Let $\chi$ be a dirichlet character mod $q$ . Does the dirichlet series for $$\frac{1}{L(s,\chi)}=\sum_{n=1}^{\infty}\frac{\mu(n)\chi(n)}{n^s}$$ converge for any $s$ on the line $\Re(s)=1$ and is this value equal to $\frac{1}{L(s,\chi)}$ ?
I can see that it does converge for the principle character using an estimate for $M(x)$ but still dont know whether this value is $\frac{1}{L(s,\chi_0)}$
(Sketch) Using partial summation and the fact that $$M(\chi,s)=\sum_{n\leq x}\mu(n)\chi(n)=\mathcal{O}\left(xe^{-c\sqrt{\log(x)}}\right)$$ for some positive constant $c$, one can compare it with the convergent sum $$\sum_{n=2}^{\infty}\frac{1}{n{(\log{n})}^2}$$ from which the convergence follows. To show that the limit is indeed $\frac{1}{L(s,\chi)}$, show that $$\sum_{n=1}^{\infty}\frac{\mu(n)\chi(n)}{n^{s_n}}\to \sum_{n=1}^{\infty}\frac{\mu(n)\chi(n)}{n^{s}}$$ for any sequence $s_n\to s$ with $\Re(s_n)>1$ and then use the fact that $\frac{1}{L(s,\chi)}$ is continuous in the half plane $\Re s\geq 1$