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原题：
设数列$\{a_n\}$满足$a_1=1,a_{n+1}=a_n+\frac{1}{a_n}(n\geq1)$.求$[a_{2019}]$
$a_{n+1}^2=a_n^2+\frac{1}{a_n^2}+2$
$a_n^2=a_1^2+\sum_{k=1}^{n-1}(2+\frac{1}{a_k^2})=2n+\sum_{k=2}^{n-1}\frac{1}{a_k^2}\gt 2n$
$a_{2019}^2=2\cdot 2019+\sum_{k=2}^{2018}\frac{1}{2k}=4038+\frac{1}{2}\sum_{k=2}^{2018}\ln (1+\frac{1}{k-1})=4038+\frac{1}{2}\ln2018(2018\lt2^{12}\lt e^{12})<4038+6\lt4096=64^2$
增强版：
求最小的正整数n,使得$[a_n]=64$ |
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