## Links forward

### 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

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

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