求证：$2\left(\sqrt{n+1}-1\right)<\displaystyle \sum\limits_{k=1}^n {\dfrac{1}{{\sqrt k }}} <2\sqrt n $
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【答案】
略
【解析】
利用积分放缩法有\[\begin{split}
\sum\limits_{k=1}^n {\dfrac{1}{{\sqrt k }}} &<\int_0^n {{x^{-\frac{1}{2}}}{\rm{d}}x}=\left. {2{x^{\frac{1}{2}}}} \right|_0^n=2\sqrt n \\
\sum\limits_{k=1}^n {\dfrac{1}{{\sqrt k }}} &> \int_1^{n+1} {{x^{-\frac{1}{2}}}{\rm{d}}x}=\left. {2{x^{\frac{1}{2}}}} \right|_1^{n+1}=2\left( {\sqrt {n+1}-1} \right).
\end{split}\]
\sum\limits_{k=1}^n {\dfrac{1}{{\sqrt k }}} &<\int_0^n {{x^{-\frac{1}{2}}}{\rm{d}}x}=\left. {2{x^{\frac{1}{2}}}} \right|_0^n=2\sqrt n \\
\sum\limits_{k=1}^n {\dfrac{1}{{\sqrt k }}} &> \int_1^{n+1} {{x^{-\frac{1}{2}}}{\rm{d}}x}=\left. {2{x^{\frac{1}{2}}}} \right|_1^{n+1}=2\left( {\sqrt {n+1}-1} \right).
\end{split}\]
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