This post is about part (b) of Stein Complex Analysis, Chapter 5, Problem 4. Part (a) has been asked and solved here: Problem on entire function of finite order [Stein, Chapter 5, Problem 4] (Conard gave a fantastic solution by not following Stein's instruction, but if you want to know how to do it following Stein's hint, I think the communication in comment would be enough.)
Part (b) is stated as follows:
Let $F(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$ be an entire function of finite order. If $$\limsup_{n\rightarrow\infty}|a_{n}|^{\frac{1}{n}}n^{\frac{1}{\rho}}<\infty,$$ then $|F(z)|\leq A_{\epsilon}e^{a_{\epsilon}|z|^{\rho+\epsilon}}$, for every $\epsilon>0$
I think I have solved this but I cannot get the exact bound. The following is my proof
Let $r>0$. Suppose $\limsup_{n\rightarrow\infty}|a_{n}|^{\frac{1}{n}}n^{\frac{1}{\rho}}=M<\infty$ and let $\epsilon>0$. Then, by definition there exists $N>0$ such that $|a_{n}|^{\frac{1}{n}}n^{\frac{1}{\rho}}<M+\epsilon,$ and thus $|a_{n}|<\frac{(M+\epsilon)^{n}}{n^{\frac{n}{\rho}}}.$
It implies that
$$|F(z)|\leq\sum_{n=0}^{\infty}|a_{n}|\cdot |z|^{n}\leq \sum_{n=0}^{\infty}\dfrac{(M+\epsilon)^{n}\cdot|z|^{n}}{n^{\frac{n}{\rho}}},$$ where the last inequality was obtained since only finitely number of $|a_{n}|^{\frac{1}{n}}n^{\frac{1}{\rho}}$ are greater than $M+\epsilon$, so it does not affect the bound on the whole infinite series.
We denote $C_{\epsilon}:=M+\epsilon$. It is well known that $n^{n}\geq n!$, and thus $$|F(z)|\leq \sum_{n=0}^{\infty}\dfrac{(M+\epsilon)^{n}\cdot |z|^{n}}{n^{\frac{n}{\rho}}}\leq \sum_{n=0}^{\infty}\dfrac{C_{\epsilon}^{n}\cdot|z|^{n}}{(n!)^{\frac{1}{\rho}}}.$$
Recall an entire function has an order of growth $\leq \rho$ if $|f(z)|\leq Ae^{B|z|^{\rho}}$, and its order $\rho_{0}$ is the $\inf$ over all such $\rho$.
By Stein, Chapter 5, Problem 3, we know that $\sum\frac{z^{n}}{(n!)^{\alpha}}$ is an entire function of order $\frac{1}{\alpha}$, and thus we have $$\sum\frac{z^{n}}{(n!)^{\alpha}}\leq \Bigg|\sum\frac{z^{n}}{(n!)^{\alpha}}\Bigg|\leq Ae^{B|z|^{\rho}}=Ae^{B^{|z|^{\frac{1}{\alpha}}}}.$$
It means that in our case ($\alpha=\frac{1}{\rho}$), $$|F(z)|\leq \sum_{n=0}^{\infty}\dfrac{C_{\epsilon}^{n}\cdot |z|^{n}}{(n!)^{\frac{1}{\rho}}}\leq \Bigg|\sum_{n=0}^{\infty}\dfrac{(C_{\epsilon}\cdot |z|)^{n}}{(n!)^{\frac{1}{\rho}}}\Bigg|\leq Ae^{B|C_{\epsilon}z|^{\rho}}\leq Ae^{B|C_{\epsilon}|^{\rho}\cdot |z|^{\rho+\epsilon}}.$$
So my question is the following:
Why in the statement of problem $A$ also depends on $\epsilon$?
I think Stein made some vague notation here, by $a_{\epsilon}$, did he mean the $a_{n}$ in the power series? If so, how could I include it? If not, then I will just $a_{\epsilon}:=B|C_{\epsilon}|^{\rho}$, is this okay?
Have I done something wrong in my argument so that I cannot let $A$ depend on $\epsilon$?
Thank you for helping me!!!