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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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