Since $\zeta$ has a single pole, at $z=1$, the radius of convergence of the Taylor series at $c$ is $r=|c-1|$. Moreover, if $$\zeta(z)=\sum_0^\infty a_n(z-c)^n$$ is the Taylor expansion at $c$, then the limit in Hadamard's formula exists $\lim|a_n|^{1/n}=1/r$ (this is an easy exercise: if a function has a single pole on its circle of convergence and no other singularitiesin a slightly bigger disk then the limit exists). Now a general theorem of Jentzsch implies that the zeros of partial sums are: a) those which tend to the zeros of $\zeta$ in this disc, and b) additional zeros which are uniformly distributed near the circle $|z-c|=r$.

I don't think that this sheds any light on the zeros of $\zeta$.

Since $\zeta$ has a single pole, at $z=1$, the radius of convergence of the Taylor series at $c$ is $r=|c-1|$. Moreover, if $$\zeta(z)=\sum_0^\infty a_n(z-c)^n$$ is the Taylor expansion at $c$, then the limit in Hadamard's formula exists $\lim|a_n|^{1/n}=1/r$ (this is an easy exercise: if a function has a single pole on its circle of convergence and no other singularities in a slightly bigger disk then the limit exists). Now a general theorem of Jentzsch implies that the zeros of partial sums are: a) those which tend to the zeros of $\zeta$ in this disc, and b) additional zeros which are uniformly distributed near the circle $|z-c|=r$.

I don't think that this sheds any light on the zeros of $\zeta$.