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Feb 14, 2017 at 0:51 comment added GH from MO @Joël: If you want, remove your last two comments, and I will remove mine as well.
Feb 13, 2017 at 18:01 comment added GH from MO @Joël: I don't see any contradiction. David Speyer shows that the Euler product for $\zeta(s)$ is divergent when $0<\Re(s)<1$. On the other hand, for a nontrivial Dirichlet character $\chi$, the Euler product of $L(s,\chi)$ converges in the half-plane $\Re(s)>1/2$ if and ony if the RH holds for $L(s,\chi)$.
Jun 18, 2015 at 8:57 vote accept Agno
Jun 18, 2015 at 7:43 comment added GH from MO @joro: Two related posts are mathoverflow.net/questions/63714/… and mathoverflow.net/questions/181241/…
Jun 18, 2015 at 7:41 comment added GH from MO @joro: Roughly speaking, the reason is the oscillation of $\chi$ (when $\chi$ is nontrivial). More precisely, while $\sum p^{-s}$ over the primes clearly diverges unless $\Re s>1$, the oscillating sum $\sum\chi(p)p^{-s}$ has a chance to converge for smaller values of $\Re s$ assuming the primes are sufficiently well-distributed mod $n$ (where $n$ is the conductor of $\chi$). And, indeed, GRH implies that the latter sum converges for $\Re s>1/2$.
Jun 18, 2015 at 7:03 comment added joro @GHfromMO Thank you. My mistake. So why the answer doesn't apply to the trivial character which is zeta AFAICT?
Jun 18, 2015 at 6:37 comment added GH from MO @joro: The only character modulo $n$ that is nonnegative is the trivial character modulo $n$.
Jun 18, 2015 at 6:36 comment added joro If $\chi$ is kronecker character it is nonnegative. Let $t=0$. Every term is real greater than one. I don't see how this will converge to the $L$ function (computing a single value above the correct is counterexample).
Jun 17, 2015 at 21:40 review First posts
Jun 17, 2015 at 22:00
Jun 17, 2015 at 21:38 history answered ABCDveve CC BY-SA 3.0