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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.
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.
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