_{2}-a_{1}=\frac{1}{1 \times 2}$;

$a_{3}-a_{2}=\frac{1}{2 \times 3}$;

$a_{4}-a_{3}=\frac{1}{3 \times 4}$;

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$a_{n}-a_{n-1}=\frac{1}{(n-1) n} ;$

以上各式累加得, $a_{n}-a_{1}=\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\cdots+\frac{1}{(n-1) n}$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdots+\frac{1}{n-1}-\frac{1}{n}$

$=1-\frac{1}{n}$.

$\therefore a_{n}+1=1-\frac{1}{n}, \therefore a_{n}=-\frac{1}{n}$.