每日一题[813]数列收敛

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


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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.

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