You probably got the sense that I like teaching mathematics, but it turns out I like doing it too. Below is a little narrative description of a number theoretical question I was fooling around with. It has minimal prerequisites for understanding — things you’ll probably know if you know about convergent series (a_1 + a_2 + a_3 + …. is finite). It’s fun, so check it out if you’re in touch with your inner mathematician, or just want to reconnect with her:

While considering strictly increasing sequences of natural numbers a bit ago, I wondered if the property that the pairwise differences be strictly increasing is enough to ensure that if you sum the reciprocals of the terms, the series converges; it turns out this is sufficient, since the largest sum that could possibly result is from the sequence (2,3,5,8,12,17,23,…) whose pairwise differences are the smallest they can possibly be (1,2,3,…). If one drops the first term (2), then the sequence (3,5,8,12,17,…) is bounded below by the triangle numbers (1,3,6,10,15,…), and consequently the series of reciprocals of the former is bounded above by the reciprocals of the latter.

It turns out that the latter’s reciprocal series converges (to 2, which can be seen by a neat little arithmetic trick using the fact that the nth triangle number can be written n(n+1)/2, and the fact that 2/[n(n+1)] = 2[1/n – 1/(n+1)]), and therefore the series associated with (2,3,5,8,…) does too. Since that sequence is the smallest (therefore having the largest reciprocal sum) strictly increasing sequence with the property that the pairwise differences are also strictly increasing, all sequences with that property have convergent reciprocal sums as well.

This made me wonder if all one needed was weakly increasing pairwise differences, with only infinitely many ‘increases in difference’, to ensure that the sum of the reciprocals would converge. Spectacular as that would be, it turns out not to be true:

Since for any fixed natural number k, the sum of the 1/kn diverges (if not, then the comparison test would imply that all the other (k – 1) such subseries converge too), one can always find finitely many n_1(k) > … > n_j(k) such that

1/k(n_1(k)) + … + 1/k(n_j(k)) > 1,

starting from any arbitrarily large n_1. Thus, we can just construct a strictly increasing sequence of natural numbers with weakly increasing pairwise differences, so that the pairwise differences only actually increase when the sum of the reciprocals of the numbers with the given pairwise differences adds up to at least 1. The associated series clearly then diverges.

A natural question is then whether having the sequence of pairwise differences be weakly increasing, *and* such that there are infinitely many constant pairwise differences (ie. infinitely many pairwise differences remain constant from one term to the next), is enough to ensure divergence. The answer is no, since we can just essentially splice two triangle number sequences into one, with pairwise differences (1,2,2,3,3,4,4,…), resulting in the sequence (1,3,5,8,11,15,19,24,29,35,…). This sequence’s reciprocal sum can be split into two series by taking every other term – one with denominator terms (1,5,11,19,29,…), and the other with (3,8,15,24,…) – each of which is bounded above by the reciprocals of the squares (which converge).

So by this point, I’d answered my original question and a couple natural follow-ups, and it’s becoming clear that what I’m really after is some condition on strictly increasing sequences of natural numbers that distinguishes those whose reciprocal sums converge from those whose reciprocal sums diverge.

More on this in a subsequent post (enough to chew on for now, yes?)…

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