已知数列$\{\dfrac{1}{n}\}$的前$n$项和为$S_n$,则下面选项正确的是( )
A.$S_{2018}-1>\ln 2018$
B.$S_{2018}-1<\ln 2018$
C.$\ln2018<S_{1009}-1$
D.$\ln2018>S_{2017}$
分析:这里主要考察$\dfrac{x}{1+x}\le\ln(1+x)\le x$
令$x=\dfrac{1}{n}$累加易得$\dfrac{1}{2}+\dfrac{1}{3}\dots+\dfrac{1}{n+1}<\ln(n+1)<1+\dfrac{1}{2}+\dfrac{1}{3}\dots+\dfrac{1}{n}$
易得答案选B
练习:证明:当$n\in N^+$时$\dfrac{1}{n+1}+\dfrac{1}{n+2}+\dots+\dfrac{1}{3n+1}<\dfrac{9}{8}$
提示:$\ln(1+x)\ge\dfrac{2x}{2+x}=\dfrac{1}{k}$
原文：https://www.cnblogs.com/mathstudy/p/10355478.html