A limit involving the exponent of convergence

Question

Let $f$ be an entire non-constant function with at least one zero. If $\{z_{j}\}_{j\in \mathbb{N}}$ are the zeros of $f$, set $$b =\inf\left\{\lambda >0 \ | \sum_{j}\frac{1}{|z_{j}|^{\lambda}}< +\infty\right\}$$

and $n(r) =$ number of zerosof $f$ in $\overline{D(0,r)}$. Then $$b =\limsup_{r \rightarrow + \infty} \frac{\log \ n(r)}{\log \ r}.$$

Any hint ?

Answer 1

Use the idea of the Cauchy condensation test to show that the series converges if

$$\frac{n(2^k)}{2^{\lambda k}}<\frac{C}{2^{k \varepsilon}}$$

and diverges if

$$\frac{n(2^k)}{2^{\lambda k}}> C \,2^{k \varepsilon}$$

---

$$\sum_{k=K_{min}}^\infty \frac{n(2^{k+1})-n(2^k)}{2^{λ(k+1)}}\le \sum_{j}\frac{1}{|z_{j}|^{\lambda}}\le \sum_{k=K_{min}}^\infty \frac{n(2^{k+1})-n(2^k)}{2^{λk}}$$

where $K_{min}$ is such that $n(2^{K_{min}})=0$.