求证：$2 \leqslant {\left( {1+\dfrac{1}{n}} \right)^n}<3$．
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无
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【答案】
略
【解析】
利用二项式定理展开为级数，有$${\left( {1+\dfrac{1}{n}} \right)^n}=1+{\rm{C}}_n^1 \cdot \dfrac{1}{n}+{\rm{C}}_n^2 \cdot \dfrac{1}{{{n^2}}}+\cdots+{\rm{C}}_n^n \cdot \dfrac{1}{{{n^n}}},$$于是$${\left( {1+\dfrac{1}{n}} \right)^n} \geqslant 2,$$而\[\begin{split}
{\left( {1+\dfrac{1}{n}} \right)^n}
&=1+{\rm{C}}_n^1 \cdot \dfrac{1}{n}+{\rm{C}}_n^2 \cdot \dfrac{1}{{{n^2}}}+\cdots+{\rm{C}}_n^n \cdot \dfrac{1}{{{n^n}}}\\
&=2+\dfrac{1}{{2 ! }} \cdot \dfrac{{n\left( {n-1} \right) }}{{{n^2}}}+\dfrac{1}{{3 ! }} \cdot \dfrac{{n\left( {n-1} \right)\left( {n-2} \right)}}{{{n^3}}}+\cdots+\dfrac{1}{{n ! }} \cdot \dfrac{{n ! }}{{{n^n}}}\\
&<2+\dfrac{1}{{1 \cdot 2}}+\dfrac{1}{{2 \cdot 3}}+\cdots+\dfrac{1}{{n\left( {n-1} \right)}}\\
&=2+\left( {1-\dfrac{1}{2}} \right)+\left( {\dfrac{1}{2}-\dfrac{1}{3}} \right)+\cdots+\left( {\dfrac{1}{{n-1}}-\dfrac{1}{n}} \right)\\
&=3-\dfrac{1}{n}\\
&<3.
\end{split}\]因此原不等式得证．
{\left( {1+\dfrac{1}{n}} \right)^n}
&=1+{\rm{C}}_n^1 \cdot \dfrac{1}{n}+{\rm{C}}_n^2 \cdot \dfrac{1}{{{n^2}}}+\cdots+{\rm{C}}_n^n \cdot \dfrac{1}{{{n^n}}}\\
&=2+\dfrac{1}{{2 ! }} \cdot \dfrac{{n\left( {n-1} \right) }}{{{n^2}}}+\dfrac{1}{{3 ! }} \cdot \dfrac{{n\left( {n-1} \right)\left( {n-2} \right)}}{{{n^3}}}+\cdots+\dfrac{1}{{n ! }} \cdot \dfrac{{n ! }}{{{n^n}}}\\
&<2+\dfrac{1}{{1 \cdot 2}}+\dfrac{1}{{2 \cdot 3}}+\cdots+\dfrac{1}{{n\left( {n-1} \right)}}\\
&=2+\left( {1-\dfrac{1}{2}} \right)+\left( {\dfrac{1}{2}-\dfrac{1}{3}} \right)+\cdots+\left( {\dfrac{1}{{n-1}}-\dfrac{1}{n}} \right)\\
&=3-\dfrac{1}{n}\\
&<3.
\end{split}\]因此原不等式得证．
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