Numerical evidence suggests that:


$$\displaystyle F(s):= \lim\_{N \to \infty}\, \ln^s\left(p\_N\right)\, \prod\_{n=1}^N \left(\dfrac{\left(p\_n-1\right)^s}{p\_n^s-1} -\frac{1}{p\_n^s}\right)$$


with $p\_n$ is the n-th prime number, converges for all $\Re(s) > \frac12$.


Note that for $s=1$ the function reduces to the well known limit for $e^{-\gamma}$ (see [here](https://en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant#e.CE.B3)).


Could this be proven?



Does the limit of this product over primes converge for all $\Re(s) > \frac12$?
As you mention, $F(1)=e^{-\gamma}$, so
$$
\begin{align\*}
e^{s\gamma} F(s)=\frac{F(s)}{F(1)^s}&=\frac{\lim\_{x\to\infty} \log^s x \prod\_{p\leq x} \frac{p^s(p-1)^s-p^s+1}{p^s(p^s-1)}}{\lim\_{x\to\infty} \log^s x \prod\_{p\leq x} \frac{(p-1)^s}{p^s}}\\
&= \prod\_{p}\frac{p^s(p-1)^s-p^s+1}{(p^s-1)(p-1)^s}\\
&=\prod\_{p}\left(1-\frac{p^s-(p-1)^s-1}{(p^s-1)(p-1)^s}\right).
\end{align\*}
$$
This product converges because
$$
\frac{p^s-(p-1)^s-1}{(p^s-1)(p-1)^s}=\frac{s}{p^{s+1}}+O\left(\frac{1}{p^{2s}}\right)+O\left(\frac{1}{p^{s+2}}\right)
$$
as $p\to\infty$.


