73.The limit of $(1+\frac{1}{n}{)}^n$

A more difficult problem which can be solved by the help of § 69 arises when $\varphi (n)=\{1+1/n{\}}^n$.

It follows from the binomial theorem¹ that $$\begin{array}{c}\begin{array}{rl}(1+\frac{1}{n}{)}^n& =1+n\cdot \frac{1}{n}+\frac{n(n–1)}{1\cdot 2}\frac{1}{n^2}+\cdots +\frac{n(n–1)\dots (n–n+1)}{1\cdot 2\dots n}\frac{1}{n^n}\\ & =1+1+\frac{1}{1\cdot 2}(1–\frac{1}{n})+\frac{1}{1\cdot 2\cdot 3}(1–\frac{1}{n})(1-\frac{2}{n})+\dots \end{array}\\ +\frac{1}{1\cdot 2\dots n}(1–\frac{1}{n})(1–\frac{2}{n})\dots (1–\frac{n–1}{n}).\end{array}$$

The $(p+1)$th term in this expression, viz. $$\frac{1}{1\cdot 2\dots p}\left(1–\frac{1}{n}\right)\left(1–\frac{2}{n}\right)\dots \left(1–\frac{p–1}{n}\right),$$ is positive and an increasing function of $n$, and the number of terms also increases with $n$. Hence ${(1+{\displaystyle \frac{1}{n}})}^n$ increases with $n$, and so tends to a limit or to $+\mathrm{\infty }$, as $n\to \mathrm{\infty }$.

But $$\begin{array}{rl}{(1+\frac{1}{n})}^n& <1+1+\frac{1}{1\cdot 2}+\frac{1}{1\cdot 2\cdot 3}+\cdots +\frac{1}{1\cdot 2\cdot 3\dots n}\\ & <1+1+\frac{1}{2}+\frac{1}{2^2}+\cdots +\frac{1}{2^{n-1}}<3.\end{array}$$

Thus ${(1+{\displaystyle \frac{1}{n}})}^n$ cannot tend to $+\mathrm{\infty }$, and so $$\underset{n\to \mathrm{\infty }}{lim}{(1+\frac{1}{n})}^n=e,$$ where $e$ is a number such that $2<e\le 3$.