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Let $\chi$ be a Dirichlet character mod q and let $$L(s,\chi)=\sum_{n\leq x} \frac{\chi(n)}{n^s}.$$ What is the value of $\displaystyle\lim_{s \rightarrow 1} \frac{L(s,\chi)}{\zeta(s)}$ for principal character $\chi_0$ and non-principal characters $\chi$?

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Hint: For $\chi \neq \chi_0$ $L(s,\chi)$ is analytic for $\text{Re}(s)>0$ (by partial summation) so that in particular $L(1,\chi)$ is finite. $\zeta(s)$ has a pole with residue $1$ at $s=1$ which can be shown by looking at $\zeta(s)-\frac{1}{s-1}$ which will be analytic for $\text{Re}(s)>1$. For $\chi_0$, $L(s,\chi_0)$ is almost the same as $\zeta(s)$. In particular by using Euler products we can write $$L(s,\chi_0)=\zeta(s)\prod_{p|q} \left(1-p^{-s}\right).$$

From this we can conclude that if $\chi $ is non principal, the limit is $0$, and if it is principal then the limit is $\frac{\phi (q)}{q}$.

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So if $\chi \neq \chi_0$, then that limit goes to $0$ and if $\chi=\chi_0$, then that limit goes to $\frac{\varphi(q)}{q}$? – Rob Dec 11 '11 at 19:47
@EricNaslund, I think Rob is right with the $\phi(q)/q$ ... after all your formula shows that $L(s,\chi_0)/\zeta(s)$ is the entire function $\prod_{p\mid q} (1-p^{-s})$, so we can just plug in $s=1$. Maybe you're thinking about logarithmic derivatives? – Greg Martin Dec 11 '11 at 21:58
@GregMartin: Ahh yes! Thanks for the correction. I was definitely mixing this up with the logarithmic derivative. – Eric Naslund Dec 12 '11 at 7:17

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