Consider a function in $\mathbb{C}$, $f(z)$, holomorphic in the half plane $\Re(z)>2$. It is also known
$$\lim_{\Re(z) \to \infty}f(z) =1$$
Prove or disprove:
$$\lim_{k \to \infty}{ \bigg|\frac{1}{k+1}\frac{f^{k+1)}(z_1)}{f^{k)}(z_1) } \Bigg| \le 1 }, \forall z_1 \mid \Re(z_1)>2$$
My approach is considering $f$ is holomorphic then it is infinitely differentiable for $\Re(z_1) > 2$ and it is equal to its own Taylor series. Then the required statement to prove is not more than the quotient of the kth and kth+1 coefficients of the Taylor series of $f$ centered on $z=z1$.
Since $\Re(z_1) > 2$ then we know the Taylor expansion of $f$ centered on $z_1$ is at least valid for $z_a= z_1 - \epsilon$. This means that the remainder $R_n(z_a)$tends to $0$ as $n$ grows.
Then, since $R_n$ tends to $0$ (It is enough to apply the limit definition, to $R_n$:
$$ \lim_{n\to\infty} \frac{|R_{n+1}(z_a)|}{|R_n(z_a)|} \le 1 $$
But also:
$$ \frac{R_{n+1}(z_a)}{R_n(z_a)} = \frac{ \frac{1}{(k+1)!}f^{k+1)}(z_1) (-\epsilon))^{k+1} + R_{n+2}}{ \frac{1}{k!}f^{k)}(z_1) (-\epsilon)^k + \frac{1}{(k+1)!}f^{k+1)}(z_1) (-\epsilon))^{k+1} + R_{n+2} } \approx \frac{ \frac{1}{(k+1)!}f^{k+1)}(z_1) (-\epsilon)}{ \frac{1}{k!}f^{k)}(z_1) } $$.
Mixing both expressions:
$$\lim_{k \to \infty}{ \bigg|\frac{1}{k+1}\frac{f^{k+1)}(z_1)}{f^{k)}(z_1) } \Bigg|} \le \frac{1}{\epsilon}$$
So this clearly proves if $\Re(z_1) \ge 3$ as we can take $\epsilon = 1$. But I wonder if there is a way to prove it in case $2<\Re(z_1)<3$.