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
We expand $\left({1 + \dfrac 1 n}\right)^n$ by the Binomial Theorem, using that $\dfrac {n - k} n = 1 - \dfrac k n$:

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:


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

Proof 2
We will assume a somewhat different position. Since it is not even initially clear that the limit exists, we will show 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:
 * $\displaystyle t_n = \frac 1 {0!} + \frac 1 {1!} + \left({1 - \frac 1 n}\right) \frac 1 {2!} + \left({1 - \frac 1 n}\right) \left({1 - \frac 2 n}\right)\frac 1 {3!} + \cdots + \left({1 - \frac 1 n}\right) \left({1 - \frac 2 n}\right) \cdots \left({1 - \frac {n-1} n}\right) \frac 1 {n!}$

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

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

Hence:
 * $\displaystyle \limsup(t_n) \le e$

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


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

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


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

Since this is true for all $m$:


 * $\displaystyle \liminf(t_n) \ge e$

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