Sales Tax and Geometric Series
It is known that, for \(\vert r \vert<1\), \(\sum_{n=0}^\infty ar^n=\frac{a}{1-r}\). The usual proof goes something as follows: let \(S=\sum_{n=0}^\infty ar^n\). Then \(rS=\sum_{n=0}^\infty ar^{n+1}=\sum_{n=1}^\infty ar^n\). Subtracting, \(S-rS=\sum_{n=0}^\infty ar^n-\sum_{n=1}^\infty ar^n=a\), so \(S=\frac{a}{1-r}\) as required. It’s very elegant, but it seems a little bit magical.