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:

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