How one infinite sum equals every number

Understanding a mathematical paradox

Photo by benjaminec on Unsplash

The rule for commutativity of addition is a + b = b + a. It makes sense, as it shouldn’t matter in what order we count certain objects. All that matters is that we count them all exactly once. If you feel extremely sadomasochistic today, you can check this formal proof.

Inductively, commutativity can be extended to

where σ is any permutation (summation order). It might seem we could also extend this to infinite sums. Well, it turns out we can’t. Not only that, but one sum can evaluate to any real number, depending on its summation order. This post is about understanding why that happens.

We define an infinite sum as a limit of partial sums:

First, we’ll show that the sum

can attain values ∞ and -∞, depending on the order of our summation. You’ll have to take my word (or read this short proof), that taking the positive and the negative parts of the sum, gets us

If for any number n, we take away the first n terms from either of the above sums, we have taken away a finite sum, therefore subtracted a finite number. Because this does not affect infinity, we get

previous sums with the first n terms taken away.

By the definition of an infinite sum being equal to ∞, we have that for arbitrarily large numbers x, we can take some finite amount of terms away from the sum and add them up into a number bigger than x.

But us taking away those terms didn’t affect infinity one bit. So we can do the exact same thing again to make our remaining terms produce yet another arbitrarily large number.

Let’s describe the order in which we sum up our initial infinite sum. We’ll iteratively sum up terms from the positive and negative parts, adding them to our partial sum.

• Add (finitely many) terms from the positive part of the sum, so that the partial sum will be larger than 1 + 1/2.
• Add the first negative term -1/2. Our partial sum is now larger than 1.
• Add (finitely many) remaining terms from the positive part of the sum, so that the partial sum will be larger than 2 + 1/4.
• Add the second negative term -1/4. Our partial sum is now larger than 2.
• Add (finitely many) remaining terms from the positive part of the sum, so that the partial sum will be larger than 3 + 1/6.
• Add the third negative term -1/6. Our partial sum is now larger than 3.
• …

See that for every added negative number, we pick a huge amount of positive numbers to balance it off. Because of this, our partial sum goes to infinity.

Python code for the above process (note that the partial sums grow at about the rate of log, so waiting for them to reach e.g. 100 will take more than a billion years).

A very important observation is that every term of our initial sum is eventually added up by this process (meaning we really shuffle the order of addition). This depends on us always having enough positive terms to produce arbitrarily large numbers.

If we wanted our sum to be -∞, the process would be completely symmetric. With a little more effort we can make the sum evaluate to any real number.

We use the fact that our terms 1 / n go to 0, so we can add up smaller and smaller numbers, which makes it possible to achieve arbitrarily precise approximations of our chosen limit.

Choose any r and proceed similarly as before:

• Add up terms from the positive part and stop immediately when the partial sum becomes (strictly) larger than r.
• Add up terms from the negative part and stop immediately when the partial sum becomes (strictly) smaller than r.
• Continue adding the remaining terms from the positive part and stop immediately when the partial sum becomes (strictly) larger than r.
• Continue adding the remaining terms from the positive part and stop immediately when the partial sum becomes (strictly) smaller than r.
• …

Python code for the above process (for |r| < 5, partial_sum should converge quickly).

In each step, we have to add at least one term to jump from strictly under r to strictly over r (or vice-versa). So, after n steps of the above process, all the remaining terms are smaller than 1 / n. Is it then possible for our partial sum to leave the interval

[r - 1 / n, r + 1 / n] ?

Let’s split two cases:

• If we are below r, then we can’t suddenly jump to something larger than r + 1 / n, as we’re adding a term smaller than 1 / n.
• If we are above r, then we can’t jump to something smaller than r - 1 / n, as we’re subtracting a term smaller than 1 / n.

Thus our partial sum will remain in the above interval for all steps after n. When we send n to infinity, we have that the partial sum will close in on r. This means that our infinite sum, with the constructed order of summation, will equal r.

The value of the sum, with terms ordered as we’ve written them, is log(2).

Conclusion

We saw that relying on intuition doesn’t always work in math. Many paradoxes show us how math is not completely aligned with our adaptation to reality. It’s a perfectionist’s view of reality. And understanding those kinds of paradoxes through subtleties is what makes it fun.