已知数列 {an} 满足 a1=1,an+1=an+12an,则 limn→+∞(an−√n)= ______.
答案 0.
解析 根据题意,有a2n+1−a2n=1+14a2n,于是a2n⩾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.