Somehow this reminds me of Zeno’s paradox but I’m not sure where my thinking goes wrong here.
Our company gives recruiting awards, and adds a “gross up” to offset the additional tax liability. I tried to approach the same problem two ways, one easy and obvious, and the other logical but not as obvious.
Assume award value A, the net award that the company wants the employee to get after taxes. Assume a tax rate R.
The easy way is to determine what gross pay G would yield a net pay of A by
G = A/(1-R)
For an award of $1000 and a tax rate of 20% we have a gross amount of exactly $1250.
OK, that’s the first way. Let’s look at it another way. You get a bonus of A, so the company adds AR to cover the taxes. But that amount is also taxable, so they have to add another AR*R, forever, for a total of
inf
[symbol]S[/symbol] A*R[sup]i[/sup] i=0
I don’t see how the sum of an infinite series can equal the result of the division shown earlier. I suspect this may be like the 0.99999… = 1 discussion. I think you can express the limit of this sum as an integral and solve it, but I haven’t done that in 30 years.
I’m probably in for a lesson on something covered in a college class on a day I was hungover…
Yes, it’s a limit-of-infinite-series thing, as you suggest. The limit only exists if the tax rate is less than 100%. But since you are talking about money, then you don’t need to take the infiniyte sum: just enough so that the remainder is less than half a cent, then round to the nearest cent (as you do in real-life payroll situations, unless you round to dollars, or some other larger currency unit).
i.e., just the first n+1 terms of the infinite series. Then:
RS_n = AR + AR^2 +…+ AR^(n+1)
Subtract the second equation from the first:
S_n - RS_n = A - AR^(n+1)
(Notice a lot of the terms on the right hand side cancelled out).
Solve for S_n:
S_n = [A - AR^(n+1)] / (1 - R)
So this gives us a “shortcut” to find the sum of the first n+1 terms.
We want the sum of the entire infinite series, so take the limit as n goes to infinity. If and only if -1 < R < 1, the limit of AR^(n+1) as n goes to infinity is zero, so the limit of the entire thing is just A/(1 - R).
As Cabbage has pointed out, your last sum is the same as A/(1-R) as long as R is between 0 and 1. Yes, the sum of an infinite number of terms can be a finite number.
Yes – though since we are talking about tax rates, it’s unlikely that you would get a negative tax rate. (Effective tax rates greater than 100% are possible though, if you include graduated means tests on social welfare payments: they exist in real life in Australia, for example).