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I know next to nothing about analytic number theory, or the theory of the Riemann $\zeta$ function in particular, so the following might be too elementary to deserve more than derision; even so it seems it wouldn't hurt to ask where the following question has been considered and what the outcome was.

$\zeta : \mathbb{C} \rightarrow \mathbb{C}$ is a meromorphic function, and therefore, locally has a Laurent series expansion, and in any disc $D(c,r)$ centered at $c$ and of radius $r$ in $\mathbb{C}$ that avoids the poles, it has a Taylor series expansion, which can be truncated at $n$-th order to obtain a polynomial approximation, $p_{\zeta, D, n}$ of degree $n$.

Let's take such a disc in the critical strip. The question is: what are the zeros of $p_{\zeta, D, n}$ and how do they behave as $n \rightarrow \infty$? What happens to the asymptotic behavior as we move the disc around? Implicit in the question, of course, is curiosity about any light zeros of $p_{\zeta, D, n}$ might shed on zeros of $\zeta$.

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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 $\zeta$.

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    $\begingroup$ While not directly related, another example where finite truncations shed no light on zeros of the zeta-function is related to a failed attempt at RH suggested by Turan (1948): the full series $\zeta(s) = \sum_{n \geq 1} 1/n^s$ has no zeros when ${\rm Re}(s) > 1$ and this suggests that the truncations $\zeta_N(s) = \sum_{n=1}^N 1/n^s$ should have "few" zeros there for large $N$. Turan showed that if $\zeta_N(s)$ for large $N$ has no zeros when ${\rm Re}(s) > 1$ then RH is a consequence, but Monach (1980) showed $\zeta_N(s)$ does have a zero with ${\rm Re}(s) > 1$ when $N \geq 31$. $\endgroup$
    – KConrad
    Commented May 5, 2019 at 6:49

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