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设$M_n$是数列{$a_n$}的前n项之积，满足$M_n+a_n=1,n\in N^*$.
(1)求数列{$a_n$}的通项公式;
(2)设$S_n=M_1^2+M_2^2+\cdots +M_n^2$,求证:$-\frac{5}{12}\le S_n-a_{n+1}<-\frac{1}{3}$.
(1)通过特殊$a_1,a_2,a_3$猜到一般$a_n$，因此换元$1-a_n=b_n,c_n=\frac{1}{b_n}$，可得{$c_n$}为等差数列.
\[(2)S_n=\frac{1}{2^2}+\frac{1}{3^2}+\cdots +\frac{1}{(n+1)^2}< \frac{1}{4}+\frac{1}{9}+\frac{1}{4^2-0.5^2}+\cdots +\frac{1}{(n+1)^2-0.5^2}=\frac{1}{4}+\frac{1}{9}+\frac{1}{3.5}-\frac{1}{n+1.5}\] |
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