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In Stein and Shakarchi's book on Complex analysis, there's a theorem (p.138) that states:

If $f$ is an entire function that has an order of growth $\leq \rho$, and if $z_1, z_2, ...$ denote the zeroes of $f$, with $z_k \neq 0$, then for all $s > \rho$, we have: $$ \sum_{k=1}^{\infty} \frac{1}{|z_k|^s} < \infty $$

We also say that if $A = \{\rho : f \text{ has order of growth} \leq \rho \}$, then the order of growth $\rho_f$ is $\inf A$.

I'm wondering if a converse exists for this theorem, namely:

If $f$ is an entire function that has order of growth $\rho_f$, and if $z_1, z_2, ...$ denote the zeroes of $f$, with $z_k \neq 0$, then: $$ \sum_{k=1}^{\infty} \frac{1}{|z_k|^{\rho_f}} = \infty $$

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  • $\begingroup$ I just realized there's a trivial counter-example where $f$ has finitely many zeroes. What about if this is not the case? $\endgroup$ – Sambo Dec 21 '17 at 22:09
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The converse does not hold, we can have

$$\sum_{k = 1}^{\infty} \frac{1}{\lvert z_k\rvert^{\alpha}} < +\infty\tag{$\ast$}$$

even for $\alpha < \rho_f$. If $(z_k)$ is any sequence of nonzero complex numbers such that $(\ast)$ holds for some $\alpha \in (0,+\infty)$, then there is always an entire function of finite order whose order is as large as we please with those zeros. For, if the canonical product $P$ with these zeros has order $\rho_P$, and $m > \rho_P$, then $f(z) = e^{z^m}\cdot P(z)$ has order $m$.

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  • $\begingroup$ Informally: the order gives an upper bound on the clustering of zeros, but not a lower bound. $\endgroup$ – Daniel Fischer Dec 22 '17 at 13:10

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