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

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## Theorem

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

$\ds e := \lim_{n \mathop \to \infty} \paren {1 + \frac 1 n}^n$

Then:

$\ds e = \sum_{k \mathop = 0}^\infty \frac 1 {k!}$

That is:

$e = \dfrac 1 {0!} + \dfrac 1 {1!} + \dfrac 1 {2!} + \dfrac 1 {3!} + \dfrac 1 {4!} \cdots$

## Proof 1

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

 $\ds \paren {1 + \frac 1 n}^n$ $=$ $\ds 1 + n \paren {\frac 1 n} + \frac {n \paren {n - 1} } 2 \paren {\frac 1 n}^2 + \cdots + \paren {\frac 1 n}^n$ $\ds$ $=$ $\ds \frac 1 {0!} + \frac 1 {1!} + \paren {1 - \frac 1 n} \frac 1 {2!} + \paren {1 - \frac 1 n} \paren {1 - \frac 2 n} \frac 1 {3!} + \cdots + \paren {1 - \frac 1 n} \paren {1 - \frac 2 n} \cdots \paren {1 - \frac {n - 1} n} \frac 1 {n!}$

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:

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

$\blacksquare$

## Proof 2

It will be shown that:

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

Let $t_n := \paren {1 + \dfrac 1 n}^n$

Then:

$t_n = \dfrac 1 {0!} + \dfrac 1 {1!} + \paren {1 - \dfrac 1 n} \dfrac 1 {2!} + \paren {1 - \dfrac 1 n} \paren {1 - \dfrac 2 n} \dfrac 1 {3!} + \cdots + \paren {1 - \dfrac 1 n} \paren {1 - \dfrac 2 n} \cdots \paren {1 - \dfrac {n-1} n} \dfrac 1 {n!}$

Now let:

$\ds s_m := \sum_{k \mathop = 0}^m \frac 1 {k!}$

We have that:

$\forall n: t_n \le s_n$

Hence:

$\limsup \paren {t_n} \le e$

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

$t_n \ge \dfrac 1 {0!} + \dfrac 1 {1!} + \paren {1 - \dfrac 1 n} \dfrac 1 {2!} + \paren {1 - \dfrac 1 n} \paren {1 - \dfrac 2 n} \dfrac 1 {3!} + \cdots + \paren {1 - \dfrac 1 n} \paren {1 - \dfrac 2 n} \cdots \paren {1 - \dfrac {m - 1} n} \dfrac 1 {m!}$

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

$\ds \liminf \paren {t_n} \ge \sum_{k \mathop = 0}^m \frac 1 {m!}$

Since this is true for all $m$:

$\liminf \paren {t_n} \ge e$

So $\ds \lim \paren {t_n}$ exists and is equal to $e$.

$\blacksquare$