convergence of $\prod_{m=2}^\infty \frac {1}{1-m^{-s}}$

Question

Is the product $$\prod_{m=2}^\infty \frac {1}{1-m^{-s}}$$ convergent for all real $s>1$$\space$ ?

Answer 1

Let $P_s = \prod_{m=2}^\infty {1 \over 1 - m^{-s}}$. Then $$\ln P_s = -\sum_{m=2}^\infty \ln(1 - m^{-s}) = \sum_{m=2}^\infty \sum_{n=1}^\infty {m^{-ns} \over n} = \sum_{n=1}^\infty {1 \over n}(\zeta(ns) - 1)$$ where $\zeta$ is the Riemann zeta function. For $s>1$ the terms in this series are all finite. Furthermore, for sufficiently large $N$ and integer $k>1$, the $(k+N)$-th term is bounded above by ${1 \over k}(\zeta(k)-1)$, and $\sum_{k=2}^\infty {1 \over k}(\zeta(k)-1) = 1 - \gamma$ where $\gamma$ is the Euler-Mascheroni constant. Therefore if $s>1$ this series converges, so $P_s$ converges.