Background
The general context for this question is the topic of rational zeta series. What I've found so far, is that it usually the case that sums of the form $$\zeta_{f} := \sum_{n=2}^{\infty} f(n) \left( \zeta(n)-1 \right) \label{1}\tag{1} $$ are more difficult to evaluate than sums of the form $$S_{f} := \sum_{n=2}^{\infty} f(n). \label{2}\tag{2} $$
Examples
- For instance, if we take $g(x) = \frac{1}{x!}$, then we have $$ S_{g} = e-2, \tag{3}\label{3} $$ but for $ \zeta_{g}$ a closed form is not known.
- Also if we consider $h(x) = \frac{1}{x^2}$, then we have $$S_{h} = \frac{\pi^2}{6} -1. \tag{4}\label{4}$$ However, an evaluation of $\zeta_{h}$ appears to be still out of reach.
- If we take $v(x) = \frac{(-1)^x}{x-1}$, we have $$ S_{v} = \ln(2), \tag{5}\label{5}$$ but an evaluation of $\zeta_{v}$ seems to be elusive -- see also equation (1.23) in corollary 4 of the following paper.
Question
I wonder whether there are any cases for which the situation is reversed. In other words, I am curious as to whether there are examples of functions such that \eqref{1} can be evaluated, but \eqref{2} cannot. Both $\zeta_{f}$ and $S_{f}$ should be finite, so an example like $q(x) = \frac{1}{x}$ -- where we have $\zeta_{q} = 1-\gamma$, but $S_{q}$ diverges -- does not count. Moreover, the function ought to be analytic on $\mathbb{R}_{\geq 2}$.
Note: in examples \eqref{3}, \eqref{4}, and \eqref{5}, the sums can be decomposed into \begin{align} \zeta_{f} &= \sum_{n=2}^{\infty} f(n) \zeta(n) - S_{f} \\ &:= \zeta'_{f} - S_{f},\end{align} where $\zeta'_{f}$ can't be evaluated (yet), so $\zeta_{f}$ can't be evaluated either.