已知数列 $\{a_n\}$ 满足 $a_1=1$，$a_{n+1}=a_n+\dfrac{1}{2a_n}$，则 $\lim\limits_{n\to +\infty}\left(a_n-\sqrt n\right)=$ ______．
答案 $0$．
解析 根据题意，有\[a_{n+1}^2-a_n^2=1+\dfrac{1}{4a_n^2},\]于是\[a_n^2\geqslant n,\]进而\[a_{n+1}^2\leqslant n+1+\sum_{k=1}^n\dfrac{1}{4k},\]因此\[\begin{split} a_n-\sqrt n&\leqslant \sqrt{n+\sum_{k=1}^{n-1}\dfrac{1}{4k}}-\sqrt n\\ &\leqslant \sqrt{n+\dfrac 14(1+\ln n)}-\sqrt n\\ &=\dfrac{\dfrac 14(1+\ln n)}{\sqrt{n+\dfrac 14(1+\ln n)}+\sqrt n}\\ &<\dfrac{1+\ln n}{8\sqrt n},\end{split}\]而\[\lim_{n\to +\infty}\dfrac{1+\ln n}{8\sqrt n}=0,\]于是\[\lim\limits_{n\to +\infty}\left(a_n-\sqrt n\right)=0.\]