MT121位差

Posted 2020-10-12 青春的记忆

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$[a_n]$_____

 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$[a_{2017}]=63$

$\\textbf{注:从上面的推导我们可以到这个数列通项的大致的估计}$

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