The Weierstrass $\zeta$-function is defined as follows for a lattice $\Lambda$, where a lattice is a discrete subgroup of $\mathbb{C}$ containing an $\mathbb{R}$-basis for $\mathbb{C}$.
$$\zeta(z) = \frac{1}{z} + \sum_{\omega\in\Lambda\setminus\{0\}}\left(\frac{1}{z-\omega}+\frac{1}{\omega}+\frac{z}{\omega^2}\right)$$
Doesn't $\sum_{\omega\in\Lambda\setminus\{0\}}\frac{1}{\omega}$ equal 0, because if $x \in \Lambda$ then $-x\in\Lambda$? I understand that the term appears when differentiating the logarithm of the Weierstrass $\sigma$-function, but why is it written anywhere if it could just as well be left out?
You can'd do that rearrangement there because as written it is not absolutely convergent. On the other hand, with a small simplification,
$$\sum_{\omega\in\Lambda\setminus\{0\}}\left(\frac{z}{(z-\omega)\omega}+\frac{z}{\omega^2}\right)$$
is absolutely convergent, because $\omega$ is quadratic is both denominators.
On a related note, an addition of some constant is often required to construct holomorphic/meromorphic functions with prescribed zeroes and poles and residues. See Weierstrass theorem and Mittag-Leffler's theorem and their proofs on more along this line of thought.