# Partial Euler product

The Riemann Zeta function defined as $$\zeta(s) = \sum_{n=1}^{\infty} n^{-s}$$ For $$\Re(s)>1$$ is convergent and admits the Euler product representation $$\zeta(s) = \prod_p (1-p^{-s})^{-1}$$ For partial Euler product $$\prod_{p we obviously will have $$\zeta_x(s) = \prod_{pwhere $$\zeta_x(s)$$ is a $$\zeta(s)$$ with "thrown out" summands with $$n$$ having in fuctorisation $$p\geq x$$ My question: How can I write with correct math notation this function as a Dirichlet series, something like: $$\zeta_x(s) = \sum_{n=1}^{\infty} \frac {a(n)}{n^{s}}$$ where $$a(n)=1, n= .... ?$$ $$a(n)=0, n= .... ?$$

I would say that $$a(n) = 1$$ if $$n$$ is $$x$$-smooth. See https://en.wikipedia.org/wiki/Smooth_number
• So $\zeta_x (s)= \sum_{n: x-smooth} {n^{-s}}$ ? – Aleksey Druggist Mar 16 at 11:15
• @AlekseyDruggist This is indeed the way we prove $\sum_{n=1}^\infty n^{-s} = \prod_p \frac{1}{1-p^{-s}}$ for $\Re(s) > 1$ – reuns Mar 16 at 12:58