Exponent of convergence $\limsup_{n\to \infty}\frac{\log n}{\log |a_n|}$ Let $(a_n)_{n\ge1}$ be the sequence of zeros on an entire function $f$.
We define the convergence exponent of $(a_n)_{n\ge1}$ as
$$\lambda=\inf\left\{\mu>0\ :\ \sum_{n=1}^{\infty}\frac{1}{|a_n|^{\mu}}<\infty\right\}$$
How to prove that the exponent of convergence  is given by $$\lambda=\limsup_{n\to \infty}\frac{\log n}{\log |a_n|}$$
Any help would be appreciated.
 A: We must require that the zeros are listed in order of increasing modulus, or something close to that. Without such an assumption, the equality need not hold. The definition of $\lambda$ as
$$\lambda := \inf \left\{\mu > 0 : \sum_{n=1}^\infty \lvert a_n\rvert^{-\mu} < +\infty \right\}$$
is invariant under arbitrary rearrangements since the summands are all non-negative, but
$$\limsup_{n\to\infty} \frac{\log n}{\log \lvert a_n\rvert}$$
is not. For an almost explicit example, consider $b_n = n$, where we know that $\lambda = 1$. If we rearrange the sequence in such a way that $a_{k^2} = b_{2k-1} = 2k-1$, and the even numbers are listed in increasing order at non-square indices, so $a_{k^2+r} = b_{2k(k-1)+2r}= 2k(k-1)+ 2r$ for $1\leqslant r \leqslant 2k$, then we have
$$\limsup_{n\to\infty} \frac{\log n}{\log \lvert a_n\rvert} = \lim_{k\to\infty} \frac{\log (k^2)}{\log (2k-1)} = 2 > \lambda.$$
Assuming that we have $\lvert a_n\rvert \leqslant \lvert a_{n+1}\rvert$ for all $n$, the equality holds, however.
Let us define
$$\kappa := \limsup_{n\to\infty} \frac{\log n}{\log \lvert a_n\rvert}.$$
The inequality $\lambda \leqslant \kappa$ holds regardless of the ordering of the $a_n$. Let $\mu > \kappa$. Then there is an $\varepsilon > 0$ such that $\mu > \kappa(1+\varepsilon)$. By the definition of $\limsup$, there is an $n_0$ such that
$$\frac{\log n}{\log \lvert a_n\rvert} < \frac{\mu}{1+\varepsilon}$$
and $\lvert a_n\rvert > 1$ for all $n \geqslant n_0$. But then we have
$$\frac{1}{\lvert a_n\rvert^\mu} < \frac{1}{n^{1+\varepsilon}}$$
for $n \geqslant n_0$, and $\sum \frac{1}{n^{1+\varepsilon}} < +\infty$ for all $\varepsilon > 0$, hence $\sum \lvert a_n\rvert^{-\mu} < +\infty$ for all $\mu > \kappa$, so $\lambda \leqslant \kappa$.
For the inequality $\kappa\leqslant \lambda$ we assume the $\lvert a_n\rvert$ monotonically increasing (non-strictly). Then we show that for all $\mu < \kappa$ we have $\sum \lvert a_n\rvert^{-\mu} = +\infty$: Fix $\mu < \kappa$. By definition of $\limsup$, there are arbitrarily large $n$ such that
$$\frac{\log n}{\log \lvert a_n\rvert} > \mu.$$
Choose an infinite sequence $(n_k)_{k\in\mathbb{N}}$ such that $n_{k+1} \geqslant 2n_k$ and
$$\frac{\log n_k}{\log \lvert a_{n_k}\rvert} > \mu.$$
Then we have
$$\frac{1}{\lvert a_{n_k}\rvert^\mu} > \frac{1}{n_k},$$
and by the monotonicity assumption
$$\sum_{n = n_{k-1}+1}^{n_k} \frac{1}{\lvert a_n\rvert^\mu} \geqslant (n_k - n_{k-1})\frac{1}{\lvert a_{n_k}\rvert^\mu} > \frac{n_k - n_{k-1}}{n_k} \geqslant \frac{1}{2},$$
which shows $\sum \lvert a_n\rvert^{-\mu} = +\infty$ for $\mu < \kappa$, whence $\lambda \geqslant \kappa$.
