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:


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

Proof 2
It will be shown that:


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

Let $t_n := \left({1 + \dfrac 1 n}\right)^n$

Then:
 * $t_n = \dfrac 1 {0!} + \dfrac 1 {1!} + \left({1 - \dfrac 1 n}\right) \dfrac 1 {2!} + \left({1 - \dfrac 1 n}\right) \left({1 - \dfrac 2 n}\right) \dfrac 1 {3!} + \cdots + \left({1 - \dfrac 1 n}\right) \left({1 - \dfrac 2 n}\right) \cdots \left({1 - \dfrac {n-1} n}\right) \dfrac 1 {n!}$

Now let:
 * $\displaystyle s_m := \sum_{k \mathop = 0}^m \frac 1 {k!}$

We have that:
 * $\forall n: t_n \le s_n$

Hence:
 * $\limsup \left({t_n}\right) \le e$

Now, for all $m$, for $n \ge m$:


 * $t_n \ge \dfrac 1 {0!} + \dfrac 1 {1!} + \left({1 - \dfrac 1 n}\right) \dfrac 1 {2!} + \left({1 - \dfrac 1 n}\right) \left({1 - \dfrac 2 n}\right) \dfrac 1 {3!} + \cdots + \left({1 - \dfrac 1 n}\right) \left({1 - \dfrac 2 n}\right) \cdots \left({1 - \dfrac {m-1} n}\right) \dfrac 1 {m!}$

Hence, for all $m$, we have the right side as being a sequence in $n$, and then:


 * $\displaystyle \liminf \left({t_n}\right) \ge \sum_{k \mathop = 0}^m \frac 1 {m!}$

Since this is true for all $m$:


 * $\liminf \left({t_n}\right) \ge e$

So $\displaystyle \lim \left({t_n}\right)$ exists and is equal to $e$.