The Riemann zeta function is can be expressed as an infinite series, as well as an infinite Euler product over primes $p$.
$$ \zeta(s) = \sum_n 1/n^s = \prod_p(1-1/p^s)^{-1} $$
Here $s=\sigma+it$ and the function is defined for $\sigma>1$.
A natural question to ask is whether there are any zeros in the region $\sigma>1$.
Question: Is the following outline proof sufficient and correct?
Step 1
We note that none of the factors $(1-1/p^s)^{-1}$ is ever zero, because $p^s = e^{s\ln(p)} \neq 0$ for $\sigma>1$.
Step 2
The previous step is insufficient. We also need to demonstrate that the infinite product doesn't converge to zero.
To do this, we make use of the following convergence criteria:
- if $\sum|a_n|$ converges, then $\prod(1+a_n)$ converges to a finite non-zero value.
Step 3
In this case $\sum |a_n| = \sum 1/p^s$, which we know converges for $\sigma > 1$.
Therefore the Euler Product converges to a non-zero finite value.
Your step 2 is overlooking some subtlety (for instance, what if $a_{17} = 1$ but the series still decays fast enough to converge?). To get around subtleties, proceed as follows.