Why are these representations of e the same?

Question

I heard that $e$ can be defined as the limit as n approaches infinity of $(1 + 1/n)^n$, but I also heard that $e$ is also defined as the sum of the reciprocals of the factorials from $0$ to $\infty$. How are these the same?

Answer 1

Hint: use the binomial theorem

Answer 2

Let's find an infinite polynomial $p(x)$ having $p(0) = 1$ and $\frac{d}{dx}p(x) = p(x)$. We know that the constant term must be 1, and the coefficient of $x^1$ must be 1. After a little reflection, it's easy to see that $$p(x) = \sum_{i=0}^{\infty} \frac{x^i}{i!}$$ This must be the only continuous function having these two properties.

Now let's look at the limit

$$L(x) = \lim_{n \rightarrow \infty} \left(1 + \frac{x}{n}\right)^n$$

$$\begin{align} \frac{d}{dx} L(x) & = \frac{d}{dx} \left\{\lim_{n \rightarrow \infty} \left(1 + \frac{x}{n}\right)^n\right\}\\ & = \lim_{n \rightarrow \infty} \frac{d}{dx} \left\{\left(1 + \frac{x}{n}\right)^n\right\}\\ & = \lim_{n \rightarrow \infty}\left(\frac{d}{dx} \left\{1 + \frac{x}{n}\right\}n\left(1 + \frac{x}{n}\right)^{n-1}\right)\\ & = \lim_{n \rightarrow \infty}\left(1 + \frac{x}{n}\right)^{n-1}\\ \end{align}$$

So:

$$\begin{align} L(x) - \frac{d}{dx}L(x) & = \lim_{n \rightarrow \infty}\left(1 + \frac{x}{n}\right)^{n}-\lim_{n \rightarrow \infty}\left(1 + \frac{x}{n}\right)^{n-1}\\ & = \lim_{n \rightarrow \infty}\left(\left(1 + \frac{x}{n}\right)^{n}-\left(1 + \frac{x}{n}\right)^{n-1}\right)\\ & = \lim_{n \rightarrow \infty}\left(\left(1 + \frac{x}{n}\right)^{n-1}\left(\left(1 + \frac{x}{n}\right)-1\right)\right)\\ & = \lim_{n \rightarrow \infty}\left(\left(1 + \frac{x}{n}\right)^{n- 1}\left(\frac{x}{n}\right)\right)\\ & = \lim_{n \rightarrow \infty}\left(\left(1 + \frac{x}{n}\right)^{n-1}\left(0\right)\right)\\ L(x) - \frac{d}{dx}L(x) & = 0 \end{align}$$ So $L(x) = \frac{d}{dx}L(x)$, $L(0) = 1$, and $L$ is a continuous function. So $L(x) = p(x)$, and specifically $L(1) = p(1) = e$.