Let $P$ be the set of all odd prime numbers. I am looking for all $s\in(1,\infty)$ for them
$ A=\prod_{p\in P} (1+\frac{1}{(p-1)^s})^{p-1} $
exists (i.e. is finite). I know that it should be somehow related to Riemann zeta function but I was not sure how can I pursue the calculations.
If I use natural logarithm I will get:
$ \ln(A)=\sum_{p\in P} (p-1) \ln(1+ \frac{1}{(p-1)^s})$
whcih I am not sure is useful of not!
The observation regarding the logarithm shows that the product exists if $s >2$, since $\ln(1+x) < x$ for $x >0$, so that the expression for $\ln(A)$ is less than $\sum_{p \in P} (p-1)^{1-s}$, which converges. However, for $1 < s \leq 2$, the product diverges, since, for a given $p$, the contribution to the product from $p$ is at least $1 + (p-1)^{1-s}$, (using the binomial theorem), so at least $\frac{p}{p-1}$. Hence the product is at least $\frac{1}{2} \left( \sum_{n=1}^{\infty} \frac{1}{n} \right)$, which diverges ( the half factor occurs since $P$ consists of only the odd primes. Since the sequence of partial sums diverges anyway, it doesn't really matter).
