求证：数列 $\left\{\left(1+\dfrac 1n\right)^n\right\}$ 收敛．
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【答案】
略
【解析】
根据伯努利不等式，有\[\begin{split}
\dfrac{\left(1+\dfrac{1}{n+1}\right)^{n+1}}{\left(1+\dfrac 1n\right)^n}&=\dfrac{n+2}{n+1}\cdot \left(\dfrac{n+2}{n+1}\cdot \dfrac{n}{n+1}\right)^{n}\\
&=\dfrac{n+2}{n+1}\cdot \left(1-\dfrac{1}{n^2+2n+1}\right)^n\\
&>\dfrac{n+2}{n+1}\cdot \left(1-\dfrac{n}{n^2+2n+1}\right)\\
&=\dfrac{n^3+3n^2+3n+2}{n^3+3n^2+3n+1}\\
&>1,
\end{split}\]因此数列 $\left\{\left(1+\dfrac 1n\right)^n\right\}$ 单调递增．另一方面，有\[\begin{split}\dfrac{\left(1+\dfrac 1n\right)^{n+1}}{\left(1+\dfrac{1}{n+1}\right)^{n+2}}&=\dfrac{n+1}{n+2}\cdot \left(\dfrac{n+1}{n}\cdot \dfrac{n+1}{n+2}\right)^{n+1}\\
&=\dfrac{n+1}{n+2}\cdot \left(1+\dfrac{1}{n^2+2n}\right)^{n+1}\\
&>\dfrac{n+1}{n+2}\cdot \left(1+\dfrac{1}{n}\right)\\
&=\dfrac{n^2+2n+1}{n^2+2n}\\
&>1,
\end{split}\]于是数列 $\left\{\left(1+\dfrac 1n\right)^{n+1}\right\}$ 单调递减，进而有\[\left(1+\dfrac 1n\right)^n<\left(1+\dfrac 1n\right)^{n+1}\leqslant 4,\]因此数列 $\left\{\left(1+\dfrac 1n\right)^n\right\}$ 有上界 $4$．综上所述，命题得证．
\dfrac{\left(1+\dfrac{1}{n+1}\right)^{n+1}}{\left(1+\dfrac 1n\right)^n}&=\dfrac{n+2}{n+1}\cdot \left(\dfrac{n+2}{n+1}\cdot \dfrac{n}{n+1}\right)^{n}\\
&=\dfrac{n+2}{n+1}\cdot \left(1-\dfrac{1}{n^2+2n+1}\right)^n\\
&>\dfrac{n+2}{n+1}\cdot \left(1-\dfrac{n}{n^2+2n+1}\right)\\
&=\dfrac{n^3+3n^2+3n+2}{n^3+3n^2+3n+1}\\
&>1,
\end{split}\]因此数列 $\left\{\left(1+\dfrac 1n\right)^n\right\}$ 单调递增．另一方面，有\[\begin{split}\dfrac{\left(1+\dfrac 1n\right)^{n+1}}{\left(1+\dfrac{1}{n+1}\right)^{n+2}}&=\dfrac{n+1}{n+2}\cdot \left(\dfrac{n+1}{n}\cdot \dfrac{n+1}{n+2}\right)^{n+1}\\
&=\dfrac{n+1}{n+2}\cdot \left(1+\dfrac{1}{n^2+2n}\right)^{n+1}\\
&>\dfrac{n+1}{n+2}\cdot \left(1+\dfrac{1}{n}\right)\\
&=\dfrac{n^2+2n+1}{n^2+2n}\\
&>1,
\end{split}\]于是数列 $\left\{\left(1+\dfrac 1n\right)^{n+1}\right\}$ 单调递减，进而有\[\left(1+\dfrac 1n\right)^n<\left(1+\dfrac 1n\right)^{n+1}\leqslant 4,\]因此数列 $\left\{\left(1+\dfrac 1n\right)^n\right\}$ 有上界 $4$．综上所述，命题得证．
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