TITLE: Does the limit of this product over primes converge for all $\Re(s) > \frac12$?
QUESTION [6 upvotes]: 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).
Could this be proven?

REPLY [8 votes]: 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 $p\to\infty$.

REPLY [3 votes]: If you write the general terms in terms of $x = 1/p_n,$ when $x\>1/2,$ the expansion in $x$ starts as $1-s x + O(x^{2 s})\dots,$ so your limit converges for $s>1/2.$ For $s=1/2,$ the expansion starts with $1+x/2,$ so your limit blows up (the product of the first $n$ terms, including the log fudge factor, should be of order of $\log n.$