## 26 October 2007

### Trickery with divergent sums

When |q| < 1, 1 - q + q2 - q3 + ... = 1/(1+q).

And (1 - q + q2 - q3 + ...)2 = 1 - 2q + 3q2 - 4q3 + ..., viewing both sides as formal power series.

Thus we have
1 - 2q + 3q2 - 4q3 + ... = 1/(1+q)2
as formal power series. Let q = 1; then
1 - 2 + 3 - 4 + ... = 1/4.
Isn't that weird?

I learned this one today; it's an example of "Abel summation". Basically, in order to sum a sequence, we take its generating function and evaluate it at 1. Nothing too controversial there; at least to me it seems like the natural thing to do. But it looks a little more mysterious if you see it as I first did, which is without the q's. It seems "obvious" that 1 - 1 + 1 - 1 + 1 - 1 + ... = 1/2, if it's going to equal anything; we define such an infinite sum as the limit of its partial sums, which are alternately zero and one, so we should take something halfway between them. But then multiply
(1 - 1 + 1 - 1 + 1 - 1 + ...) (1 - 1 + 1 - 1 + 1 - 1 + 1 ...)
and you get 1 - 2 + 3 - 4 + 5 - 6 ..., in the following sense: let
(a0 + a1 + a2 + ...) and (b0 + b1 + b2 + ...)
be formal sums. Then we define
(a0 + a1 + a2 + ...) (b0 + b1 + b2 + ...) = c0 + c1 + c2 + ...
where
c0 = a0b0, c1 = a1b0 + a0b1, c2 = a2b0 + a1b1 + a0b2
and so on. (The definition is of course inspired by the definition on formal power series.) Then here we have ai = bi = (-1)i, and so each term in the sum defining cn is (-1)n; there are n+1 such terms. So cn = (-1)n(n+1), which gives that sum. Substituting the value 1 - 1 + 1 - 1 + ... = 1/2 into the above equation gives

1 - 2 + 3 - 4 + 5 - 6 + ... = 1/4.

Suppressing the q's makes this seem a whole lot more mysterious.

There's a Wikipedia article that shows how you can write out that sum four times and rearrange the terms to get 1; that doesn't seem all that satisfying to me, though, because with similar methods one can show that 1 - 1 + 1 - 1 + 1 - ... equals either zero or one.

John Baez has reproduced Euler's "proof" that 1 + 2 + 3 + 4 + ... = ζ(-1) = -1/12, which to me seems even crazier. But here's another reason why ζ(-1) = 1/12 "makes sense", if you buy 1 - 2 + 3 - 4 + ... = 1/4.

Let S = 1 + 2 + 3 + 4 + ...; add this "term by term" to 1/4 = 1 - 2 + 3 - 4 + ... . Then you get

S + 1/4 = 2 + 0 + 6 + 0 + 10 + 0 + ...

Then divide through by 2 to get

S/2 + 1/8 = 1 + 0 + 3 + 0 + 5 + 0 + 7 + 0 + 9 + ...

but it's clear that

2S = 2 + 4 + 6 + 8 + ...

and we might as well just stick some zeroes in there to get

2S = 0 + 2 + 0 + 4 + 0 + 6 + 0 + 8 + ...

(If we were still doing this with the q-analogue, this would amount to replacing q with q1/2.) Adding these two sums together, term-by-term, gives

5S/2 + 1/8 = 1 + 2 + 3 + 4 + 5 + ...

but we know the sum on the right-hand side is just S. So we have 5S/2 + 1/8 = S; solving for S gives S = -1/12. (I make no guarantee that this is the "most efficient" way to get ζ(-1).)
A basically identical calculation, starting with
and thus (letting q = 1) -1/8 = 1 - 8 + 27 - 64 + ... yields ζ(-3) = 1/120.

Similarly, we can "compute" ζ(-2); begin with the generating function
and letting q = 1, we get 0 = 1 - 4 + 9 - 16 + ...; let S = 1 + 4 + 9 + 16 + ... and add the two sums. Then S = 2 + 0 + 18 + 0 + ...; divide by two to get S/2 = 1 + 0 + 9 + 0 + ...

But remember that S = 1 + 4 + 9 + 16 + ...; thus 4S = 0 + 4 + 0 + 16 + 0 + 36 + ... if we let ourselves stick in zeros arbitrarily. Adding these two expressions, we get

9S/2 = 1 + 4 + 9 + 16 + 25 + ...

or 9S/2 = S; thus S = 0, which is in line with the standard result that ζ(-2n) = 0 for all integers n. This is a consequence of the functional equation for the zeta function,
since when s = -2n for some positive integer n, the right-hand side is zero.

#### 1 comment:

Theo said...

The problem with this argument is that 1+2+3+\dots is more simply a product:

1+2+3+\dots = (1+1+1+\dots)^2.

And the geometric sum 1+1+1+\dots must, by Abel's method, be 1/(1-1) = 1/0 = \infty, so \zeta(-1) = \infty^2. Not exactly a useful number.

What's going on here is that the "gaps" between the numbers matter: just as conditionally convergent sums are not commutative, divergent sums are not associative. For instance, if 0<m<n,

1 - 1 + 1 - 1 + \dots = (1-x^m)/(1-x^n) at x=1 (expand in power series).

But this converges to m/n. This is a lack of associativity in a very sensitive sense: really (1-x^m)/(1-x^n) = 1 + (m-1 0s) - 1 + (n-m-1 0s) + 1 + \dots.

Associativity says that arithmetic only depends on the topology of the expression; divergent sums depend on the geometry.