For $\left|z\right|<1$ we have
$$S_2(z):=\dfrac{1}{(z-1)^2}+\dfrac{1}{z-1}= 0\,z^0+1\,z^1+2\,z^2+3\,z^3+\dots$$
and then
$$\lim_{z\,\to\,1}\left(S_2(z)-\frac{1}{\log(z)^2}\right)=-\dfrac{1}{12}.$$
A connection to the Euler-MacLaurin formula can be motivated by comparing the $z\frac{\partial}{\partial z}$-derivatives of
$\sum_{n=0}^\infty z^n=-\frac{1}{z-1}$
$\int_0^\infty z^t\,{\mathrm d}t=\int_0^\infty {\mathrm e}^{t\log(z)}\,{\mathrm d}t=-\frac{1}{\log(z)}$.
For a generalization coming from higher derivatives I want to know the closed form expressions
$S_n(z)=\sum_{k=0}^{n-1}\dfrac{s_{n,k}}{(z-1)^{n-k}}$,
so that
$\lim_{z\to 1}\left(S_n(z)-\frac{1}{\log(z)^n}\right)$
is finite. This function is the divergent part of the expansion of $\frac{1}{\log(z)^n}$.
Mathematica tells me the coefficients start out as $1,\, \frac{n}{2},\, \frac{n}{2}\frac{1}{12}(3n-5),\dots$. So indeed, for $n=2$ this means we have $s_{2,0}=1$ for a double pole, $s_{2,1}=1$ for a single pole and then the finite part of the log-expansion is $\frac{1}{12}$.
Thoughts:
I could use $\log(z)=1-\left(1-\sum_{k=1}^\infty\,(-1)^{k}\,\frac{1}{k}\,(z-1)^k\right)$
and the expansion for $\frac{1}{1-x}$, but then I end up with two levels of taking arbitrary powers. I know there is this formula for exponentiation of power series, but using it twice on the alternating coefficients on a modified log seems intractable.
Playing around makes me think expanding $\left(\dfrac{\log(z)}{1-z}\right)^n=\left(\sum_{k=1}^\infty\,(-1)^{k}\,\frac{1}{k}\,(z-1)^{k-1}\right)^n$
is easier and leads me to the desired result, but I still end up with an expression where I must use the above formula on a sum and then I have iterated sum.
Is there an approachable way of deriving the expansion of $\frac{1}{\log(z)^n}$ up to the first positive order in $z-1$?
I'm now thinking maybe one can also obtain the result by by taking n'th derivatives.