Prove that there exists a constant $C$ such that $[z^n]\exp(z/(1-z)) = O(\exp(C\sqrt{n})) $

Question

Prove that there exists a constant $C$ such that:

$$[z^n]\exp(z/(1-z)) = O(\exp(C\sqrt{n})).$$

The bound of $z$ is $\vert z \vert<\frac14$

Answer 1

Clearly, $f(z)=\exp\left(\frac{z}{1-z}\right)$ is analytic in the open unit disk $D(0,1)$. Thus, $$ \forall\,z\in D(0,1),\qquad f(z)=\sum_{n=0}^\infty a_nz^n $$ and we are seeking a bound on $a_n$.

Consider $r$ a positive real from $(0,1)$ then, by Cauchy's formula applied to the circle $C^+(0,r)$, we have $$ a_n=\frac{1}{2\pi r^n}\int_0^{2\pi}f(re^{i\theta})e^{-in\theta}d\theta $$ Thus, $$ \vert a_n\vert\leq\frac{1}{ r^n}\sup_{\theta\in\Bbb{R}} \vert f(re^{i\theta})\vert \tag{1} $$ But, $$\eqalign{\vert f(re^{i\theta})\vert&=\exp\left(\hbox{Re}\frac{re^{i\theta}}{1-re^{i\theta}}\right) =\exp\left(\frac{r(\cos\theta-r)}{\vert r-e^{i\theta}\vert^2}\right)\cr &\leq \exp\left(\frac{r(1-r)}{(1-r)^2}\right)=\exp\left(\frac{r}{1-r}\right)}$$ Hence, from $(1)$ we get, for $0<r<1$ $$ \vert a_n\vert\leq\frac{1}{ r^n}\exp\left(\frac{r}{1-r}\right) \tag{2} $$ Now, taking $r=\dfrac{\sqrt{n}}{1+\sqrt{n}}$ so that $\frac{r}{1-r}=\sqrt{n}$, we obtain $$ \vert a_n\vert\leq\left(1+\frac{1}{ \sqrt{n}}\right)^n\exp\left(\sqrt{n}\right) \tag{3} $$ finally, using the well-known inequality $1+t\leq e^t$ with $t=1/\sqrt{n}$ we get $\vert a_n\vert\leq e^{2\sqrt{n}}$, which is the desired conclusion.