I am currently studying analysis with Rudin's PMA myself without looking at proofs for theorems stated in the book. I'm now at the stage where I should prove
$e = \lim_{n\to\infty} (1 + \frac{1}{n})^{n}$
when $e$ is defined as follows:
$e = \sum\limits_{n = 0}^\infty \frac{1}{n!}$.
So here is my proof and can you tell me whether I got it right or wrong?
Proof of the statement)
We firstly know that
$(1 + \frac{1}{n})^{n} \leq \sum\limits_{i = 0}^{n} \frac{1}{i!}$ for all $n \in \mathbb{N}$
since
$(1 + \frac{1}{n})^{n} = \sum\limits_{i = 0}^{n} \frac{n!}{i! \cdot (n - i)!} \cdot (\frac{1}{n})^{i} \leq 1 + \sum\limits_{i = 1}^{n} \frac{n!}{i! \cdot (n - i)!} \cdot \frac{1}{n \times (n - 1) \times \cdots \times (n - i + 1)} = \sum\limits_{i = 0}^{n} \frac{n!}{i! \cdot (n - i)!} \cdot \frac{(n - i)!}{n!} = \sum\limits_{i = 0}^{n} \frac{1}{i!}$
Now for any given integer $n \geq 2$, we know that
$0 \leq (\sum\limits_{i = 0}^n \frac{1}{i!}) - (1 + \frac{1}{n})^{n} = \sum\limits_{i = 0}^{n} (\frac{1}{i!} - \frac{n!}{i! \cdot (n - i)!} \cdot (\frac{1}{n})^{i}) = \sum\limits_{i = 2}^{n} (\frac{1}{i!} - \frac{n!}{i! \cdot (n - i)!} \cdot (\frac{1}{n})^{i}) = \sum\limits_{i = 2}^{n} \frac{n^{i} \cdot (n - i)! - n!}{i! \cdot (n - i)! \cdot n^i} = \sum\limits_{i = 2}^{n} \frac{(n - i)! \cdot (n^{i} - n \times \cdots \times (n - i + 1))}{i! \cdot (n - i)! \cdot n^i} \leq \sum\limits_{i = 2}^{n} \frac{(n - i)! \cdot (n^{i} - (n - i + 1)^{i})}{i! \cdot (n - i)! \cdot n^i} \leq \sum\limits_{i = 2}^{n} \frac{(n - i)! \cdot i \cdot (i - 1) \cdot n^{i - 1}}{i! \cdot (n - i)! \cdot n^i} = \frac{1}{n} \cdot \sum\limits_{i = 2}^n \frac{1}{(i - 2)!}$
The last inequality can be deriven if we let
$a = n - i + 1$
and
$b = n$
so that
$b^{i} - a^{i} = (b - a) \cdot (b^{i - 1} + a \cdot b^{i - 2} + \cdots a^{i - 1}) \leq (b - a) \cdot i \cdot b^{i - 1} = i \cdot (i - 1) \cdot n^{i - 1}$
By applying the squeeze theorem we get
$\lim_{n\to\infty} (1 + \frac{1}{n})^{n} = \sum\limits_{n=0}^\infty \frac{1}{n!}$