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

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

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

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

That is, $$e = \frac 1 {0!} + \frac 1 {1!} + \frac 1 {2!} + \frac 1 {3!} + \frac 1 {4!} \cdots$$

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

$$ $$

Take one of the terms in the above:

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

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

From the Combination Theorem for Sequences:


 * $$\forall \lambda \in \mathbb{R}: \frac \lambda n \to 0$$ as $$n \to \infty$$;


 * $$\forall \lambda \in \mathbb{R}: 1 - \frac \lambda n \to 1$$ as $$n \to \infty$$;


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

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