Zeta function to Euler product

Question

I heard that the progress zeta function to Euler product is like below $$\zeta(s) = 1+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+...$$ $$\frac{1}{2^s}\zeta(s) = \frac{1}{2^s}+\frac{1}{4^s}+\frac{1}{6^s}+\frac{1}{8^s}+...$$ $$(1-\frac{1}{2^s})\zeta(s) = 1+\frac{1}{3^s}+\frac{1}{5^s}+\frac{1}{7^s}+...$$ $$\frac{1}{3^s}(1-\frac{1}{2^s})\zeta(s) = \frac{1}{3^s}+\frac{1}{9^s}+\frac{1}{15^s}+\frac{1}{21^s}+...$$ $$(1-\frac{1}{3^s})(1-\frac{1}{2^s})\zeta(s) = 1+\frac{1}{5^s}+\frac{1}{7^s}+\frac{1}{11^s}+...$$ and then repeat this process $$...(1-\frac{1}{5^s})(1-\frac{1}{3^s})(1-\frac{1}{2^s})\zeta(s) = 1$$ $$\zeta(s) = \prod_{p = prime}(1-p^{-s})^{-1}$$ I got some questions, isn't it true that last formula is true where $s \in \mathbb{C}$? even if right side(Euler product) diverges, left side(zeta function) also diverges and I think the last formula is true for all complex number s. And I'm wondering about what error is in my thought.

Answer 1

The error is that $\zeta(s)$, in its traditional summation or Euler's product forms, only converges (and does so absolutely) for $s \in \Bbb C$ such that $\text{Re}(s) > 1$. Throughout this entire process, our manipulations are only justified as a result of that: that is, in transforming $\zeta$ from a sum to a product, you implicitly have to assume absolute convergence, and thus $\text{Re}(s)>1$.