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记 ${\displaystyle t_n={\displaystyle 1+\frac{1}{2}+\cdots +\frac{1}{n}}}$, 证明: ${\displaystyle t_n-{\displaystyle \frac{5}{6}<\ln (n+1)<t_n}}$.

首先利用 ${\displaystyle \ln x\le x-1}$, 有$$\begin{array}{r}\ln (n+1)=\sum _{i=1}^n\ln \frac{i+1}{i}<\sum _{i=1}^n(\frac{i+1}{i}-1)=t_n.\end{array}$$
再利用${\displaystyle \ln x\ge 1-{\displaystyle \frac{1}{x}}}$, 有$$\begin{array}{r}\ln (n+1)=\sum _{i=2}^n\ln \frac{i+1}{i}+\ln 2>\sum _{i=2}^n(1-\frac{i}{i+1})+\ln 2=t_n+\frac{1}{n+1}-\frac{3}{2}+\ln 2>t_n-\frac{5}{6},\end{array}$$
其中依据熟知的${\displaystyle \ln 2>{\displaystyle \frac{2}{3}}}$(可由${\displaystyle \ln x\ge {\displaystyle \frac{2(x-1)}{x+1},x\ge 1}}$推得)