已知数列$\{a_n\}$中$a_1=2$，$a_{n+1}=\left(\sqrt 2-1\right)(a_n+2)$，$n=1,2,3,\cdots $．
(1) 求$\{a_n\}$的通项公式；
(2) 若数列$\{b_n\}$中，$b_1=2$，$b_{n+1}=\dfrac{3b_n+4}{2b_n+3}$，证明：$\sqrt 2<b_n\leqslant a_{4n-3}$，$n=1,2,3,\cdots$．
分析与解 (1) 递推公式即$$a_{n+1}-\sqrt 2=\left(\sqrt 2-1\right)\left(a_n-\sqrt 2\right),$$进而不难求出$a_n=\sqrt 2\cdot \left[\left(\sqrt 2-1\right)^n+1\right]$，$n=1,2,3,\cdots $．
(2) 设$f(x)=\dfrac{3x+4}{2x+3}$，则当$x\in \left(\sqrt 2,2\right]$时，$f(x)\in \left(\sqrt 2,2\right]$，且函数$f(x)$在区间$\left(\sqrt 2,2\right]$上单调递增．因此不难证明$\{b_n\}$单调递减（由$b_{2}<b_{1}$，可以得到$f(b_{2})=b_{3}<f(b_{1})=b_{2}$，所以由数学归纳法容易证得$\{b_{n}\}$单调递减），且$b_n\in \left(\sqrt 2,2\right]$．利用$f(x)=x$的不动点$x=\sqrt 2$将递推公式改造为\[\begin{split} b_{n+1}-\sqrt 2&=\dfrac{3-2\sqrt 2}{2b_n+3}\cdot \left(b_n-\sqrt 2\right)\\
&<\dfrac{3-2\sqrt 2}{2\sqrt 2+3}\cdot \left(b_n-\sqrt 2\right)\\
&=\left(\sqrt 2-1\right)^4\cdot \left(b_n-\sqrt 2\right),\end{split} \]于是取$n=1,2,3,\cdots ,n-1$累乘可得$$b_{n}-\sqrt 2\leqslant \left(\sqrt 2-1\right)^{4n-4}\cdot \left(2-\sqrt 2\right)=a_{4n-3}-\sqrt 2,$$因此原命题得证．
(2)另法 由$f(x)=\dfrac {3x+4}{2x+3}=x$解得不动点$x=\pm\sqrt 2$．通过不动点改造数列得$$\begin{split} b_{n+1}-\sqrt 2=(3-2\sqrt 2)\cdot\dfrac {b_n-\sqrt 2}{2b_n+3},\\b_{n+1}+\sqrt 2=(3+2\sqrt 2)\cdot\dfrac {b_n+\sqrt 2}{2b_n+3}\end{split} $$两式相比得$$\dfrac {b_{n+1}-\sqrt 2}{b_{n+1}+\sqrt 2}=\dfrac {3-2\sqrt 2}{3+2\sqrt 2}\cdot\dfrac {b_n-\sqrt 2}{b_n+\sqrt 2}.$$于是得$$\dfrac {b_n-\sqrt 2}{b_n+\sqrt 2}=\dfrac {2-\sqrt 2}{2+\sqrt 2}\cdot\left(\dfrac {3-2\sqrt 2}{3+2\sqrt 2}\right )^{n-1}=(\sqrt 2-1)^{4n-2}.$$从而解得$b_n=\dfrac {1+m^{4n-2}}{1-m^{4n-2}}\cdot\sqrt 2$，其中$m=\sqrt 2-1$．直接可以得到$b_n>\sqrt 2$，右边不等式等价于证明$$\dfrac {1+m^{4n-2}}{1-m^{4n-2}}\leqslant 1+m^{4n-3}.$$整理得$$m^{4n-3}-2m^{4n-2}-m^{8n-5}=m^{4n-3}(1-2m-m^{4n-2})\geqslant 0,$$也即证明$$m^{4n-2}=(\sqrt 2-1)^{4n-2}\leqslant 1-2m=(\sqrt 2-1)^2.$$因为$4n-2\geqslant 2$，所以此不等式显然成立．