Consider the analytic continuation of
$$f(s) = \prod_{n>1} \frac{n^s}{n^s - 1}$$
This is an analogue to the Riemann zeta function and generalization of the Euler product $$ \zeta(s) = \prod_{p = \text{prime}} \frac{p^s}{p^s - 1} $$ and hence both product definitions of $f(s)$ and $\zeta(s)$ are valid for $\operatorname{Re}(s)>1$.
I wonder where are the zeros $z_i$ of the analytic continuation of $f(s)$?
$$f(z_i) = \prod_{n>1} \dfrac{n^{z_i}}{n^{z_i} - 1} \overset{?}{=} 0.$$
Motivation
Imo, it is naturally interesting because the counting function is simply $n-1$ rather than $\pi(n)$ since we replaced primes by the naturals $> 1$. So we have a closed form of the counting function, but not yet info about the zero's. This seems like a potential interesting insight to come for zeta like functions.
Related
This Q&A might be related: Explicit formula for floor(x)?
Some analysis
In analogue to the prime zeta function or the trick performed in this Q&A: Roots and analytic continuation of $T(s)=\sum_{n>0} (n^s + n^{-s})^{-1} $?
(and many other proofs related to the Riemann zeta function or primes).
It seems a good idea to once again take the logarithm.
As user Conrad suggested.
$$
\begin{align}
f(s) &= \prod_{n>1} \frac{n^s}{n^s - 1}\\
f(s) &= \prod_{n>1} 1 + \frac{1}{n^s - 1}\\
\ln(f(s)) &= \sum_{n>2} \ln\left(1 + \frac{1}{n^s - 1}\right)
\end{align}
$$
Attempt 1 $$ \ln\left(1 + \frac{1}{x-1}\right) = [ - \ln(x-1) ] + \sum_{k = 1}^{+\infty} \frac{(-1)^k}{k} (x-1)^{- [-1]k}$$
where the terms $[- \ln(x-1)]$ and $[-1]$ exist when $|x-1| < 1$ and vanish otherwise.
For $Re(s) >1$ we have $|n^s - 1|>1$ so the term $[- \ln(x-1) ]$ vanishes, and this implies that for $Re(s)>1$ we can continue the expansion, therefore:
$$
\ln(f(s)) = \sum_{n>2} \ln\left(1 + \frac{1}{n^s - 1}\right) = \sum_{n>1} \sum_{k = 1}^{+\infty} \frac{(-1)^k}{k} (n^s-1)^{-k}
$$
and for $0 < Re(s) < 1$ we have
$$
\ln(f(s)) = \sum_{n>2} \ln\left(1 + \frac{1}{n^s - 1}\right) = \sum_{n>1} -\ln(n^s - 1) + \sum_{n>1} \sum_{k = 1}^{+\infty} \frac{(-1)^k}{k} (n^s-1)^{k}
$$
However the $\log$ term and the complicated double sum have issues of convergence or continuation that may be solvable, but another simpler series expansions, the following one, seems better.
Attempt 2
$$
\ln\left(1 + \frac{1}{x-1}\right) = \ln(-x) + x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \frac{x^5}{5} + \ldots .
$$
Now by putting $x=n^s$ we get
$$
\ln(f(s)) = \sum_{n>1} \ln\left(1 + \frac{1}{n^s-1}\right) = \sum_{n>1 }\ln(-n^s) + n^s + \frac{n^{2s}}{2} + \frac{n^{3s}}{3} + \frac{n^{4s}}{4} + \frac{n^{5s}}{5} + \ldots .
$$
This reduces further to
$$
\ln(f(s)) = \left[\sum_{n>1} \ln(-n^s)\right] + \zeta(s) + \frac{\zeta(2s)}{2} + \frac{\zeta(3s)}{3} + \frac{\zeta(4s)}{4} + \frac{\zeta(5s)}{5} + \ldots .
$$
The zeta sums are fine but the $\ln$ part is problematic ...
$$
\sum_{n>1} \exp(\ln(-n^s)) = \sum_{n>1} - n^s.
$$
So
$$
f(s) = \pm \zeta(s) \exp\left(\zeta(s) + \frac{\zeta(2s)}{2} + \frac{\zeta(3s)}{3} + \frac{\zeta(4s)}{4} + \frac{\zeta(5s)}{5} + \ldots\right)
$$
So for $Re(s) > 0$ we have the zero's of $\zeta(s)$ (RH!) and the cases where
$$
g(s) = \zeta(s) + \frac{\zeta(2s)}{2} + \frac{\zeta(3s)}{3} + \frac{\zeta(4s)}{4} + \frac{\zeta(5s)}{5} + \ldots\text{ is }- \infty.
$$
These cases occur IFF $s = 1/m$ for integer $m>1$ due to the pole in $\zeta(1)$.
So attempt 2 is pretty successful, but nevertheless one wonders if there are more zero's or continuations possible for $Re(s) \le 0$: is this the case?
I am not sure if in that case ($Re(s) \le 0$) it is still correct to say that
$$
f(s) = \pm \zeta(s) \exp\left(\zeta(s) + \frac{\zeta(2s)}{2} + \frac{\zeta(3s)}{3} + \frac{\zeta(4s)}{4} + \frac{\zeta(5s)}{5} + \ldots \right),
$$
just as it was dubious with this Q&A: Roots and analytic continuation of $T(s)=\sum_{n>0} (n^s + n^{-s})^{-1} $?
It is clear that $f(s)$ has a singularity at $s=0$ but what happens on the rest of that line $R(s)=0$ is unclear to me.
Special thanks to user Conrad.
How could I proceed further?
edit
both attempt 1 and attempt 2 have convergeance issues.
So I am back to square 1.
any ideas ?
