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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<x}(1-p^{-s})^{-1}$ we obviously will have $$\zeta_x(s) = \prod_{p<x} (1-p^{-s})^{-1}$$where $\zeta_x(s)$ 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 $$a(n)=1, n= .... ?$$ $$a(n)=0, n= .... ?$$

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I would say that $a(n) = 1$ if $n$ is $x$-smooth. See https://en.wikipedia.org/wiki/Smooth_number

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  • $\begingroup$ So $ \zeta_x (s)= \sum_{n: x-smooth} {n^{-s}}$ ? $\endgroup$ – Aleksey Druggist Mar 16 '19 at 11:15
  • $\begingroup$ Yes that's right $\endgroup$ – Esteban Crespi Mar 16 '19 at 11:57
  • $\begingroup$ @AlekseyDruggist This is indeed the way we prove $\sum_{n=1}^\infty n^{-s} = \prod_p \frac{1}{1-p^{-s}}$ for $\Re(s) > 1$ $\endgroup$ – reuns Mar 16 '19 at 12:58

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