If $f(z)$ is analytic for $|z|<1$ and $f(z)≤\frac{1}{1－|z|}$,find the best estimate of $|f^{(n)}(0)|$ that Cauchy inequality will yield

Question

There have a solution that Fix $n≥0$ and let $R＝\frac{n}{n＋1}$ so that $f(z)$ is analytic on$|z|＝R$. Then by Cauchy inequality $$|f^{(n)}(0)|≤\frac{n!}{R^n} \max_{z＝R}|f(z)|＝\frac{n!}{R^n(1－R)}＝\frac{n!(n＋1)^{n＋1}}{n^n}$$

The answer to this question is$(n＋\frac{1}{2})e$ But I don't know how to continue to solve this problem

Answer 1

I believe there may be an error in the provided solution. Consider the function

$$f(z) = \frac{1}{1-z}$$

It does satisfy the given conditions and its derivatives at $z=0$ are not bounded by $(n+\frac{1}{2})e$.

On the contrary, if we examine

$$(n+1)!\left(\frac{n+1}{n}\right)^n = (n+1)!\left(1+\frac{1}{n}\right)^n \leq (n+1)!e$$

This is due to the fact that $\left(1+\frac{1}{n}\right)^n$ approaches $e$ from below. This observation might be the source of the appearance of $e$ in the solution.