裂项相消

已知数列 { a n } \{a_n\} {an​} 的前 $n$ 项和为 $S_n$ ，且满足 $S_n=a_n+n^2+1,(n\in N^{\ast })$.
(1) 求 { a n } \{a_n\} {an​} 的通项公式；
(2) 求证： $\frac{1}{S_1}+\frac{1}{S_2}+\cdots +\frac{1}{S_n}<\frac{3}{4}$.

[解析]
(1) 由$$S_n=a_n+n^2+1.$$得$$S_{n+1}=a_{n+1}+(n+1{)}^2+1.$$两式相减得$$a_n=2n+1.$$
(2)由(1)得 $S_n=n(n+2)$，所以

$\frac{1}{S_1}+\frac{1}{S_2}+\cdots +\frac{1}{S_n}$

$=\frac{1}{1\times 3}+\frac{1}{2\times 4}+\frac{1}{3\times 5}+\cdots +\frac{1}{(n-1)\times (n+1)}+\frac{1}{n\times (n+2)}$

$=\frac{1}{2}[(1-\frac{1}{3})+(\frac{1}{2}-\frac{1}{4})+(\frac{1}{3}-\frac{1}{5})+\cdots +(\frac{1}{n-1}-\frac{1}{n+1})+(\frac{1}{n}-\frac{1}{n+2})]$

$=\frac{1}{2}(1+\frac{1}{2}-\frac{1}{n+1}-\frac{1}{n+2})$

$=\frac{3}{4}-\frac{1}{2(n+1)}-\frac{1}{2(n+2)}<\frac{3}{4}.$