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Let $\{z_j\}$ be the sequence of zeros on an entire function $f$. We define the convergence exponent of $\{z_j\}$ as $$b=\inf\left\{\lambda>0\ \text{s.t.}\ \sum_{j=1}^{+\infty}\frac{1}{|z_j|^{\lambda}}<+\infty\right\}$$ Let $n(r)$ be the number of $z_j$'s with $|z_j|\leq r$. Then the following identity holds: $$b=\limsup_{r\rightarrow +\infty}\frac{\log{\ n(r)}}{\log{r}}$$ Do you think i should use Jensen formula to prove this?

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I think this should be possible to show straightforwardly. If $n(r)>r^\lambda$ infinitely often, then ... – Hagen von Eitzen Jan 10 '13 at 20:51
Is this a homework problem? – Greg Martin Jan 10 '13 at 21:34
@haigen von eitzen suppose i know $\lim_{r\rightarrow +\infty}\frac{n(r)}{r^{\lambda}}=0$. What could i deduce from this? – Federica Maggioni Jan 10 '13 at 21:35
@greg martin a part of it – Federica Maggioni Jan 10 '13 at 21:36
up vote 0 down vote accepted

Hint: note that $$ \sum_{j\colon U\le |z_j|<2U} \frac1{|z_j|^\lambda} \le \sum_{j\colon U\le |z_j|<2U} \frac1{U^\lambda} \le \frac{n(2U)}{U^\lambda}. $$ Therefore \begin{align*} \sum_{j=1}^\infty \frac1{|z_j|^\lambda} &\le \sum_{j\colon |z_j|<1} \frac1{|z_j|^\lambda} + \sum_{k=1}^\infty \sum_{j\colon 2^{k-1}\le |z_j|<2^k} \frac1{|z_j|^\lambda} \\ &\le \text{(finite number of terms)} + \sum_{k=1}^\infty \frac{n(2^k)}{(2^{k-1})^\lambda}. \end{align*} In this way, knowledge about the growth of $n(r)$ can be converted into bounds on the sum in question. Lower bounds can be handled similarly.

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thank you very much, but i can't understand the second line in your answer. we have $\sum_{j=1}^{+\infty}\frac{1}{|z_j|^{\lambda}}=$ sum of finite number of terms +$\sum_{U=1}^{+\infty}(\displaystyle\sum_{U\leq|z_j|<2U}\frac{1}{|z_j|^{\lambda}‌​})\leq\sum_{U=1}^{+\infty}\frac{n(2U)}{U^{\lambda}}$ and how can i get your second line? – Federica Maggioni Jan 12 '13 at 13:45
sorry, I had a typo in there - and I added another step while I was at it, hopefully better now – Greg Martin Jan 13 '13 at 7:55

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