The exponent of convergence of the zeros of an entire function does not exceed its order

Let $m$ be the order of the zero of $f$ at $0$, with $m=0$ if $f(0)\mathrm{\ne }0$. By [F4], there is an entire function $g$ with $g(0)\mathrm{\ne }0$ and $f(z)=z^{m}g(z),$ so $g$ has exactly the same nonzero zeros as $f$, with the same multiplicities.

If that nonzero zero multiset is finite, every reciprocal power sum over it is finite and its exponent is $0\le \rho$, so the conclusion holds. Hence assume from now on that it is infinite and enumerate it as $(a_n{)}_{n\ge 1}$.

For $r\ge 1$ and $\mathrm{\mid }z\mathrm{\mid }=r$, step 1.1 gives $\mathrm{\mid }g(z)\mathrm{\mid }=\mathrm{\mid }f(z)\mathrm{\mid }\mathrm{/}{r}^m\le \mathrm{\mid }f(z)\mathrm{\mid }$, hence $M_g(r)\le M_f(r)$. Therefore $g$ has order at most $\rho$ by [F1].

Fix real numbers $\sigma ,s$ with $\rho <\sigma <s$. By [F1] and step 2.1, for all sufficiently large $r$ one has $\log M_g(2r)\le (2r{)}^{\sigma }$. Applying [F2] to $g$, whose value at $0$ is nonzero by step 1.1, yields $n(r)\log 2\le \frac{1}{2\pi }{\int }_0^{2\pi }\log \mathrm{\mid }g(2r{e}^{it})\mathrm{\mid }dt-\log \mathrm{\mid }g(0)\mathrm{\mid }\le (2r{)}^{\sigma }-\log \mathrm{\mid }g(0)\mathrm{\mid },$ where $n(r)$ counts the nonzero zeros of $f$ in $\mathrm{\mid }z\mathrm{\mid }\le r$. Thus $n(r)\le C{r}^{\sigma }$ for all large $r$.

Split the nonzero zeros into dyadic shells $2^j\le \mathrm{\mid }{a}_n\mathrm{\mid }<2^{j+1}$. The number of zeros in the $j$th shell is at most $n(2^{j+1})$, so for large $j$ one has ${\sum }_{2^j\le \mathrm{\mid }{a}_n\mathrm{\mid }<2^{j+1}}\mathrm{\mid }{a}_n{\mathrm{\mid }}^{-s}\le n(2^{j+1})2^{-js}\le C{2}^{(j+1)\sigma }{2}^{-js}=C{2}^{\sigma }{2}^{-j(s-\sigma )}\mathrm{.}$ Since $s-\sigma >0$, the dyadic majorant is summable.

Therefore ${\sum }_{n\ge 1}\mathrm{\mid }{a}_n{\mathrm{\mid }}^{-s}<\mathrm{\infty }$ for every $s>\rho$. By [F3], this means the exponent of convergence $\lambda$ of the nonzero zero sequence satisfies $\lambda \le \rho$.