To show $\zeta(s)$ is convergent (click here and go to theorem 16.2 on page 1 and 2), it was shown that-
Step 1. $\zeta(s) = \prod_p(1 − p^{−s})^{−1}$, thus $\zeta_m(s) = \prod_{p \leq m}(1 − p^{−s})^{−1}$.
Step 2. $\zeta_m(s)$ converges to $\zeta(s)$, by using the definition of uniform converges, thus, $\prod_{p \leq m}(1 − p^{−s})^{−1}$ converges to $\prod_{p}(1 − p^{−s})^{−1}$.
Step 3. $\prod_{p}(1 − p^−s)^{−1}$ is convergent (a series is convergent if the sequence of its partial sums $(S_{1},S_{2},S_{3},\dots )$ tends to a limit; that means that the partial sums become closer and closer to a given number when the number of their terms increases).
The problem is, I don't see the necessity of step 2, why it is required?
Isn't it enough to show that $\zeta(s) = \prod_p(1 − p^{−s})^{−1}$, since $\prod_{p}(1 − p^−s)^{−1}$ is convergent, $\zeta(s)$ is also convergent?
The formal statement: For $Re(s) > 1$ we have $\zeta(s) = \sum_{n\geq 1}n^{-s}= \prod_p(1 − p^{−s})^{−1}$where the product converges absolutely. In particular, $\zeta(s)\neq 0$ for $Re(s) > 1$. Plz, break down the motivation, objective and inference obtained by each step above which might not be written/said explicitly, I am learning the subject, did little calculus many years ago, thanks.
For the purposes of proving Theorem 16.2, talking about uniform convergence is indeed completely unnecessary. The notes show that for any $s$ with $Re(s) > 1$, for any $\epsilon > 0$, $|\zeta_m(s)-\zeta(s)| < \epsilon$ holds for $m$ large enough. This shows $\zeta_m(s) \to \zeta(s)$ pointwise, and since $\zeta_m(s) = \prod_{p \le m} (1-p^{-s})^{-1}$, we have $\zeta(s) = \prod_p (1-p^{-s})^{-1}$ for each $s$ with $Re(s) > 1$ (recall that the definition of $\prod_p (1-p^{-s})^{-1}$ is $\lim_{m \to \infty} \prod_{p \le m} (1-p^{-s})^{-1}$).
I think the author is using the definition: $\prod_n a_n$ is "absolutely convergent" if $\sum_n |\log a_n| < \infty$. And he shows that for any $s$ with $Re(s) > 1$, $\sum_p |\log (1-p^{-s})^{-1}| < \infty$ just directly (i.e. no need for uniform convergence). With this definition of absolute convergence, it is true that any absolutely convergent product is nonzero, so that gives all of Theorem 16.2.
However, knowing that $\prod_{p \le m} (1-p^{-s})^{-1}$ converges locally uniformly to $\zeta(s)$ is important for other purposes. For example, local uniform convergence implies that $\zeta$ is holomorphic in $\{Re(s) > 1\}$ (since obviously each $\prod_{p \le m} (1-p^{-s})^{-1}$ is holomorphic).