China Mathematical Competition

As $$\begin{array}{rl}{a}_{n+1}& =S_{n+1}-S_n\\ & =\frac{n}{(n+1)(n+2)}-a_{n+1}-\frac{n-1}{n(n+1)}+a_n,\end{array}$$ we have $$\begin{array}{rl}2a_{n+1}& =\frac{n+2-2}{(n+1)(n+2)}-\frac{1}{n+1}+\frac{1}{n(n+1)}+a_n\\ & =\frac{-2}{(n+1)(n+2)}+a_n+\frac{1}{n(n+1)}.\end{array}$$ 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, $$a_n=b_n-\frac{1}{n(n+1)}=\frac{1}{2^n}-\frac{1}{n(n+1)}.$$