Euler's Number: Limit of Sequence implies Limit of Series/Proof 1

Proof
We expand $\paren {1 + \dfrac 1 n}^n$ by the Binomial Theorem, using that $\dfrac {n - k} n = 1 - \dfrac k n$:

Take one of the terms in the above:


 * $x = \paren {1 - \dfrac 1 n} \paren {1 - \dfrac 2 n} \cdots \paren {1 - \dfrac {k - 1} n} \dfrac 1 {k!}$

From Sequence of Powers of Reciprocals is Null Sequence, $\dfrac 1 n \to 0$ 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 $n \to \infty$


 * $x = \paren {1 - \dfrac 1 n} \paren {1 - \dfrac 2 n} \cdots \paren {1 - \dfrac {k - 1} n} \dfrac 1 {k!} \to \dfrac 1 {k!}$ as $n \to \infty$

Hence:


 * $\ds \lim_{n \mathop \to \infty} \paren {1 + \frac 1 n}^n = \frac 1 {0!} + \frac 1 {1!} + \frac 1 {2!} + \frac 1 {3!} + \cdots = \sum_{k \mathop = 0}^\infty \frac 1 {k!}$