Implication of $\zeta(s)\ne 0$

Question

I could not understand why $\lim_{m\to\infty}\zeta(s)-\prod_{p\le m} \frac{1}{1-p^{-s}}=0$ does not imply $\zeta(s)\ne 0$, can anyone explain elaborately please? The source of the question is a comment of forum member reuns from this post, click here, I didn't get a clear answer of the post either.

Answer 1

Referring to that post,

It's been proved that $\lim_{m\to\infty} ($$\zeta(s )-\prod_{p\leq m}\frac{1}{1-p^{-s}})=0 $. This means $\lim_{m\to\infty} ($$\zeta(s)$- $\zeta_m(s)) =0$. So, $\lim_{m\to\infty} $$\zeta_m(s)$ = $\zeta(s)$.

Now, we need to prove that $\lim_{m\to\infty} $$\zeta_m(s)$ exists. Hence, considering $\prod_{p}\frac{1}{1-p^{-s}}$ and it has been proved that the product converges (to some positive value) for $R(s)>1$ (considering the logarithm of the product and turn it into a summation). So, $\lim_{m\to\infty} $$\zeta_m(s)$ exists and $\lim_{m\to\infty} $$\zeta_m(s)$=$\prod_{p}\frac{1}{1-p^{-s}}$.

So, $\zeta(s)\neq0$ since $\prod_{p}\frac{1}{1-p^{-s}}\neq0$ for $R(s)>1$.

Answer 2

I gave all the clues in the comment. There are two steps :

For $\Re(s) > 1$, with $Lpf(n)$ the largest prime factor of n.

• $$\zeta(s)-\prod_{p\le m} \frac{1}{1-p^{-s}}= \zeta(s)-\sum_{n\ge1,Lpf(n)\le m}n^{-s}=\sum_{n\ge 1, Lpf(n) > m} n^{-s}$$ thus $$\lim_{m\to\infty}\zeta(s)-\prod_{p\le m} \frac{1}{1-p^{-s}}=0$$

• $$\prod_p \frac{1}{1-p^{-s}}=\lim_{m \to \infty}\prod_{p\le m} \frac{1}{1-p^{-s}}\ne 0$$ because its logarithm is the series $-\sum_p \log(1-p^{-s})$ which converges absolutely (why ?), to a finite value $F(s)$.

• Thus $\zeta(s) = \exp(F(s))\ne 0$.

We say an infinite product $\lim_{m \to \infty} \prod_{k \le m} a_k$ converges if $A_m=\prod_{k \le m} a_k$ is a multiplicative Cauchy sequence in $\Bbb{C}^*$, that's why most people say that the Euler product converges implies $\zeta(s)\ne 0$

Of course we do also prove the analyticity of those things.

And you are supposed to get the relevant textbooks, you can't ask every single step about the theory of $\zeta(s)$ on a math forum.