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:


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

That is:


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