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

From ProofWiki
Jump to navigation Jump to search

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 $\dfrac {n - k} n = 1 - \dfrac k n$:

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


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


Hence:

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

$\blacksquare$