From products of infinite sums to subtraction and division

In some of the research I'm doing, I have occasion to write expressions like

A + ABA + ABABA + ABABABA + \dots = (A^{-1} - B)^{-1}

with no consideration of convergence.

Although in my setting I have (multiplicative) inverses and (additive) negatives, it nonetheless becomes interesting to ask what sorts of operations we can do with just plus and times, provided that we can do such things infinitely.

For example, with just positive integers and some questionable summations, we can get all negative fractions:

-1 = 1 + 2 + 4 + 8 + 16 + \dots
-m/n = m + (n+1)\times m + (n+1)^2\times m + \dots

and some irrationals:

\sqrt{2\pi} = 1 \times 2 \times 3 \times 4 \times 5 \times \dot

The first two sums are from the fact that 1+r+r^2+\dots=1/(1-r). Indeed, this comes from multiplying and shifting. If, e.g., we add 1 to the first equation, and then telescope the sum, we have

1+1 + 2 + 4 + 8 + 16 + \dots = 2+2 + 4 + 8 + 16 + \dots = 4+4 + 8 + 16 + \dots = 8+8 + 16 + \dots = etc.

The product is from \zeta-function regularization: we recognize the right hand side as the exponential of d\zeta(s)/ds at s=0.

Since we can multiply power series, we can get all fractions; for any given fraction, there are lots of ways of writing it as a series. For example,

1 = (-1)^2 = (1+2+4+8+\dots)^2 = 1 + (2+2) + (4+4+4) + (8+8+8+8) + \dots
= \sum (n+1)2^n.

Since \sum 2^n = -1, we clearly have \sum n2^{n-1} = 1 as well, which can be checked directly (it is d[\sum 2^n]/d2 = d[1/(1-2)]/d2 = 1/(1-2)^2.).

How about inverses of variables? Especially if you're not sure of the 2-adic sum, perhaps you'd rather use

-1/X = 1 + (1+X) + (1+X)^2 + (1+X)^3 + \dots

to get your negative and inverse Xs. This formula is hard to verify, since it still requires the same shifting tricks:

1 + ( X + X(1+X) + X(1+X)^2 + \dots ) = 1 + X + X(1+X) + X(1+X)^2 + \dots
= 1+X + X(1+X) + X(1+X)^2 + \dots
= (1+X)(1+X) + X(1+X)^2 + \dots
= (1+X)(1+X)^2 + \dots
= etc.

See, infinite sums are not associative --- the distance between terms matters --- but these manipulations are justified since infinite sums are finitely associative.

We get, either by squaring or differentiating,

1/X^2 = 1 + 2(1+X) + 3(1+X)^2 + 4(1+X)^3 + \dots

and hence all powers of X. Well, we get \pm 1/X^n; for, say, -X, the shortest is to use X^2(-1/X) = X^2 + X^2(1+X) + X^2(1+X)^2 + X^2(1+X)^3 + \dots, which telescopes in the same way.

These manipulations require that there be a number 1 which multiplies with X to X and which we can add to X. Which is true is X is, say, a matrix, or a CW complex, but not if X is a two-zero-tensor. If X is such a tensor, then X^{-1} is the tensor which convolves on both sides to the identity matrix. This is the setting of my original line (A is two-zero, and B is zero-two). In this situation, I don't have a good construction of X^{-1}.