已知数列$\{a_n\}$满足$a_1=1$，$a_{n+1}a_n-1=a_n^2$．
(1) 求证：$\sqrt{2n-1}\leqslant a_n\leqslant\sqrt{3n-2}$；
(2) 求整数$m$，使得$\left|a_{2005}-m\right|$的值最小．
分析与解 (1) 根据题意，有$a_{n+1}=a_n+\dfrac{1}{a_n}$，于是$$a_{n+1}^2-a_n^2=2+\dfrac{1}{a_n^2}.$$显然$\{a_n\}$单调递增，于是$a_n^2\geqslant 1$，从而$$2\leqslant a_{n+1}^2-a_n^2\leqslant 3,$$进而结合$a_1^2=1$，可得$$2n-1\leqslant a_n^2\leqslant 3n-2,$$因此命题得证．
(2) 在$a_{n+1}^2-a_n^2=2+\dfrac{1}{a_n^2}$中分别令$n=2,3,\cdots ,2004$，叠加可得$$a_{2005}^2-a_2^2=2(2005-2)+\dfrac{1}{a_2^2}+\dfrac 1{a_3^2}+\cdots +\dfrac{1}{a_{2004}^2}.$$根据(1)的结果，有$$\dfrac{1}{a_2^2}+\dfrac 1{a_3^2}+\cdots +\dfrac{1}{a_{2004}^2}<\dfrac 13+\dfrac 15+\cdots +\dfrac{1}{2\cdot 2004-1}.$$而\[\begin{split} \dfrac 13+\dfrac 15+\cdots +\dfrac{1}{2\cdot 2004-1}&<\dfrac 12+\dfrac 14+\cdots +\dfrac{1}{4006}\\
&=\dfrac 12\left(1+\dfrac 12+\dfrac 13+\dfrac 14+\dfrac 15+\cdots +\dfrac 1{1024}+\dfrac{1}{1025}+\cdots +\dfrac{1}{2003}\right)\\
&<\dfrac 12\left(1+\dfrac 12+\dfrac 12+\dfrac 14+\dfrac 14+\cdots +\dfrac 1{1024}+\dfrac 1{1024}+\cdots +\dfrac 1{1024}\right)\\
&<\dfrac {11}2,\end{split} \]于是可得$$4009=2\cdot 2005-1\leqslant a_{2005}^2<2(2005-2)+\dfrac{11}2+2^2=4015.5,$$而$63^2=3969$，$63.5^2=4032.25$，于是所求的正整数为$63$．
注 利用已经得到的界进行反复估计可以得到更加精确的界．