I encountered this problem from "Classical complex analysis" by L.S. Hahn and B. Epstein, p.133.
"Let $f(z)$ be analytic in the unit disc $D$ and suppose that $$\left|f(z)\right| \leq \dfrac{1}{1-|z|}$$ for all $z \in D$. Show that $$\left|f^{(n)}(0)\right| \leq (n+1)!e.$$
The hint is to use the Cauchy integral formula with a path of integration suitably chosen for each $n$."
My attempt is:
By the hypothesis, we have $|f(0)| \leq 1$ . Now let $r \in (0;1)$ and consider the circle $C(0,r)$. By the Cauchy integral formula and taking modulus of both sides, we have $$|f^{(n)}(0)| = \left|\dfrac{n!}{2\pi i} \oint_{C(0,r)} \dfrac{f(z)}{z^{n+1}}\,dz \right| \leq \dfrac{n!}{2\pi } \oint_{C(0,r)} \left|\dfrac{f(z)}{z^{n+1}} \right|\,\left|dz\right|= \dfrac{n!}{(1-r)r^n}.$$ Now we have to choose $r$ (depend on $n$) so that $$\dfrac{1}{(1-r)r^n} \leq (n+1)e$$ and I got stuck at this step. I tried to remove the power $n$ of $r$ in the denumerator by using Bernoulli inequality, $$r^n = (1+(r-1))^n \geq 1+n(r-1)$$ and made the number $e$ appear by using the estimation $$1-r \geq e^{-2r}$$ but still couldn't work things out.
Do you have any idea to find $r$ (depend on $n$)? Any help would be highly appreciated.
Thanks in advance!
Let $f(r) = (1-r)r^n = r^n-r^{n+1}$. Then, $f'(r) = nr^{n-1}-(n+1)r^n = [n-(n+1)r]r^{n-1}$.
Clearly, $f'(r) \ge 0$ for $0 \le r \le \dfrac{n}{n+1}$ and $f'(r) \le 0$ for $\dfrac{n}{n+1} \le r \le 1$.
Hence, $f$ attains its maximum over $r \in [0,1]$ at $r = \dfrac{n}{n+1}$.
Using the well known result that $\left(1-\dfrac{1}{n+1}\right)^n$ monotonically decreases towards $\dfrac{1}{e}$ as $n \to \infty$, we have that $f\left(\dfrac{n}{n+1}\right) = \dfrac{1}{n+1}\left(1-\dfrac{1}{n+1}\right)^n \ge \dfrac{1}{n+1} \cdot \dfrac{1}{e}$.
Therefore, by selecting $r = \dfrac{n}{n+1}$, we have $|f^{(n)}(0)| \le \dfrac{n!}{(1-r)r^n} \le (n+1)!e$, as desired.