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Dec 8, 2019 at 23:09 comment added Greg Martin Sorry, I misread—I was thinking of the general situation of Dirichlet series convergence, but you are indeed postulating an actual pole of $1/\zeta(s)$.
Dec 8, 2019 at 13:38 comment added Wojowu @GregMartin What partial sums do you mean? The point of the argument is to deduce divergence of a sum from existence of a pole. Also I thought it is possible for a Dirichlet series to converge at a point and have a logarithmic singularity on the same abscissa - doesn't $\log\zeta(s)$ have that property?
Dec 8, 2019 at 12:07 comment added Greg Martin I think "pole" is a bit too specific in ths last phrase; the value grows more slowly than it could if the partial sums didn't converge.
Sep 18, 2019 at 17:29 history answered Wojowu CC BY-SA 4.0