每日一题[813]数列收敛

求证：数列$\left\{\left(1+\dfrac 1n\right)^n\right\}$收敛．

证明　法一　二项式定理

根据二项式定理，有 $$\begin{split} \left(1+\dfrac 1n\right)^n&=1+{\rm C}_n^1\cdot \dfrac 1n+{\rm C}_n^2\cdot \dfrac 1{n^2}+\cdots+{\rm C}_n^n\cdot\dfrac{1}{n^n}\\ &=1+1+\dfrac 1{2!}\cdot \left(1-\dfrac 1n\right)+\cdots+\dfrac{1}{n!}\cdot \left(1-\dfrac 1n\right)\left(1-\dfrac 2n\right)\cdots\left(1-\dfrac {n-1}{n}\right), \end{split}$$ 因此数列$\left\{\left(1+\dfrac 1n\right)^n\right\}$单调递增．另一方面，有 $$\begin{split} \left(1+\dfrac 1n\right)^n&<2+\dfrac{1}{2!}+\dfrac{1}{3!}+\cdots+\dfrac{1}{n!}\\ &<2+\dfrac{1}{1\cdot 2}+\dfrac{1}{2\cdot 3}+\cdots+\dfrac{1}{(n-1)\cdot n}\\ &<3, \end{split}$$ 因此数列$\left\{\left(1+\dfrac 1n\right)^n\right\}$有上界$3$．综上所述，命题得证．

法二　均值不等式

根据均值不等式，有 $$\begin{split} \left(1+\dfrac 1n\right)^n&=\underbrace{\left(1+\dfrac 1n\right)\cdot\left(1+\dfrac 1n\right)\cdots \left(1+\dfrac 1n\right)}\cdot 1\\ &<\left[\dfrac{\left(1+\dfrac 1n\right)\cdot n+1}{n+1}\right]^{n+1}\\ &=\left(1+\dfrac{1}{n+1}\right)^{n+1}, \end{split}$$ 因此数列$\left\{\left(1+\dfrac 1n\right)^n\right\}$单调递增．另一方面，有 $$\begin{split} \dfrac 14\left(1+\dfrac 1n\right)^n&=\underbrace{\left(1+\dfrac 1n\right)\cdot\left(1+\dfrac 1n\right)\cdots \left(1+\dfrac 1n\right)}\cdot \dfrac 12\cdot \dfrac 12\\&<\left[\dfrac{\left(1+\dfrac 1n\right)\cdot n+\dfrac 12+\dfrac 12}{n+2}\right]^{n+2}\\ &=1, \end{split}$$ 因此数列$\left\{\left(1+\dfrac 1n\right)^n\right\}$有上界$4$．综上所述，命题得证．

法三　伯努利不等式

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 $$\begin{split} \dfrac{\left(1+\dfrac 1n\right)^{n+1}}{\left(1+\dfrac{1}{n+1}\right)^{n+2}}&=\dfrac{n+1}{n+2}\cdot \left(\dfrac{n+1}{n}\cdot \dfrac{n+1}{n+2}\right)^{n+1}\\ &=\dfrac{n+1}{n+2}\cdot \left(1+\dfrac{1}{n^2+2n}\right)^{n+1}\\ &>\dfrac {n+1}{n+2}\cdot\left(1+\dfrac {n+1}{n^2+2n}\right)\\ &=\dfrac {(n+1)(n^2+3n+1)}{n(n+2)^2}\\ &>1, \end{split}$$ 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．

法四　辅助不等式

我们熟知 $$a^{n+1}-b^{n+1}=(a-b)\left(a^n+a^{n-1}b+a^{n-2}b^2+\cdots +b^n\right),$$ 设$a>b>0$，则有 $$a^{n+1}-b^{n+1}<(a-b)\cdot (n+1)a^n,$$ 即 $$b^{n+1}>a^n\cdot \left[(n+1)b-na\right].$$ 令$a=1+\dfrac 1n$，$b=1+\dfrac{1}{n+1}$，则 $$\left(1+\dfrac{1}{n+1}\right)^{n+1}>\left(1+\dfrac 1n\right)^n,$$ 从而数列$\left\{\left(1+\dfrac 1n\right)^n\right\}$单调递增．另一方面，令$a=1+\dfrac{1}{2n}$，$b=1$，可得 $$1>\left(1+\dfrac 1{2n}\right)^n\cdot \dfrac 12,$$ 即 $$\left(1+\dfrac 1{2n}\right)^{2n}<4,$$ 因此数列$\left\{\left(1+\dfrac 1n\right)^n\right\}$有上界$4$．综上所述，命题得证．