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$a_{n+1}=a_n+dfrac{1}{a_n},nin N^*,$两边平方得 $a_{n+1}^2=a_n^2+2+dfrac{1}{a_n^2},$于是$a_{n+1}^2-a_{n}^2ge2,$从而$a_{n+1}^2ge2n+1,$即$a_{n+1}gesqrt{2n+1}.$
又由于$a_{n+1}^2-a_n^2=2+dfrac{1}{a_n^2},$ 于是
[egin{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$$a_{n+1}lesqrt{2n+ln n+2}.$$
综上$$sqrt{2n+1}le a_{n+1}lesqrt{2n+ln n+2},$$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$[a_{2017}]=63$
$ extbf{注:从上面的推导我们可以到这个数列通项的大致的估计}$
$limlimits_{n o+infty}dfrac{a_n}{sqrt n}=sqrt 2.$