Extend Euler Product Convergence Over Primes: Basic Qs

[benorin]

I would like to extend the convergence of the Euler product over primes, and I tried to do so in the exact manor it was done for the Dirichlet series, namely, given a completely multiplicative sequence $a( {kj} ) =a(k) \cdot a(j)\text{ and }a(1)=1$, the Dirichlet series $\xi (s) := \sum_{k=1}^{\infty} \tfrac{a(k)}{k^s}$ can be shown (by adding the alternating version of itself to itself and simplifying) to be equal to $\hat \xi (s) := \left( 1-2^{1-s} a(2) \right) ^{-1} \sum_{k=1}^{\infty} (-1)^k \tfrac{a(k)}{k^s}$, and this series is an analytic continuation of the former.
Tried to do this to the Euler product

$$\sum_{k=1}^{\infty} \tfrac{a(k)}{k^s} = \prod_{k=1}^{\infty} \left( 1+a( {p_k} ) p_k^{-s} + a ( {p_k^2} ) p_k^{-2s}+\cdots \right)=\prod_{k=1}^{\infty} \tfrac{1}{1-a( {p_k} ) p_k^{-s}}$$

(where $p_k$ is the $k^{th}$ prime and the later equality holds only for completely multiplicative sequences $a(k)$ taking a cue from the analytic continuation of the Dirichlet series I set $a(k) :=(-1)^{k-1}a^{\prime}(k)$ where $a^{\prime}(k)$ is a completely multiplicative sequence and the product becomes

$$\sum_{k=1}^{\infty} (-1)^{k-1}\tfrac{a^{\prime}(k)}{k^s} =\prod_{k=1}^{\infty} \tfrac{1}{1-(-1)^{p_k -1}a^{\prime}( {p_k} ) p_k^{-s}} = \tfrac{1}{1+a^{\prime}( {2} ) 2^{-s}}\prod_{k=2}^{\infty} \tfrac{1}{1-a^{\prime}( {p_k} ) p_k^{-s}}$$

I was hoping for an analytic continuation of the product over primes but this product differs from the original Euler product by only one term hence converges in the same region. I had hoped to follow this up with the rest of the steps to globally analytically continue the zeta function and wind up with a product over primes converging for all complex $s\neq 1$ but effecting the convergence of the Dirichlet series the Euler product is equal to didn't effect the convergence of the product itself. I think my problem may be that I need to be working with absolute convergence? Do you understand what I'm trying to do? How can I do that?

Edit: sorry if this is an easy one but I just started on these and I have no text on it.

 

[jvicens]

Hello,

Thank you for sharing your ideas on extending the convergence of the Euler product over primes. It seems like you have a good understanding of the topic and have made some interesting observations.

To answer your question, I think you are correct in thinking that absolute convergence may be necessary for your method to work. In order to globally analytically continue the zeta function, it is important to have absolute convergence for all values of s. This means that the series must converge for all complex numbers s, not just for certain values.

One way to ensure absolute convergence is to use the absolute values of the terms in the product, rather than alternating them as you have done. This would give you a product of the form

$$\prod_{k=1}^{\infty} \left( \frac{1}{1+a( {p_k} ) |p_k|^{-s}} \right)$$

where |p_k| denotes the absolute value of the kth prime. This product should converge absolutely for all complex s, and you can then continue with the rest of your steps to globally analytically continue the zeta function.

I hope this helps and good luck with your research!