已知数列$\{a_n\}$满足$a_n\geqslant 1$，$a_1=1$，$a_2=1+\dfrac{\sqrt 2}2$，$\left(\dfrac{a_n}{a_{n+1}-1}\right)^2+\left(\dfrac{a_n-1}{a_{n-1}}\right)^2=2$，求证：$$\dfrac 23<\dfrac{a_n}n<\dfrac 23\left(1+\dfrac 1n\right).$$
证明 记$b_n=\left(\dfrac{a_{n-1}}{a_n-1}\right)^2$，则$b_{n+1}+\dfrac{1}{b_n}=2$，从而利用不动点改造递推公式，可得当$n\geqslant 2$时，有$$\dfrac{1}{b_{n+1}-1}=\dfrac{1}{b_n-1}+1,$$又$b_2=2$，于是$\dfrac{1}{b_n-1}=n-1$，即$b_n=\dfrac{n}{n-1}$($n\geqslant 2$)．这样就有$$\dfrac{a_n}{a_{n+1}-1}=\dfrac{\sqrt{n+1}}{\sqrt n},$$也即$$\sqrt{n+1}\cdot a_{n+1}-\sqrt n\cdot a_n=\sqrt{n+1},$$累加可得$$\sqrt n\cdot a_n=1+\sqrt 2+\cdots +\sqrt n,$$于是$a_n=\displaystyle\sum\limits_{i=1}^{n}\sqrt{\dfrac{i}n}$．由于函数$f(x)=\sqrt x$单调递增，于是有$$\sum_{i=1}^n\left(\sqrt{\dfrac in}\cdot \dfrac 1n\right)>\int_{0}^1\sqrt{x}{\ \rm d}x=\dfrac 23.$$又由于函数$f(x)=\sqrt x$上凸，于是有$$\sum_{i=1}^n\left(\sqrt{\dfrac in}\cdot \dfrac 1n\right)<\int_{0}^1\sqrt{x}{\ \rm d}x+\dfrac {f(1)-f(0)}{2n}=\dfrac 23+\dfrac{1}{2n}.$$综上所述，原命题得证．