e=2.718⋯: Napier's constant

We have defined the constant known as Napier's constant $e$ in a previous post, as $e=\exp (1)$ where $\exp (x)$ is the inverse of $\log (x)$. 

See also: $\log$ and $e$

Here we provide an alternative definition of this constant. That is,

$$e=\underset{n\to \mathrm{\infty }}{lim}{(1+\frac{1}{n})}^n.$$

For this definition to be valid, we need to show that the sequence $\{a_n\}$ defined by

$$a_n={(1+\frac{1}{n})}^n$$

converges.

Example. The first several terms of the above sequence are:

$$\begin{array}{rl}{a}_1& =2,\\ a_2& =2.25,\\ a_3& =2.370\cdots ,\\ a_4& =2.441\cdots ,\\ a_5& =2.488\cdots ,\\ & ⋮\\ a_{10}& =2.593\cdots ,\\ a_{100}& =2.704\cdots ,\\ a_{1000}& =2.716\cdots ,\\ & ⋮\end{array}$$

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Lemma

The sequence $\{a_n\}$ defined above is monotone increasing.

Proof. We show that $a_n<a_{n+1}$ for all $n\in \mathbb{N}$.

By the Binomial Theorem, we can write

$$a_n=\sum _{k=0}^n\frac{s_k}{k!}$$

where

$$\begin{array}{rcl}{s}_k& =& \frac{n(n-1)\cdots (n-k+1)}{n^k}\\ & =& \frac{n}{n}\cdot \frac{n-1}{n}\cdot \cdots \cdot \frac{n-k+1}{n}\\ & =& 1\cdot (1-\frac{1}{n})\cdot \cdots \cdot (1-\frac{k-1}{n}).\end{array}$$

Similarly,

$$a_{n+1}=\sum _{k=0}^{n+1}\frac{t_k}{k!}$$

where

$$t_k=1\cdot (1-\frac{1}{n+1})\cdot \cdots \cdot (1-\frac{k-1}{n+1}).$$

For each $i=1,2,\cdots ,k-1$, we have

$$1-\frac{i}{n}<1-\frac{i}{n+1}$$

so that

$$s_k<t_k(k=1,2,\cdots ,n).$$

Therefore, for $n\ge 1$,

$$a_n=\sum _{k=0}^n\frac{s_k}{k!}<\sum _{k=0}^n\frac{t_k}{k!}<\sum _{k=0}^n\frac{t_k}{k!}+\frac{t_{n+1}}{(n+1)!}=a_{n+1}.$$

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Lemma

For all $n\in \mathbb{N}$, the following holds:

$$2\le {(1+\frac{1}{n})}^n<3.$$

In particular, the sequence $\{a_n\}$ defined by $a_n={(1+\frac{1}{n})}^n$ is bounded.

Proof. $a_1=2$ and $\{a_n\}$ is strictly monotone increasing. Therefore $a_n\ge 2$ for all $n\in \mathbb{N}$. Using $s_k$ defined in the previous lemma,

$$s_k=1\cdot (1-\frac{1}{n})\cdot (1-\frac{2}{n})\cdot \cdots \cdot (1-\frac{k-1}{n})\le 1$$

for $k=1,\cdots ,n-1$. Therefore

$$\begin{array}{rcl}{a}_n& =& 1+1+\frac{s_2}{2!}+\frac{s_3}{3!}+\cdots +\frac{s_n}{n!}\\ & \le & 1+1+\frac{1}{2!}+\frac{1}{3!}+\cdots +\frac{1}{n!}.\end{array}$$

Noting that $n!=1\cdot 2\cdot 3\cdot \cdots \cdot n>1\cdot 2\cdot 2\cdot \cdots \cdot 2=2^{n-1}$ for $n\ge 3$,

$$\begin{array}{rcl}{a}_n& \le & 1+(1+\frac{1}{2}+\frac{1}{2^2}+\cdots +\frac{1}{2^{n-1}})\\ & =& 1+(2-\frac{1}{2^{n-1}})<3\end{array}$$

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The next theorem follows immediately from these lemmas.

Theorem

The sequence $\{a_n\}$ defined by $a_n={(1+\frac{1}{n})}^n$ converges.

Now we can define

$$e=\underset{n\to \mathrm{\infty }}{lim}{(1+\frac{1}{n})}^n=2.71828\cdots .$$