求证：$\displaystyle\sum\limits_{k=1}^n {\dfrac{1}{{n+k}}}<\dfrac{{\sqrt 2 }}{2}$．
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【答案】
略
【解析】
利用不等式处理\[\begin{split}
\sum\limits_{k=1}^n {\dfrac{1}{{n+k}}} &< \sqrt {\sum\limits_{k=1}^n 1 \cdot \sum\limits_{k=1}^n {\dfrac{1}{{{{\left( {n+k} \right)}^2}}}} } \\
&= \sqrt {n \cdot \sum\limits_{k=n+1}^{2n} {\dfrac{1}{{{k^2}}}} } \\
&<\sqrt {n \cdot \sum\limits_{k=n+1}^{2n} {\left( {\dfrac{1}{{k-1}}-\dfrac{1}{k}} \right)} } \\
&= \sqrt {n \cdot \left( {\dfrac{1}{n}-\dfrac{1}{{2n}}} \right)} \\
&=\dfrac{{\sqrt 2 }}{2}\end{split}\]于是原不等式得证．
\sum\limits_{k=1}^n {\dfrac{1}{{n+k}}} &< \sqrt {\sum\limits_{k=1}^n 1 \cdot \sum\limits_{k=1}^n {\dfrac{1}{{{{\left( {n+k} \right)}^2}}}} } \\
&= \sqrt {n \cdot \sum\limits_{k=n+1}^{2n} {\dfrac{1}{{{k^2}}}} } \\
&<\sqrt {n \cdot \sum\limits_{k=n+1}^{2n} {\left( {\dfrac{1}{{k-1}}-\dfrac{1}{k}} \right)} } \\
&= \sqrt {n \cdot \left( {\dfrac{1}{n}-\dfrac{1}{{2n}}} \right)} \\
&=\dfrac{{\sqrt 2 }}{2}\end{split}\]于是原不等式得证．
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