As
$$\begin{aligned} a_{n+1} &= S_{n+1} - S_n \\ &= \frac{n}{(n+1)(n+2)} - a_{n+1} - \frac{n-1}{n(n+1)} + a_n, \end{aligned}$$
we have
$$\begin{aligned} 2a_{n+1} &= \frac{n+2-2}{(n+1)(n+2)} - \frac{1}{n+1} + \frac{1}{n(n+1)} + a_n \\ &= \frac{-2}{(n+1)(n+2)} + a_n + \frac{1}{n(n+1)}. \end{aligned}$$
Therefore,
$$a_{n+1} + \frac{1}{(n+1)(n+2)} = \frac{1}{2} \left( a_n + \frac{1}{n(n+1)} \right).$$
Define $b_n = a_n + \frac{1}{n(n+1)}$. It is easy to see that $b_n = \frac{1}{2^{n-1}}b_1$, $b_1 = a_1 + \frac{1}{2}$. On the other hand, from $S_1 + a_1 = 2a_1 = 0$ we get $a_1 = 0$. So $b_1 = \frac{1}{2}$, $b_n = \frac{1}{2^n}$. Therefore,
$$a_n = b_n - \frac{1}{n(n+1)} = \frac{1}{2^n} - \frac{1}{n(n+1)}.$$