MT【121】耐克数列的估计

 已知$\{a_n\}$满足$a_1=1,a_2=2,\dfrac{a_{n+2}}{a_n}=\dfrac{a_{n+1}^2+1}{a_n^2+1}$, 求$[a_{2017}]$_____

解：容易用累乘法得到$a_{n+1}=a_n+\dfrac{1}{a_n},n\in N^*,$两边平方得 $a_{n+1}^2=a_n^2+2+\dfrac{1}{a_n^2},$于是$a_{n+1}^2-a_{n}^2\ge2,$从而$a_{n+1}^2\ge2n+1,$即$a_{n+1}\ge\sqrt{2n+1}.$

又由于$a_{n+1}^2-a_n^2=2+\dfrac{1}{a_n^2},$ 于是
\[\begin{split} a_{n+1}^2-a_1^2&=2n+\dfrac{1}{a_1^2}+\dfrac{1}{a_2^2}+\cdots +\dfrac{1}{a_n^2}\\ &\le2n+1+\dfrac 13+\cdots +\dfrac{1}{2n-1}\\ & \leqslant 2n+1+\dfrac 12+\cdots +\dfrac 1n \\ &\le 2n+\ln n+1,\end{split} \]因此$$a_{n+1}\le\sqrt{2n+\ln n+2}.$$
综上$$\sqrt{2n+1}\le a_{n+1}\le\sqrt{2n+\ln n+2},$$进而可得
$\sqrt{4033}\le a_{2017}\le \sqrt{4034+ln(2016)}$

注意到$ln(2016)<ln(2^{11})<ln(e^{11})=11,63^2=3969,64^2=4096$则$[a_{2017}]=63$

$\textbf{注:从上面的推导我们可以到这个数列通项的大致的估计}$

$\lim\limits_{n\to+\infty}\dfrac{a_n}{\sqrt n}=\sqrt 2.$