I’m writing in response to an apparently deleted blog post that was submitted to Reddit about conditional convergence.  The post was incorrect… well, perhaps nonsensical would be a better phrase.  But even though I’m more of an algebraist, conditional convergence is still an interesting idea.  So I’m taking my response there, and turning it into a new post.

Most people probably know that the infinite series (1 + 1/2 + 1/3 + 1/4 + …) diverges.  Even though the numbers in the sum keep getting smaller and approach zero the further you get out the sequence, you can still add up enough terms of this sequence to surpass any number you like.  It just might take a long time.  Now here are a couple other simple consequences of that.

• Even if you throw away a bunch of terms from the beginning of the sequence, it still diverges.  Hopefully, this is obvious.
• Even if you only add up every other term, it still diverges.  (Proving this by contradiction is fun if you haven’t done a lot of basic analysis.  All you need to know is that if each term of one series A_n is closer to zero than the corresponding term of a series B_n, and B_n converges, then A_n converges, too.)

Okay, now lets consider the sequence we’re really interested in:

1 – 1/2 + 1/3 – 1/4 + 1/5 – 1/6 …

The difference between this sequence and the original is that every other term is negative.  Now, it’s not hard to see that this series actually converges.  But what does it converge to?  Well, in the order it’s given here, it converges to something between 1/2 and 1.  But the interesting point here is that I can rearrange the terms of this sequence to give it any convergence behavior I want.  I can make it converge to any real number x. I can make it diverge toward positive infinity.  I can make it diverge toward negative infinity.  I can even make it jump around back and forth between 0 and 1 but without actually converging anywhere.  And with all of these different convergence behaviors, I am still going to use the same terms of this series… just add them in different orders.

Making The Series Converge to x: Suppose I choose a real number x, and I want my series to converge to x.  Easy.  For simplicity, I’ll assume x is positive, but it’s easy (and obvious) to modify this for x negative, as well.  I’ll start out just taking the positive terms of the sequence… 1 + 1/3 + 1/5 + 1/7 + …, and I’ll keep going that way until the sum is strictly greater than x.  (How do I know that it will eventually become strictly greater than x?  Because I am just adding up every other term of that first sequence, and we agreed that sum diverges to infinity.  So I’ll reach x in some finite number of terms.  As soon as the sum surpasses x, I’m going to start including negative terms… – 1/2 – 1/4 – 1/6 … and so on.  I’ll continue that way until the sum drops back down below x again.  Then I’ll swap back to the positive terms, and so on.

I claim this uses all the same terms as the original series… but instead of converging to something between 1/2 and 1, it converges to any real number x.  (If x is negative, the process is the same, except you start with negative terms.  And if x is zero, you start with whichever you prefer.)  Here’s why that is.  Suppose you’re looking for term n of the original series.  In the rearranged series, it won’t occur at position n.  But every time I switch directions, I’m taking at least one term from that direction… so term number n is guaranteed to occur somewhere in the first n direction changes.  Each direction change is only finitely long (as we saw above; this is a consequence of the fact that the strictly positive series diverges), so term n occurs at some finite index in the sum.  So every term appears, and by the way I chose terms, I know I didn’t introduce any new ones, or duplicate any of them.  The terms are the same; only the order differs!

But this new series definitely converges to x.  Right at each direction change, we overshoot x by a little bit.  But the amount that we overshoot x in the sum is always no more than the last term picked in the sequence… and as we choose more terms both on the positive and negative sides, the largest unpicked term is decreasing.  So over time, we start overshooting by smaller and smaller amounts, and the sum converges.

Making the Series Diverge to Positive Infinity: I can also make the same series, with the same terms, increase without bound, thereby diverging to positive infinity.  The trick it to use the negative terms, but just use them much less often.  So add positive terms until you get to, say, 2.  Then negative terms until you get back down to one.  Then positive terms until you reach 3.  Then negative terms until you’re back down to 2.  Keep that up and you get a divergent series.  Change the numbers a bit, and your series diverges to negative infinity instead.

You can even keep the sum from approaching infinity, while also preventing it from converging to any particular finite sum.  Add positive terms up to 1.  Add negative terms until you get back to zero.  Positive terms back up to 1.  Repeat.  The details of proving that you’ve got the same terms, and that the series actually shows the desired behavior, are similar to the first case.

So that’s conditional convergence.  You really can take a series that conditionally converges, and make it show any behavior you like simply by rearranging the terms.

1. Gen Zhang / Apr 20 2009 2:29 am Nice writeup. A passing thought however: the fact that re-ordering of the summation gives different answers can be seen as an issue of choosing a canonical ordering. The usual understanding of infinite sums as converging sequences is then only one possible way. Abel summation will give a consistent answer to many apparently divergent sums, and I’ve never worked out which ordering it effectively corresponds to.

2. dysfunctor / Apr 20 2009 4:37 am You said, “Now, it’s not hard to see that this series actually converges.” Well it’s hard for me! Why does it converge?

3. cdsmith / Apr 20 2009 8:02 am Dysfunctor, a more complete argument that the series in the original order converges would go something like this: we prove, first, that every time we add a positive term, the result is a maximum for the rest of the series; and conversely, every time we add a negative term, the result is a minimum of the rest of the series. Suppose we’ve just added a positive term to get one value of the series… you can see this is a maximum because the future behavior of the sequence consists of repeatedly subtracting a number, and then adding a slightly smaller number. No matter how many times we do this, we’ll never get larger than the original. Conversely, if we’ve just subtracted, then the future of the series is built by repeatedly adding a number, and then subtracting a slightly smaller number. So every other term of the sequence of partial sums are maxima, and the remaining terms are minima, for the remainder of the sequence starting there. Now, for any epsilon, you can go far enough out to find two consecutive terms that differ by less than epsilon… but those terms are, of course, the maxima and minima of the entire rest of the sequence of partial sums… so the sequence is Cauchy, and therefore it converges.

4. cdsmith / Apr 20 2009 9:24 am Gen Zhang, Abel summation will not be equivalent to choosing a canonical ordering of the sum. Indeed, all of the limits here except the next to last paragraph either converge, or diverge toward positive or negative infinity. In these cases, Abel summation will give the same answer as the original series; the limit in Abel’s formula converges to a finite number when the series converges, and it diverges to infinity when the series diverges to infinity. Abel summation can give an answer for series like the one I build in the next to last paragraph, which oscillates between approximately 0 and 1, but that’s really only one of many possible ways you can make this series behave. I can also make this sequence oscillate in other ways, and the result from Abel summation will change. Abel summation is just as dependent on the order of terms as ordinary converges of series.

5. dysfunctor / Apr 20 2009 3:24 pm Thanks for explaining that point. It perfect sense.

6. Anonymous / May 3 2012 8:04 pm It seems to me like you could be skipping a lot of terms. Lets say x=1.7 it takes 5 positive terms to overshoot x, then you subtract 1/2, which will brings the sum below x, then you’d have to add another 7 positive terms to get above x, subtract1/3, add another ten positive terms. You can see that you’ll be using a lot more positive than negative terms.

• Anonymous / May 3 2012 10:30 pm Sure, you might use positive terms more quickly than negative terms. But that’s not the same thing as skipping terms. If you claim that a term is skipped, then which one got skipped? There is no possible answer, for the reasons explained in the answer.