### A series for $$e^x$$

Consider the series

$f(x) = 1 + \dfrac{x^1}{1!} + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dfrac{x^4}{4!} + \dfrac{x^5}{5!} + \dotsb.$

An infinite sum like this, with increasing powers of $$x$$, is like an infinite version of a polynomial and known as a power series. The denominators are all factorials, such as $$4! = 4 \cdot 3 \cdot 2 \cdot 1 = 24$$. We can write the series succinctly in summation notation, as

$f(x) = \sum_{n=0}^\infty \dfrac{x^n}{n!}.$ (In the $$n=0$$ term, we set $$x^0 = 1$$ and follow the convention that $$0! = 1$$.) Substituting $$x=0$$ gives $$f(0) = 1$$. It turns out that, for any $$x$$, this series converges, and the function $$f$$ is continuous and differentiable. And we can see that, if we differentiate the series term-by-term, we obtain \dots\ the same series! \begin{align*} f'(x) &= 1 + \dfrac{2x}{2!} + \dfrac{3x^2}{3!} + \dfrac{4x^3}{4!} + \dotsb \\ &= 1 + \dfrac{x}{1!} + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dotsb = f(x). \end{align*}

Hence the series $$f(x)$$ converges to $$e^x$$. In particular, substituting $$x=1$$, we obtain an amazing formula for $$e$$:

$e = 1 + \dfrac{1}{1!} + \dfrac{1}{2!} + \dfrac{1}{3!} + \dfrac{1}{4!} + \dotsb = \sum_{n=0}^\infty \dfrac{1}{n!}.$

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