已知 $f\left( x \right)=\dfrac{2}{3}x+\dfrac{1}{2}$，$h\left( x \right)=\sqrt x $，试比较 $\displaystyle f\left( {100} \right)h\left( {100} \right)-\sum_{k=1}^{100} {h\left( k \right)} $ 与 $\dfrac{1}{6}$ 的大小关系．
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【答案】
$f(100)h(100)-\displaystyle\sum_{k=1}^{100}{h(k)}>\dfrac 16$
【解析】
令 $\displaystyle {S_n}=f\left( n \right) \cdot g\left( n \right)-\sum\limits_{k=1}^n {h\left( k \right)} $，则$$S_n =\left( {\dfrac{2}{3}n+\dfrac{1}{2}} \right) \cdot \sqrt n-\sum\limits_{k=1}^n {\sqrt k } ,$$于是$${S_1}=\left( {\dfrac{2}{3}+\dfrac{1}{2}} \right) \cdot 1-1=\dfrac{1}{6}$$事实上，$${S_2}=\left( {\dfrac{2}{3} \cdot 2+\dfrac{1}{2}} \right) \cdot \sqrt 2-1-\sqrt 2=\dfrac{{5\sqrt 2 }}{6}-1>\dfrac{7}{6}-1=\dfrac{1}{6}$$考虑证明 ${S_n}$ 单调递增．\[\begin{split}
{S_{n+1}}-{S_n} &= \left[ {\dfrac{2}{3}\left( {n+1} \right)+\dfrac{1}{2}} \right]\sqrt {n+1}-\sum\limits_{k=1}^{n+1} {\sqrt k }-\left( {\dfrac{2}{3}n+\dfrac{1}{2}} \right)\sqrt n+\sum\limits_{k=1}^n {\sqrt k } \\
&=\left( {\dfrac{2}{3}n+\dfrac{1}{6}} \right)\sqrt {n+1}-\left( {\dfrac{2}{3}n+\dfrac{1}{2}} \right)\sqrt n \\
&=\dfrac{1}{6}\left[ {\left( {4n+1} \right)\sqrt {n+1}-\left( {4n+3} \right)\sqrt n } \right]\\
&=\dfrac{1}{6}\left( {\sqrt {16{n^3}+24{n^2}+9n+1}-\sqrt {16{n^3}+24{n^2}+9n} } \right)\\
&>0
\end{split}\]因此当 $n \geqslant 2$ 时，${S_n}>{S_1}=\dfrac{1}{6}$，从而$$f(100)h(100)-\sum_{k=1}^{100}{h(k)}>\dfrac 16.$$
{S_{n+1}}-{S_n} &= \left[ {\dfrac{2}{3}\left( {n+1} \right)+\dfrac{1}{2}} \right]\sqrt {n+1}-\sum\limits_{k=1}^{n+1} {\sqrt k }-\left( {\dfrac{2}{3}n+\dfrac{1}{2}} \right)\sqrt n+\sum\limits_{k=1}^n {\sqrt k } \\
&=\left( {\dfrac{2}{3}n+\dfrac{1}{6}} \right)\sqrt {n+1}-\left( {\dfrac{2}{3}n+\dfrac{1}{2}} \right)\sqrt n \\
&=\dfrac{1}{6}\left[ {\left( {4n+1} \right)\sqrt {n+1}-\left( {4n+3} \right)\sqrt n } \right]\\
&=\dfrac{1}{6}\left( {\sqrt {16{n^3}+24{n^2}+9n+1}-\sqrt {16{n^3}+24{n^2}+9n} } \right)\\
&>0
\end{split}\]因此当 $n \geqslant 2$ 时，${S_n}>{S_1}=\dfrac{1}{6}$，从而$$f(100)h(100)-\sum_{k=1}^{100}{h(k)}>\dfrac 16.$$
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