数列\(\{a_n\}\)满足\(a_1=1\)，且\(a_2=\sqrt 5\)，且当\(n\geqslant 2\)时，\(\dfrac{a_{n+1}}{a_n}=\dfrac{a_n^2}{a_{n-1}^2}-2\)，求证：\[\dfrac{1}{a_1}+\dfrac{1}{a_2}+\dfrac{1}{a_3}+\cdots+\dfrac{1}{a_n} <\dfrac{1+\sqrt 5}{2}.\]
解 记\(\dfrac{a_{n+1}}{a_n}=b_n\)，则\[b_n=b_{n-1}^2-2,n\geqslant 2.\]
由于\(b_1=\dfrac{a_2}{a_1}=\sqrt 5\)，于是容易证明\[\forall n\in\mathcal N^*,b_n>2,\]因此可以设\(b_n=x_n+\dfrac{1}{x_n}\)，且\(x_n>1\)，并记\(\{x_n\}\)的首项为\(\alpha=\dfrac{1+\sqrt 5}{2}\)．
于是递推式变为\[x_n+\dfrac{1}{x_n}=x_{n-1}^2+\dfrac{1}{x_{n-1}^2},n\geqslant 2,\]即\[x_n=x_{n-1}^2,\]可得\[x_n=\alpha^{2^{n-1}},\]进而\[b_n=\alpha^{2^{n-1}}+\dfrac{1}{\alpha^{2^{n-1}}}.\]
进而计算\[\begin{split}a_n&=a_1\cdot b_1\cdot b_2\cdots b_{n-1}\\&=\left(\alpha+\dfrac{1}{\alpha}\right)\cdot\left(\alpha ^2+\dfrac{1}{\alpha ^2}\right)\cdots\left(\alpha ^{2^{n-2}}+\dfrac{1}{\alpha ^{2^{n-2}}}\right)\\&=\dfrac{1}{\alpha-\dfrac{1}{\alpha}}\cdot\left(\alpha^{2^{n-1}}-\dfrac{1}{\alpha^{2^{n-1}}}\right)\\&=\alpha^{2^{n-1}}-\dfrac{1}{\alpha^{2^{n-1}}}.\end{split}\]
注意到\[\begin{split}\dfrac{1}{a_n}&=\dfrac{1}{\alpha^{2^{n-1}}-\dfrac{1}{\alpha^{2^{n-1}}}}\\&=\dfrac{\alpha^{2^{n-1}}}{\alpha^{2^n}-1}\\&=\dfrac{1}{\alpha^{2^{n-1}}-1}-\dfrac{1}{\alpha^{2^n}-1},\end{split}\]
显然有\[\dfrac{1}{a_1}+\dfrac{1}{a_2}+\dfrac{1}{a_3}+\cdots+\dfrac{1}{a_n}<\dfrac{1}{\alpha -1}=\dfrac{1+\sqrt 5}{2},\]因此原不等式得证．
注 对于\(a_{n+1}=a_n^2-2\)型的递推数列，可以改写为\[\dfrac{1}{2}a_{n+1}=2\cdot\left(\dfrac 12a_n\right)^2-1,\]当初值在\([-1,1]\)内时，可以令\[\dfrac 12a_n=\cos\theta_n,\]当初值在\([-1,1]\)外时，可以令\[\dfrac 12a_n={\mathrm {ch}} x_n.\]