Yes, I thought of that but was not sure exactly how to deal with it. Upon finding the residues, one manages to get the series in question among them. Then, solving for it would do the trick. But, of course, it does not work due to the branch. It sure looks like it should though. I wonder if there is a way to deal with it in this manner?.

i.e.

The series at the integers n:

$$\frac{(-1)\ln(n)}{(z-n)^{2}}+\frac{(-1)^{n}(1/n+H_{n}\ln(n))}{z-n}+.....$$

some interesting series are involved. such as, $$\sum_{n=1}^{\infty}\frac{(-1)^{n}\ln(n)}{n^{2}}=1/2\zeta'(2)+\frac{\pi^{2}}{12}\ln(2)$$

Of course, the negative integers have complex involved due to the negative.

anyway, I just played with it a little. It looks like it should work if handled properly.

Maybe not though.

sorry, gotta go to work now. Be back later.