证明：$\displaystyle\sum\limits_{k=1}^{2^n} {\dfrac{1}{n}} >1+\dfrac n2$．
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【答案】
略
【解析】
适当分组后放缩\[\begin{split}
\sum\limits_{k=1}^{{2^n}} {\dfrac{1}{k}}
&=1+\dfrac{1}{2}+\left( {\dfrac{1}{3}+\dfrac{1}{4}} \right)+\left( {\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}} \right)+\cdots+\left( {\dfrac{1}{{{2^{n-1}}+1}}+\cdots+\dfrac{1}{{{2^n}}}} \right)\\
&>1+\dfrac{1}{2}+\left( {\dfrac{1}{4}+\dfrac{1}{4}} \right)+\left( {\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}} \right)+\cdots+\left( {\dfrac{1}{{{2^n}}}+\dfrac{1}{{{2^n}}}+\cdots+\dfrac{1}{{{2^n}}}} \right)\\
&=1+\underbrace {\dfrac{1}{2}+\dfrac{1}{2}+\cdots+\dfrac{1}{2}}_{n \text{个}}\\
&>1+\dfrac{n}{2},\end{split}\]于是级数 $\displaystyle \sum\limits_{n=1}^\infty {\dfrac{1}{n}} $ 发散．
\sum\limits_{k=1}^{{2^n}} {\dfrac{1}{k}}
&=1+\dfrac{1}{2}+\left( {\dfrac{1}{3}+\dfrac{1}{4}} \right)+\left( {\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}} \right)+\cdots+\left( {\dfrac{1}{{{2^{n-1}}+1}}+\cdots+\dfrac{1}{{{2^n}}}} \right)\\
&>1+\dfrac{1}{2}+\left( {\dfrac{1}{4}+\dfrac{1}{4}} \right)+\left( {\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}} \right)+\cdots+\left( {\dfrac{1}{{{2^n}}}+\dfrac{1}{{{2^n}}}+\cdots+\dfrac{1}{{{2^n}}}} \right)\\
&=1+\underbrace {\dfrac{1}{2}+\dfrac{1}{2}+\cdots+\dfrac{1}{2}}_{n \text{个}}\\
&>1+\dfrac{n}{2},\end{split}\]于是级数 $\displaystyle \sum\limits_{n=1}^\infty {\dfrac{1}{n}} $ 发散．
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