Euler's Number: Limit of Sequence implies Limit of Series

Theorem

Let Euler's number $e$ be defined as:

$\displaystyle e := \lim_{n \to \infty} \left({1 + \frac 1 n}\right)^n$

Then:

$\displaystyle e = \sum_{k \ge 0} \frac 1 {k!}$

That is:

$\displaystyle e = \frac 1 {0!} + \frac 1 {1!} + \frac 1 {2!} + \frac 1 {3!} + \frac 1 {4!} \cdots$

Proof

We expand $\left({1 + \frac 1 n}\right)^n$ by the Binomial Theorem:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left({1 + \frac 1 n}\right)^n\)	\(=\)	\(\displaystyle 1 + n \left({\frac 1 n}\right) + \frac {n \left({n-1}\right)} 2 \left({\frac 1 n}\right)^2 + \cdots + \left({\frac 1 n}\right)^n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac 1 {0!} + \frac 1 {1!} + \left({1 - \frac 1 n}\right) \frac 1 {2!} + \left({1 - \frac 1 n}\right) \left({1 - \frac 2 n}\right)\frac 1 {3!} + \cdots + \left({1 - \frac 1 n}\right) \left({1 - \frac 2 n}\right) \cdots \left({1 - \frac {n-1} n}\right) \frac 1 {n!}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Take one of the terms in the above:

$\displaystyle x = \left({1 - \frac 1 n}\right) \left({1 - \frac 2 n}\right) \cdots \left({1 - \frac {k-1} n}\right) \frac 1 {k!}$

From Power of Reciprocal, $\dfrac 1 n \to 0$ as $n \to \infty$.

From the Combination Theorem for Sequences:

• $\forall \lambda \in \R: \dfrac \lambda n \to 0$ as $n \to \infty$

• $\forall \lambda \in \R: 1 - \dfrac \lambda n \to 1$ as $n \to \infty$

• $\displaystyle x = \left({1 - \frac 1 n}\right) \left({1 - \frac 2 n}\right) \cdots \left({1 - \frac {k-1} n}\right) \frac 1 {k!} \to \frac 1 {k!}$ as $n \to \infty$

Hence:

$\displaystyle \lim_{n \to \infty} \left({1 + \frac 1 n}\right)^n = \frac 1 {0!} + \frac 1 {1!} + \frac 1 {2!} + \frac 1 {3!} + \cdots = \sum_{k=0}^\infty \frac 1 {k!}$

Sources

• W.A. Sutherland: Introduction to Metric and Topological Spaces (1975): $\S 1.2$ (passim)