# Dirichlet series and Dirichlet $L$ function

For $a, b$ relatively prime, I would like to express $$\sum_{n \equiv a\mod b} \frac{\mu(n)}{n^s}$$ in term of Dirichlet L function.

Not sure how to handle this kind of question. Do not find any Theorems in my books state something useful about it either.

$\sum_{\substack{n = 1 \\ n \equiv a \pmod{q}}}^{\infty} \frac{\mu(n)}{n^s} = \frac{1}{\varphi(q)} \sum_{\chi \pmod{q}} \overline{\chi}(a) \sum_{n = 1}^{\infty} \frac{\mu(n) \chi(n)}{n^s}.$ Then via multiplicativity, we find that $\sum_{n = 1}^{\infty} \frac{\mu(n) \chi(n)}{n^s} = \prod_p \left(1 - \frac{\chi(p)}{p^s}\right) = \frac{1}{L(s,\chi)}.$ So $\sum_{\substack{n = 1 \\ n \equiv a \pmod{q}}}^{\infty} \frac{\mu(n)}{n^s} = \frac{1}{\varphi(q)} \sum_{\chi \pmod{q}} \frac{\overline{\chi}(a)}{L(s,\chi)}.$ This is initially valid for $\Re(s) > 1$ (because then both sides converge absolutely), but this then gives a meromorphic extension to all of $\mathbb{C}$.
• Thank you !!! And is that the final form ? I mean $$\sum_{n \equiv a \mod b} \mu(n)n^{-s} = \frac{1}{\phi(k)} \sum_\chi \frac{\overline \chi (k) }{L(s, \chi)}.$$ For $\Re(s) > 0$, $$\sum_{n \equiv a \mod b} \mu(n)n^{-s} = \frac{1}{\phi(k)L(s, \chi_0)} + \frac{1}{\phi(k)} \sum_{\chi, \chi \neq \chi_0} \frac{\overline \chi (k) }{L(s, \chi)} = \frac{1}{\phi(k)L(s,\chi_0)} + O(1).$$ – user117375 Dec 5 '16 at 19:34
• See my edit. I don't know what you mean by your last line, with regards to $O(1)$: what is tending to a limit, and why do the Dirichlet $L$-functions associated to nonprincipal characters contribute less than $L(s,\chi_0)$? – Peter Humphries Dec 5 '16 at 19:39