As is known, Cauchy's estimate states that for a function $f$ that's analytic in a disk $B_R$ and bounded by the inequality $|f(z)|< M$ for all $z$ in $B_R$, then the n-th derivtive can be estimated as $$|f^{(n)}(0)|\leq \frac{n!M}{R^n}$$ For $z$ on a circle centered at $0$ with radius $r<R$ the estimate can be written as $$|f^{(n)}(z)|<\frac{n!M}{(R-r)^n}$$ All proofs I have seen for this estimate use the representations of the coefficients in the power series of $f$ which are given by contour integrals. However, I was wondering if it's possible to get the same result (at least for the first derivative) by direct calculations using the power series. My approach was the following: Since $f$ is bounded by $M$, I know that for $f(z) = \sum_{k = 0}^{\infty} f_kz^k$ the coefficients can be estimated as $|f_k| < M R^{-k}$ on $B_R$. The derivative is given by $f'(z) = \sum_{k = 0}^{\infty} k f_k z^{k-1}$, so if I take $z$ to be on a smaller circle, I get the estimate $$|f'(z)|\leq\sum_{k = 0}^{\infty}k|f_k||z|^{k-1}<M\sum_{k= 0}^{\infty}k R^{-k}r^{k-1} = \frac{M}{R}\frac{1}{(1-r/R)^2} = \frac{MR}{(R-r)^2}$$ I don't know if there is a better way to do it like this, because this estimate is more or less of the form that I want, but not accurate enough.
Thanks for any help!