求证：$\dfrac{1}{{n+1}}+\dfrac{1}{{n+2}}+\cdots+\dfrac{1}{{2n}}<\dfrac{3}{4}$．
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【答案】
略
【解析】
倒序相加，有$$\begin{cases}
S=\dfrac{1}{{n+1}}+\dfrac{1}{{n+2}}+\cdots+\dfrac{1}{{2n}} \\
S=\dfrac{1}{{2n}}+\dfrac{1}{{2n-1}}+\cdots+\dfrac{1}{{n+1}} \\
\end{cases}$$从而\[\begin{split}2S &=\left( {\dfrac{1}{{n+1}}+\dfrac{1}{{2n}}} \right)+\cdots+\left( {\dfrac{1}{k}+\dfrac{1}{{3n+1-k}}} \right)+\cdots+\left( {\dfrac{1}{{2n}}+\dfrac{1}{{n+1}}} \right)\\
&<n \cdot \left( {\dfrac{1}{{n+1}}+\dfrac{1}{{2n}}} \right)\\
&<\dfrac{3}{2}\
,\end{split}\]于是原不等式得证．
S=\dfrac{1}{{n+1}}+\dfrac{1}{{n+2}}+\cdots+\dfrac{1}{{2n}} \\
S=\dfrac{1}{{2n}}+\dfrac{1}{{2n-1}}+\cdots+\dfrac{1}{{n+1}} \\
\end{cases}$$从而\[\begin{split}2S &=\left( {\dfrac{1}{{n+1}}+\dfrac{1}{{2n}}} \right)+\cdots+\left( {\dfrac{1}{k}+\dfrac{1}{{3n+1-k}}} \right)+\cdots+\left( {\dfrac{1}{{2n}}+\dfrac{1}{{n+1}}} \right)\\
&<n \cdot \left( {\dfrac{1}{{n+1}}+\dfrac{1}{{2n}}} \right)\\
&<\dfrac{3}{2}\
,\end{split}\]于是原不等式得证．
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