Given a divergent series $\sum_n a_n$, one can sometimes define a zeta function: $$\zeta_a(s) = \sum_n a_n^{-s}$$ if the sum converges for large enough $s$ and admits an analytic continuation. My question is: supposing this is possible, can the large-cutoff asymptotics of the "mollified" version of this sum be extracted from $\zeta_a$? In other words, is there a large-$\Lambda$ expansion like: $$\sum_n \frac{1}{a_n^s}\, f(a_n/\Lambda) = \zeta_a(s)+\sum_\rho K_\rho\,\Lambda^{\rho-s},$$ (where $f(0)=1$ and $f$ is nice), and if so, can we express the $K_\rho$ in terms of $\zeta_a$ and $f$?
Motivation
In this post, Terry Tao gives exactly what I'm after but for a different notion of zeta-function regularization. He defines:
$$D(s) = \sum_{n=1}^\infty \frac{a_n}{n^s},$$
in terms of which he says:
$$\sum_{n=1}^\infty \frac{a_n}{n^s}\,\eta(n/N) = D(s) + \sum_\rho C_{\eta,\rho-s-1}r_\rho N^{\rho-s}+\cdots,\tag{!}$$
where $\rho$ ranges over the poles of $D$, $r_\rho$ is the residue at $\rho$, $\eta$ is a cutoff function, and:
$$C_{\eta,s} = \int_0^\infty x^s\,\eta(x)\,dx.$$
The formula (!) is very neat because it shows that the coefficient of $N^0$ in the $1/N$ expansion is independent of $\eta$, and it explicitly shows how the other coefficients do depend on $\eta$.
These two methods of zeta-regularization are distinct; for example, the former one can't sum $a_n=1$ while the latter one can. It seems like the former one does appear in the wild (especially in the study of functional determinants e.g. the Minakshisundaram–Pleijel zeta function). Thus it would be nice to have a version of (!) for the former procedure.