已知数列 $\{a_{n}\}$ 满足递推关系式：$2a_{n+1}=1-a_{n}^{2}(n\geqslant 1,n\in\mathbb N)$，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．