I was having dinner with a visiting colleague this week when talk turned to what we were teaching this term.
He mentioned the part of calculus dealing with infinite series (the bane of many students) and how really he just mentally compares everything with the harmonic series: 1 + (1/2) + (1/3) + (1/4) + … This series diverges; that is, the sum is infinite (contrast this with the convergent series (1/2) + (1/4) + (1/8) + … = 1). I then casually mentioned that if you take the harmonic series and throw out the terms whose denominators contain a 9 then the resulting series converges.
“I don’t believe that. How can that be true?”
That was my first reaction to hearing this fact, too, because we get so used to the idea that the harmonic series diverges that we can’t believe that throwing out a few terms, even infinitely many, will make a difference. And, there’s nothing special about 9; you can toss out terms containing any particular digit. In fact, you can pick any finite string of digits, toss out the terms containing those, and the result converges. With that set-up, let’s talk about what all this means and how we can prove it.
The first thing to figure out is the correct definition. After all, I cannot add up infinitely many numbers (there’s not enough time), but my intuition about the series (1/2) + (1/4) + (1/8) + … = 1 tells me that I should be able to make sense of this idea. By the way, I hope it’s clear why the sum of this series is 1–imagine you are trying to walk across the room; you first must walk halfway, then half that distance (1/4), then half that distance (1/8), and you eventually get where you’re going so all those fractions add up to 1. Zeno would disagree, of course, since it’s precisely this example that forms his paradoxical assertion that motion is impossible.
What to do? Well, I can add up finitely many numbers, look for a pattern, and then see what happens if I let that finite number of terms get arbitrarily large. In general, given an infinite series ∑ ai = a1 + a2 + a3 + … we form the sequence of partial sums sn = a1+ a2 + … + an and then define the sum of the infinite series to be the limit of this sequence of partial sums, if it exists. If the limit doesn’t exist we say the series diverges. So, the powers of 2 series has partial sums that look like
And since the 1/2n bit gets really small as n goes off to infinity we say the sum of the series is 1. But what about the harmonic series? There is no obvious formula for its partial sums, but we can argue that they go off to infinity in a couple of ways. One way is to look at the 2nth partial sum:
We can make this as large as we like by taking n to be large, so the series diverges. Here’s another way to see it. Each term of the harmonic series may be thought of as the area of a rectangle of base 1 and height 1/n. Consider the graph of the function y=1/xand compute the area under the curve from x=1 to n+1:
Since all these rectangles lie above the curve and the sum of their areas is the nth partial sum of the harmonic series we obtain the inequality
Since the natural logarithm function increases without bound, so do the partial sums and the harmonic series diverges. As an aside, this formulation is very useful for estimating the partial sums. By a similar analysis, we can get an upper bound on the partial sums by drawing rectangles under the curve y=1/x to approximate the area and we then discover that
This observation leads to the old joke that the harmonic series is known to diverge but it has never been observed to do so. Suppose we began at the Big Bang, 13.8 billion years ago, and added one term of the harmonic series per second. The partial sum today would be between 40.6 and 41.6; that’s one slow trip to infinity.
But what about the assertion I made above that if you pick your favorite digit, 9 say, and removed all the terms whose denominators have a 9 in them, then the resulting series converges? This may not seem plausible at first, but here’s a heuristic. When we have a large number of digits, a random number contains a 9 as one of the digits with very high probability. Put another way, random numbers with a large number of digits, none of which is a 9, are rare. Let’s try to count this closely. The first term we remove from the harmonic series is 1/9. We then remove 1/19, 1/29,…,1/89, but then we remove the next ten terms 1/90, 1/91,…,1/99. We do this for the next few hundred terms until we get to 1/900 where we remove 100 consecutive terms. In general, if we call the series with all the terms involving 9 deleted D, then we have D = D1 + D2 + … where Di is the sum of the terms whose denominators have exactly i digits, none of which is a 9. The number ofi-digit numbers that do not contain a 9 is 8·9i-1 since the first digit cannot be 0 or 9 and the remaining digits cannot be a 9. Each term in Di is of the form 1/r, where r is an i-digit number. So each r≥10i-1 and as a result Di ≤ 8·(9/10)i-1. It follows that we have the inequality
and since D is a series of positive terms which is bounded above it must converge.
Note that there was nothing special about 9, and if you want to choose 0 only a small modification is required. Moreover, it should be pretty clear that it’s possible to choose any finite string like 3456789, remove all the terms in which this appears, and the result is still a convergent series, albeit with a much larger sum.
Series like this are called Kempner series, and there is a nearly 100-year-old paper about them here. Finding their exact sum is generally an intractible problem and we must be content with numerical estimates. We know the sum is less than 80 (for any digit except 0), but with the 9′s removed the actual sum is approximately 22.92067.
Series are great fun and they will make an appearance again when I resume my collection of articles about π. There are lots more surprises in store so stay tuned.
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