The Help Needed To Get to One

This is a view from a lower vantage point of the Disaster Transport roller coaster car.

So, it’s established that my little series, representing the number of rides we could expect to get if we based re-riding on a fair coin flip, is convergent. So trying to figure out the sum will get a meaningful answer. The question is, how do we calculate it?

My first impulse is to see if someone else solved the problem first, for exactly the reasons you might guess. This is a case where mathematics textbooks can have an advantage over the web, really, since an introduction to calculus book is almost certain to have page after page of Common Series Sums. Figuring out the right combination of keywords to search the web for it can be an act of elaborate guesswork. Mercifully, Wikipedia has a List of Mathematical Series which covers my problem exactly. Almost.

In the section of “low-order polylogarithms” the second entry tells us that the finite series \sum_{k = 1}^{n} k z^k = z\frac{1 - (n + 1)z^n + n z^{n + 1}}{(1 - z)^2}. The series I wanted the solution for was \sum_{k = 0}^{\infty} k \left(\frac12\right)^{k + 1}, which … you kind of have to squint to see a resemblance there. Let me explain why that looks like the same pattern to me.

The k’s at least carry through. The symbol labelled z in the Wikipedia series is meant to be a fixed number, and that’s pretty clearly the one-half in my series. My series starts out evaluating an expression at k = 0, while Wikipedia’s starts at k = 1. On the other hand, when k is zero, k times one-half raised to the first power is zero. I could just as easily say my series was \sum_{k = 1}^{\infty} k \left(\frac12\right)^{k + 1} and I wouldn’t be missing any parts of it.

Wikipedia’s series has k multiplied by z raised to the k power. My series has k multiplied by one-half raised to the k plus one power. That’s a little more serious; even if I let z be one-half, the terms in my series don’t match those in Wikipedia’s. But one-half to the power k plus one is equal to one-half times one-half raised to the power k. I can think of my series as \sum_{k = 1}^{\infty} \frac12 k \left(\frac12\right)^{k}. That may not seem like much of an improvement, but, since my series converges, and every term in the series is multiplied by the same one-half, I can pull that one-half out of the summation. That is, I can look at one-half times what I would get from \sum_{k = 1}^{\infty} k \left(\frac12\right)^{k}, which is a lot more like the formula Wikipedia offered.

It’s not quite there, yet. Wikipedia’s formula talks of a finite sum, of just n terms, while I need infinitely many terms. Ah, but, remember that convergence proved last time around. It means, among other things, that if we add up enough terms we can get as close to the sum of infinitely many terms as we like. The formal calculus trick I’d be using here is to take the limit of what Wikipedia’s formula tells me, as the number of terms n grows infinitely large. We can put this in reasonably plain English, though. Let me put one-half in for z in Wikipedia’s formula, first:

\sum_{k = 1}^{n} k \left(\frac12\right)^{k} = \left(\frac12\right) \frac{1 - (n + 1)\left(\frac12\right)^n + n \left(\frac12\right)^{n + 1} }{(1 - \left(\frac12\right))^2}

I want to look at that numerator on the right-hand side and think about what happens in it if n is a really big number. That 1 doesn’t have an n in it at all, so, it’s not going to change any as n gets larger. Now, (n + 1)\left(\frac12\right)^{n}, when n is really large … is going to be a big number multiplied by a small number. That might be anything. If you try it with a couple particular numbers you’ll see that it turns out to be zero. (If you want to prove this, L’Hopital’s rule is probably the way to go, but you’ll take my word, won’t you?) And the third part, n\left(\frac12\right)^{n + 1}, is just the second term over again. (If n is really big, n and n+1 might as well be the same number. And again, L’Hopital’s rule is the way to prove this. Or, think of it as \frac14 n\left(\frac12\right)^{n - 1} instead.) So, if n is getting really big, then the numerator there is going to be just 1.

So: if n is growing infinitely large, then Wikipedia’s formula tells me \sum_{k = 1}^{\infty} = \frac12\frac{1}{(1 - \left(\frac12\right))^2} = \left(\frac12\right)\frac{1}{\frac14} = \frac12\frac41 = 2. And my series was one-half of this sum, so, the expected value of the number of rides — remember, that was what I was interested in? — is one-half of two, or 1.

I go through all this because it’s pretty common to find in mathematics that there’s a formula which is almost what you want, often with some factor or a summation term or something like that a bit off. Usually the formulas can be used anyway, since we can make corrections for things like here where my series had an extra power of one-half that the Wikipedia formula didn’t. It’s one of the needed mathematical skills to spot that a formula is usable, and to make whatever modifications — to your series or to the printed formula’s — are needed to use it.

Yet even though we’ve got the right answer there’s still something not quite satisfying about it. I started out trying to do things by hand because I wanted to know the magic by which Maple figured out the answer to be 1, and I resorted to a formula which might even be right. I have no particular reason to think Wikipedia is mistaken here, but, I didn’t derivate the formula, I didn’t double-check it, and to be honest I don’t remember hearing of “polylogarithms” before looking at this page. As a method of double-checking the work I let the computer do, this hasn’t been very useful.

So: can I work out my series using only stuff I can actually check? (Yes, I can; with a little background material, so can you.)

Author: Joseph Nebus

I was born 198 years to the day after Johnny Appleseed. The differences between us do not end there. He/him.

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