已知数列 {an} 满足递推关系式：2an+1=1−a2n(n⩾，且 0<a_{1}<1．
1、求 a_{3} 的取值范围．
2、用数学归纳法证明：当 n\geqslant 3 且 n\in\mathbb N 时，\left|a_{n}-\left(\sqrt 2-1\right)\right|<\dfrac{1}{2^{n}}．
3、若 b_{n}=\dfrac{1}{a_{n}}，求证：当 n\geqslant 3 且 n\in\mathbb N 时 \left|b_{n}-\left(\sqrt 2+1\right)\right|<\dfrac{12}{2^{n}}．
解析
1、因为a_{n+1}=\dfrac{1-a_{n}^{2}}{2},所以 0<a_{2}<\dfrac{1}{2}，\dfrac{3}{8}<a_{3}<\dfrac{1}{2}．
2、当 n=3 时，a_{3}-\left(\sqrt 2-1\right)\in\left(\dfrac{11}{8}-\sqrt 2,\dfrac{3}{2}-\sqrt 2\right)\supset \left(-\dfrac{1}{8},\dfrac{1}{8}\right),所以命题成立．
假设当命题对 n 成立，即\left|a_{n}-\left(\sqrt 2-1\right)\right|<\dfrac{1}{2^{n}}. 当在 n+1 时，\begin{split}\left|a_{n+1}-\left(\sqrt 2-1\right)\right|&=\left|\dfrac{1-a_{n}^{2}}{2}-\left(\sqrt 2-1\right)\right|\\&=\dfrac{1}{2}\left|a_{n}^{2}-\left(\sqrt 2-1\right)^{2}\right|\\&=\dfrac{1}{2}\left|a_{n}-\left(\sqrt 2-1\right)\right|\cdot \left|a_{n}+\left(\sqrt 2-1\right)\right| \\&<\dfrac{1}{2^{n+1}}\left|a_{n}+\left(\sqrt 2-1\right)\right|\end{split}因为-\dfrac{1}{2^{n}}<a_{n}-\left(\sqrt 2-1\right)<\dfrac{1}{2^{n}},所以\begin{split}-\dfrac{1}{2^{n}}+2\left(\sqrt 2-1\right)&<a_{n}+\left(\sqrt 2-1\right)\\ &<\dfrac{1}{2^{n}}+2\left(\sqrt 2-1\right).\end{split}因此 \left|a_{n}+\left(\sqrt 2-1\right)\right|<1，于是\left|a_{n+1}-\left(\sqrt 2-1\right)\right|<\dfrac{1}{2^{n+1}}. 综上，原命题得证．
3、以下讨论中 n\geqslant 3，由 (2)，\left|a_{n}-\left(\sqrt 2-1\right)\right|<\dfrac{1}{2^{n}} 即为\left|\dfrac{1}{\sqrt 2-1}-\dfrac{1}{a_{n}}\right|<\dfrac{1}{2^{n}}\cdot \dfrac{1}{\left(\sqrt 2-1\right)a_{n}}=\dfrac{\sqrt 2+1}{2^{n}}\cdot \dfrac{1}{\left|a_{n}\right|},而由\sqrt 2-1-\dfrac{1}{2^{n}}<a_{n}<\sqrt 2-1+\dfrac{1}{2^{n}},有\sqrt 2-1-\dfrac{1}{8}<a_{n}<\sqrt 2+1+\dfrac{1}{8}恒成立，于是\dfrac{1}{a_{n}}<\dfrac{1}{\sqrt 2-1-\dfrac{1}{8}}, \dfrac{\sqrt 2+1}{a_{n}}<\dfrac{\sqrt 2+1}{\sqrt 2-1-\dfrac{1}{8}}<12,所以\left|\dfrac{1}{a_{n}}-\left(\sqrt 2+1\right)\right|<\dfrac{12}{2^{n}},从而原不等式 \left|b_{n}-\left(\sqrt 2+1\right)\right|<\dfrac{12}{2^{n}}(n\geqslant 3,n\in\mathbb N) 得证．