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.
I came up with an alternative proof a while ago which, while not quite so nice, makes a little bit of a better narrative in my admittedly biased opinion. Let \(a\) be the price of some item, without tax. Now, a dastardly government imposes sales tax on the item at a rate \(r\) – so, if the item sells for a total price \(P\), the government will take \(rP\) of that. (In order to make it possible for commerce to continue, we can assume \(\vert r \vert<1\).) The shop manager wants to pass the cost on to the customer, so she adjusts the price: from \(a\) to \(a+ra\).
But now that the price is updated, that will change the total tax charged on the sale! As well as the tax \(ra\) charged on the original price \(a\), the government will also charge \(r\cdot ra=r^2a\) on the added amount \(ra\). So the shop manager has no choice but to charge \(a+ra+r^2a\). But as soon as she makes this change, suddenly the tax increases by \(r^3a\), and so on. To pass the full cost of the tax onto the purchaser, the price must be the sum of the infinite series \(P=a+ra+r^2a+r^3a+\cdots\).
Or, we could calculate the price another way. If the eventual price is \(P\), and the government takes \(rP\), then the shop is left with \((1-r)P\). In the scenario where the consumer pays the entire tax, this must be exactly the original price \(a\). So \(P=\frac{a}{1-r}\) and the proof is complete.