Definition. A entire function $f$ has order of grow $\leq \rho$ if $\exists A,B>0$ such than $$ |f(z)|\leq Ae^{B|z|^\rho} \quad \forall z \in \mathbb{C} $$ And has order $\rho_0$ if $\rho_0 =\inf \rho$ such than $\rho$ verifies last inequality.
My question is: if $f$ has order of grow $\rho_0$ and $\rho<\rho_0$ then $\forall A,B>0$ exists a sequence $z_n$ such than $f$ doesn't verify the inequality or exists a sequence $z_n$ such than $\forall A,B>0$ $f$ doesn't verify the inequality?
For each $A,B>0$, there is such a sequence. Indeed, if the order of growth of $f$ is $\rho_0$ and $\rho<\rho_0$, then the order of growth of $f$ is not smaller than or equal to $\rho$. Therefore, for all $A,B>0$, there is some $z\in\mathbb C$ such that$$\bigl|f(z)\bigr|>Ae^{B|z|^\rho}.$$So, take $z_n=z$ for each $n\in\mathbb N$ and the sequence $(z_n)_{n\in\mathbb N}$ will be such that$$(\forall n\in\mathbb{N}):\bigl|f(z_n)\bigr|>Ae^{B|z_n|^\rho}.$$