This is alike choosing starting point (lower limit) for the summation. If the start is on the right of the singularity, you have contnous function on the right but discontinous once you jumped over the singularity (as you sum it up in each unit it is natural that as you add up infinity, it should appear now on on each your step).
If you start left of the zero, you'll have continous function until you cross zero where you add up singularity again.
So it would be more precise to say that we choose not a constant term but lower limit for the sum. If
\( \sum_u^{z-1} \frac1z=\psi(z) \)
then it means we fixed u at the point solution of
\( \psi (u+1)=\frac{1}{u} \)
u =1.461632144...
For \( \psi(-z+1) \) there is no such real u.
If you start left of the zero, you'll have continous function until you cross zero where you add up singularity again.
So it would be more precise to say that we choose not a constant term but lower limit for the sum. If
\( \sum_u^{z-1} \frac1z=\psi(z) \)
then it means we fixed u at the point solution of
\( \psi (u+1)=\frac{1}{u} \)
u =1.461632144...
For \( \psi(-z+1) \) there is no such real u.