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The context is the proof of Hadamard's theorem (Chapter 5, Theorem 8).

The setup is the following: $f(z)$ is entire, $f(0) \neq 0$, $f(z)$ is of finite order $\lambda$.

The paragraph that is confusing me:

Let us denote by $\nu(\rho)$ the number of zeros $a_n$ with $|a_n| \leq \rho$. In order to find an upper bound for $\nu(\rho)$ we apply (44) with $2\rho$ in the place of $\rho$ and omit the terms $\log(2\rho/|a_n|)$ with $|a_n| \geq \rho$. We find $$ \nu(\rho) \log2 \leq \frac{1}{2\pi}\int_0^{2\pi} \log|f(2\rho e^{i\theta})| d\theta - \log|f(0)|. $$ In view of (47) it follows that $\lim_{\rho \to \infty} \nu(\rho)\rho^{-\lambda - \epsilon} = 0$ for every $\epsilon > 0$.

Here, (44) is Jensen's formula, $$ \log|f(0)| = - \sum_{i = 1}^n \log\left(\frac{\rho}{|a_i|}\right) + \frac{1}{2\pi} \int_0^{2\pi} \log|f(\rho e^{i \theta})|d\theta$$ and (47) is the definition of the order $\lambda$, $$M(r) \leq e^{r^{\lambda + \epsilon}}$$ for any given $\epsilon > 0$ as soon as $r$ is sufficiently large, where $M(r) = \sup_{|z| = r} |f(z)|$.

I don't understand why the limit in the last sentence is $0$. Substituting (47), I'm getting $$\nu(\rho) \log 2 \leq (2\rho)^{\lambda + \epsilon} - \log|f(0)|, $$ and I don't see how this implies $\nu(\rho) \rho^{-\lambda - \epsilon} \to 0$ as $\rho \to \infty$.

It might just be late, but I'm not seeing it... Thanks for your help.

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Multiply both sides of your last inequality by $\rho^{-\lambda - 2 \epsilon}$, you get that $\nu(\rho) \rho^{-\lambda - 2 \epsilon} \to 0$ as $\rho \to \infty$. And as this is true for all $\epsilon >0$ (in particular for $\epsilon/2$) you're done.

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