(1)当\(|a_1|\leq 2\),此时\(2^{n-1}(|a_1|-2)<0<|a_n|\),得证
当\(|a_1|>2\),\(|a_n-\frac{a_{n+1}}{2}|\leq 1,2a_n-2\leq a_{n+1}\leq 2a_n+2\)
使用数学归纳法,假设\(2^{n-1}(|a_1|-2)<|a_n|,-a_n<2^{n-1}(|a_1|-2)<a_n\),证明\(-a_{n+1}<2^n(|a_1|-2)<a_{n+1}\)
\(-a_n<2^{n-1}(|a_1|-2),-2a_n<2^{n}(|a_1|-2)\),和\(2a_n-2\leq a_{n+1}\)相加得到\(-2-a_{n+1}\leq 2^{n}(|a_1|-2)\)
\(2^{n-1}(|a_1|-2)<a_n,2^n(|a_1|-2)<2a_n\)就是\(-2a_n<-2^n(|a_1|-2)\),和\(2a_n-2\leq a_{n+1}\)相加得到\(-2<-2^n(|a_1|-2)+a_{n+1},2^n(|a_1|-2)<a_{n+1}+2\)
所以\(-2-a_{n+1}\leq 2^{n}(|a_1|-2)<2+a_{n+1}\),得证
(2)