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

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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 1

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

\(\displaystyle \left({1 + \frac 1 n}\right)^n\) \(=\) \(\displaystyle 1 + n \left({\frac 1 n}\right) + \frac {n \left({n - 1}\right)} 2 \left({\frac 1 n}\right)^2 + \cdots + \left({\frac 1 n}\right)^n\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {0!} + \frac 1 {1!} + \left({1 - \frac 1 n}\right) \frac 1 {2!} + \left({1 - \frac 1 n}\right) \left({1 - \frac 2 n}\right)\frac 1 {3!} + \cdots + \left({1 - \frac 1 n}\right) \left({1 - \frac 2 n}\right) \cdots \left({1 - \frac {n - 1} n}\right) \frac 1 {n!}\) $\quad$ $\quad$


Take one of the terms in the above:

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

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


Hence:

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

$\blacksquare$


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 $e$.

$\blacksquare$


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