If you’re like me, you may have suffered through sequences and series in Calculus class with amazement: it makes sense, but how did anyone come up with this in the first place? Why would they think to even do this? Clearly smarter people exist in the world to bring such strange things to us.

But I was looking at this problem recently…

A customer with a large invoice (let’s say, \$1000) wants to pay with a credit card — but credit cards charge serious merchant fees, a percentage of the total billed amount.

So, for that \$1000 charge, the credit card company might take 3%, or \$30. At larger dollar amounts or over many transactions, this can take a serious chunk out of revenue.

Since credit cards are offered as a courtesy rather than a necessity here, often we agree to pass that cost on to the customer in exchange for the convenience^{†}, but that introduces an interesting problem:

Each time you calculate the credit card fee and add it to the bill, that amount is *also* subject to the credit card fee! So if I charge the customer \$1030, the credit card company takes 3%, which is \$30.90, and now I’m still \$0.90 short!

You can keep going, adding the \$0.90 to the total, for \$1030.90, but this leaves you with a fee of \$30.92, still two cents short. Surely two cents isn’t going to upset anyone, but this is not an accurate answer, just an acceptable approximation. Also, the number of iterations required to get a fixed amount is different for different starting amounts.

An engineer might be inclined to just stop here, but this is not something that will fit nicely into an Excel formula. It works, but it’s not great.

Fortunately, despite getting wrapped around ourselves iterating an answer, there’s a much simpler way — the “right” way involves taking a step back and asking a simpler question: *What is the unknown total amount I need to charge?*

$$A=T-tT,$$

where $A$ is the known amount we want to have after fees (i.e. \$1000), $T$ is the unknown total amount to charge, and $tT$ is the fee (where $t$ is the fee percent).

The solution to this is simple. Collect $T$ terms and divide to get

$$T = \frac{A}{1-t}$$

so that

$$T = \frac{\$1000}{1-0.03}=$1030.93$$

or, if separating the surcharge out from the original amount as a separate line-item,

$$\text{surcharge}=T-A=A\left({\frac{1}{1-t}-1}\right)=A\frac{t}{1-t}$$

This is the reasonable way to approach this problem.

*(Note to the reader: If you just wanted the answer, this was it. Dicey waters lay ahead.)*

### The OTHER way

Going back to the iterative approach, for many people this seems like the more ‘natural’ way to think about it, at least initially.

We can put it more mathematically like this:

If I calculate the additional amount $tA$ to charge, I *also* need to charge for the fee on that amount, which is $ttA$. The fee on that is $tttA$ and so on forever. It’s only at this point we can say we’ve determined the *true total*

$$T=A + tA + ttA + tttA + \cdots$$

Reaching back into the darkest recesses, you might say to yourself “aha! I’ve seen this before.. this is an *infinite series*!”

And you are right. But do you remember how to deal with that, like, at all? No, you don’t. For we are only mortal and Calculus class was (way) more than a year ago.

When you write this in notation, it does look a little more manageable and may even jog a memory or two:

$$T=\sum_{n=0}^\infty{At^n}$$

There are a couple of things we might intuit here that are helpful. One is that this series probably *converges* since we know that the fee on the fee (on the fee on the fee on the fee) gets smaller and smaller forever toward zero, or

$$\left.\lim_{n\rightarrow\infty}{t^n}=0\right|t<1,$$

so that eventually the sums stop having any effect on the result at a sufficiently high $n$.

Expanding the notation doesn’t seem to get us very far:

$$T= At^0 + At^1 + At^2 + At^3 + \cdots + At^\infty$$

$$=A(t^0 + t^1 + t^2 + t^3 + \cdots + t^\infty)$$

There isn’t any obvious place to go from here… but we can cheat a bit because we already determined a very simple and likely answer:

$$T = \frac{A}{1-t}$$

Smashing these together,

$$A(t^0 + t^1 + t^2 + t^3 + \cdots + t^\infty) = T = \frac{A}{1-t}$$

$$t^0 + t^1 + t^2 + t^3 + \cdots + t^\infty = \frac{1}{1-t}$$

A very strange, unintuitive looking assertion. But multiplying $(1-t)$ across,

$$t^0(1-t) + t^1(1-t) + t^2(1-t) + t^3(1-t) + \cdots + t^\infty(1-t) = 1$$

$$1(1-t) + t(1-t) + t^2(1-t) + t^3(1-t) + \cdots + t^\infty(1-t) = 1$$

$$(1-t) + (t-t^2) + (t^2-t^3) + (t^3-t^4) + \cdots + (t^\infty-t^{\infty+1}) = 1$$

$$1-t + t-t^2 + t^2-t^3 + t^3-t^4 + \cdots + t^\infty-t^{\infty+1} = 1$$

There’s a pretty noticeable pattern emerging where $t$ cancels, and $t^2$ cancels, and $t^3$ cancels… even $(t^\infty-t^{\infty+1})=(t^\infty-t^{\infty})$ cancels (which is *not* the way to do this!), leaving

$$1{=}1\checkmark$$

### Summary

Somehow we managed to use a practical credit card fee problem to discover that the infinite geometric series is simply

$$\sum_{n=0}^\infty{At^n}=\frac{A}{1-t}$$

Notably, what we did is not a formal proof, but this demonstrates how interesting mathematics can arise from a natural question about something very common and *very* boring.

This was certainly an unexpected turn in my day.

### Postscript

The simplest^{††} formal proof uses a more generic solution of the geometric series, which with a little effort could be sussed out from the work above:

$$1+t+t^2+t^3+\cdots+t^{n-1}=\frac{1-t^n}{1-t}$$

which is true much more generally for any $t\in\mathbb{C}|t\neq1$. ($|t|<1$ only needs to be true to make the infinite series bounded!)

† This is illegal in some states! Check your local laws.

†† That I can find.