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I would like to prove that

$$e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = \sum_{k=0}^{+\infty}\frac{1}{k!}.$$

Among the several attempts I've performed to prove the previous, I report the one I judge the most promising one, where I've made use of the Newton's binomial theorem:

$$\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = \lim_{n \to \infty} \frac{(n+1)^n}{n^n} = \\ = \lim_{n \to \infty} \frac{\displaystyle\sum_{k=0}^n \binom{n}{k}n^{n-k}1^k}{n^n} = \lim_{n \to \infty} \displaystyle\sum_{k=0}^n \frac{n!}{n^{k} \cdot (n-k)!k!} = \ldots $$

Anyway, I'm stuck.

Any hints?

the_candyman
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  • @Sebastiano Thanks for this link. Anyway, I've already read it, but the (most voted and accepted) answer is somehow confusing for me. – the_candyman Jan 11 '23 at 19:43
  • Try to adapt my answer here to help answer your question: https://math.stackexchange.com/a/4452536/29156 But basically this is the prooff from Rudin's PMA. – Adam Rubinson Jan 11 '23 at 19:56

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