I want to know about when the infinite product form of the Riemann Zeta Function converges.

It is easy to show that $$\sum_{n=1}^\infty \frac 1{n^x}$$ converges for $\Bbb R(x)>1$.

When does $$\prod_{p=primes} \frac {p^x}{p^x-1}$$ converge? And how do you show that?


Let $$A_k = \{ n \in \mathbb{N}, \text{ largest prime factor of } n \le k\}$$ Using the fundamental theorem of arithmetic, or by induction $$\prod_{p \in \mathcal{P},p \le k} (1+p^{-s}+(p^2)^{-s}+(p^3)^{-s}+\ldots) = \sum_{n \in A_k} n^{-s}$$

For $s > 1$ the sequence $\{ n^{-s}\}_{n\in \mathbb{N}}$ is non-negative and summable, which means the order of summation doesn't matter, thus $$\zeta(s) = \lim_{N \to \infty} \sum_{n \in [1,N]} n^{-s} = \lim_{k \to \infty} \sum_{n \in A_k} n^{-s}$$

And for $s \le 1$, since each term is positive, if $\lim_{k \to \infty} \sum_{n \in A_k} n^{-s}$ converges then $\lim_{N \to \infty} \sum_{n \in [1,N]} n^{-s}$ converges, a contradiction since we know it diverges by comparison with $\sum_{n=1}^N n^{-1} > \int_1^N x^{-1}dx = \log N$

  • $\begingroup$ Everything works the same way for $s \in \mathbb{C}$, except when $s = 1+it$ which needs a special treatment $\endgroup$ – reuns Jun 23 '17 at 13:15

The product converges for $\mathbb{R}(s)>1$ (from wikipedia page). This product approaches $1$ from the right hand side, which follows from the Dirichlet series for $\zeta(s)$, which converges for $\mathbb{R}(s)>1$.


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