\(e = 2.718\cdots\): Napier's constant

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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,

\[e = \lim_{n\to\infty}\left(1 + \frac{1}{n}\right)^n.\]

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

\[a_n = \left(1 + \frac{1}{n}\right)^n\]

converges.



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

\[\begin{align*}a_1 &= 2,\\ a_2 &= 2.25,\\ a_3 &= 2.370\cdots,\\ a_4 &= 2.441\cdots,\\ a_5 &=2.488\cdots,\\ &\vdots\\a_{10}&=2.593\cdots,\\a_{100} &= 2.704\cdots,\\a_{1000} &= 2.716\cdots,\\&\vdots\end{align*}\]

□

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

\[ \begin{eqnarray*} s_k &=& \frac{n(n-1)\cdots(n-k+1)}{n^k}\\ &=& \frac{n}{n}\cdot\frac{n-1}{n}\cdot\cdots\cdot\frac{n-k+1}{n}\\ &=&1\cdot\left(1 - \frac{1}{n}\right)\cdot\cdots\cdot\left(1 - \frac{k-1}{n}\right). \end{eqnarray*} \]

Similarly,

\[a_{n+1} = \sum_{k=0}^{n+1}\frac{t_k}{k!}\]

where

\[t_k = 1\cdot\left(1 - \frac{1}{n+1}\right)\cdot\cdots\cdot\left(1 - \frac{k-1}{n+1}\right). \]

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 \(n\geq 1\),

\[a_n = \sum_{k=0}^{n}\frac{s_k}{k!} < \sum_{k=0}^{n}\frac{t_k}{k!} < \sum_{k=0}^{n}\frac{t_k}{k!} + \frac{t_{n+1}}{(n+1)!} = a_{n+1}.\]

■

Lemma

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

\[2 \leq \left(1 + \frac{1}{n}\right)^n < 3.\]

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

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

\[s_k = 1\cdot\left(1 - \frac{1}{n}\right)\cdot\left(1 - \frac{2}{n}\right)\cdot\cdots\cdot\left(1 - \frac{k-1}{n}\right) \leq 1\]

for \(k = 1, \cdots, n-1\). Therefore

\[ \begin{eqnarray*} a_n &=& 1 + 1 + \frac{s_2}{2!} + \frac{s_3}{3!} + \cdots + \frac{s_n}{n!}\\ &\leq& 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots + \frac{1}{n!}. \end{eqnarray*} \]

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 \(n\geq 3\),

\[ \begin{eqnarray*} a_n &\leq& 1 + \left(1 + \frac{1}{2} + \frac{1}{2^2} + \cdots + \frac{1}{2^{n-1}}\right)\\ &=& 1 + \left(2 - \frac{1}{2^{n-1}}\right) < 3 \end{eqnarray*} \]

■

The next theorem follows immediately from these lemmas.

Theorem

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

Now we can define

\[e = \lim_{n\to\infty}\left(1 + \frac{1}{n}\right)^n = 2.71828\cdots.\]


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