Validity of proof $\zeta(s) \neq 0$ for $\sigma\gt1$, where $\sigma$ is the real part of $s$

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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.

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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.

  1. If $\prod |1-a_n|$ converges, then so does $\prod (1-a_n).$
  2. Hence, instead of looking at a product of possibly complex numbers as in the factorization of $\zeta(s),$ look at their magnitudes, each of which we know is nonzero.
  3. Prove that if $b_n$ are strictly positive reals, then $\prod b_n$ converges to a non-zero real if and only if $\sum \log b_n$ converges, which is fairly immediate by taking logarithms of partial products/exponentials of partial sums (to do this directly with the complex numbers requires more effort since the logarithm requires a branch cut to be defined, so we just take absolute values to reduce to positive reals and avoid that).
  4. Now conclude as in your step 2 above. (It's helpful to use the limit comparison test to change the log series into a series not involving logs, getting closer to your original formulation of step 2.)