In Stein and Shakarchi's book on Complex analysis, there's a theorem (p.138) that states:
Theorem: Let $f$ be an entire function that has an order of growth $\leq \rho$. If $z_1,z_2,\dots$ denotes the zeroes of $f$, with $z_k \neq 0$, then for all $s \geq \rho$, we have
$$\sum_{k=1}^{\infty}\frac{1}{|z_k|^s} < \infty.$$
Now i'm currently reading the proof of the statement
The entire function $(s-1)^m F(s)$ is of order $1$. (Here $F(s)$ is in the Selberg Class
Proof: First recall that, for each $j$, $F(s)$ has equally spaced trivial zeroes at $s=-\frac{\mu_j + l}{\lambda_j}$ where $l \in \mathbb{Z}_{\geq 0}$, which are nothing but poles of Gamma. Now, if the order of $(s-1)^m F(s)$ is strictly less than $1,$ then by the Theorem above we would have
$$\sum_{\substack{s\neq 0\\ F(s)=0}} \frac{1}{|s|^{1-\epsilon}}<\infty.$$ Which is a contradiction. (See lemma 4.2 in Article).
Now I'm not sure if my understanding of the contradiction there is correct.
First by the Theorem for all $\delta > \rho$ we have the inequality $$\sum_{\substack{s\neq 0 \\ F(s)=0}} \frac{1}{|s|^{\delta}} \leq\sum_{\substack{s\neq 0 \\ F(s)=0}} \frac{1}{|s|} \leq \sum_{\substack{s\neq 0\\ F(s)=0}} \frac{1}{|s|^{1-\epsilon}}<\infty .$$ Now note that the sum on the left is divergent. Therefore the sum on the right cannot converge, because it is bigger than or equal to that sum over nonzero nontrivial zeros. But what we deduced from the assumption that the order $<1$, the sum converges which is a contradiction?
But a friend of mine told me that The inequality is only true for $|s| > 1$. This series is convergent if $\epsilon$ is chosen such that $1 - \epsilon > \rho$. With all of these considerations, there appears to be no contradiction. Now I'm quite confused here since I don't get his argument. Would someone care to enlighten me regarding this comment. Thank you in advance!
Any help would so much appreciated!