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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

$\lim_{n\to\infty} (1 + \frac{1}{n})^{n} = \sum\limits_{n=0}^\infty \frac{1}{n!}$

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  • $\begingroup$ This seems ok to me, the method of bounding the difference is a bit unusual but I think it is working well. I took another approach in this post math.stackexchange.com/questions/828320/…, but your method only involve simple inequalities so it' quite interesting. $\endgroup$ – zwim Jun 15 '18 at 22:17
  • $\begingroup$ While I applaud you for trying to prove the theorems yourself before reading Rudin's proofs, there is nothing wrong with looking and how Rudin does it afterward.... Anyway, is argument is also based on the squeeze theorem. His algebra is quite a bit different. But, what I see here appears to be correct. $\endgroup$ – Doug M Jun 15 '18 at 22:26

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