I'd like to know how fast the infinite product $$A_{s,k}=\prod_{p}\left(1-p^{-1}\right)^{s-k+1}\sum_{m=0}^{k-1}{s\choose m}\left(1-p^{-1}\right)^{k-1-m}p^{-m}$$ converges, where the product is taken over all primes $p$. In particular, I'd like to know how many primes I'd need to multiply over to get $d$ digits of accuracy in terms of $s,k$. This comes from Jerry Hu's paper The probability that random positive integers are $k$-wise relatively prime.
How would I go about doing this? My first guess would be to start with something like $$A_{s,k}\left(N\right)=\prod_{i=1}^N\left(1-p_i^{-1}\right)^{s-k+1}\sum_{m=0}^{k-1}{s\choose m}\left(1-p_i^{-1}\right)^{k-1-m}p_i^{-m},$$ where $p_i$ is the $i$th prime, and look at $$\varepsilon_N:=\left|A_{s,k}-A_{s,k}\left(N\right)\right|,$$ but I'm not sure where to go.
In a similar paper of László Tóth, The probability that $k$ positive integers are pairwise relatively prime (pdf link), the author made use of the AM-GM inequality, the Bernoulli inequality, and the fact that $p_n>2n$ for $n\geq5$ to analyze $-\log A_k$ for $$A_k=\prod_{p}\left(1-p^{-1}\right)^{k-1}\left(1+\frac{k-1}{p}\right),$$ obtaining $r$ exact decimals for the value of $A_k$ from $$\prod_{n=1}^N\left(1-p_n^{-1}\right)^{k-1}\left(1+\frac{k-1}{p_n}\right)$$ for $N\geq\frac{1}{2}\left(\left(k-1\right)^210^r+k-3\right)$.
So, as a second candidate for the estimates, we may look at something like $$-\log A_{s,k}=\sum_{p}\log\left(\left(1-p^{-1}\right)^{k-1-s}\left(\sum_{m=0}^{k-1}{s\choose m}\left(1-p^{-1}\right)^{k-1-m}p^{-m}\right)^{-1}\right)=\sum_{i=1}^\infty\log\left(\left(1-p_i^{-1}\right)^{k-1-s}\left(\sum_{m=0}^{k-1}{s\choose m}\left(1-p_i^{-1}\right)^{k-1-m}p_i^{-m}\right)^{-1}\right),$$ though this looks to be a much wilder beast than in Tóth's paper.
I have almost no experience in determining rates of convergence. If anyone can help and/or provide references for study, that would be very much appreciated. Thank you.
EDIT: Upon further inspection of Tóth's estimations, I seem to have found a small typo on page 18. On the line "Now using $p_n<2n$, valid for $n\geq5$ . . .," I believe this should say $p_n>2n$, which is indeed valid for $n\geq5$. The subsequent estimates still work properly, as it was this property that we needed, in the first place.