How can the index of a sum contain $-\infty$?

Question

I have come across an interesting identity on an online forum. A user posted this identity as a challenge for the community to prove: $$\prod_{n=1}^\infty \left( 1-q^n\right)\left(1-z q^n \right)\left( 1-z^{-1}q^{n-1}\right)\left( 1-z^2 q^{2n-1}\right)\left( 1-z^{-2}q^{2n-1}\right)=\sum_{m=-\infty}^{\infty} \left(z^{3m}-z^{-3m-1} \right)q^{m(3m+1)/2}$$ A proof of this identity would be way above my paygrade, but I have a question regarding the index of the summation. I have never seen a sum that begins with $-\infty$. I have experience with improper integrals containing infinities, but that is usually taken care of with a limit. I am not sure whether the limit approach can be applied to sums of this type. I assume that we are about to split the sum into things like: $$\sum_{m=-\infty}^{\infty} = \sum_{m=-\infty}^{a} + \sum_{m=a}^{\infty}$$ However, I do not understand what it means to start a sum at $-\infty$. Is this an abuse of notation? How can a sum begin at $-\infty$?

Answer 1

$\displaystyle\sum_{n=-\infty}^{\infty} f(n)$ is shorthand for $\displaystyle\lim_{j \to -\infty, \,k \to \infty} \,\sum_{n=j}^{k}f(n)\,$, just as $\displaystyle \sum_{n=a}^{\infty} f(n)$ is shorthand for $\displaystyle\lim_{k \to \infty} \,\sum_{n=a}^{k}f(n)\,$.

Answer 2

You can rewrite $$\sum_{m=-\infty}^{a} f(m)$$ as $$(\sum_{m=0}^{\infty} f(-m))+(\sum_{m=1}^{a}f(m))$$ since the first term takes care of the sum over all the negative integers, and the second term takes care of the positive integers ≤ a. This is assuming $f$ is defined on the negative integers. From there you rewrite this in the usual way involving a limit that you are used to.