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 $$a(n)=1, n= .... ?$$ $$a(n)=0, n= .... ?$$


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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.