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 $e$.
$\blacksquare$