estimate the coefficients of the power series. If a holomorphic function on unit disk satisfying that $|f'(z)|<(1-|z|)^{-1}$, show that for $f(z)=\sum_{n=0}^{\infty}a_nz^n$ we have $|a_n|<e$ for $n>0$.
It seems that this question have something to do with Schwarz lemma. But I don't know how to construct the function we need for proving this.
 A: One has $f‘(z)=\sum_{n=1}^\infty n\,a_n z^{n–1}$. Consider now any $n\in{\mathbb N}$, $n\geq1$. Then by Cauchy's formula  $$na_n=\frac1{2\pi i}\int_{|z|=r}\frac{f‘(z)}{z^n}dz$$ for any $r$, $0<r<1$, where the integral is taken counterclockwise over the circle of radius $r$ with center at the origin. The classical estimate of integrals implies using the hypothesis $|f'(z)|\leq(1-|z|)^{-1}$ that $$\label{eq1}\tag1n|a_n|\leq \frac1{2\pi}\cdot\max\{|f'(z)||z|^{-n}\,\mid\,|z|=r\}\cdot2\pi r\leq{(1–r)^{-1}r^{-(n–1)}}$$ for all $r$, $0<r<1$. We can choose any $r$ between 0 and 1 we want. Therefore we choose it such that the expression on the right hand side is minimised. This is equivalent to maximising $(1-r)r^{n-1}$ for $0<r<1$. This maximum is reached where the derivative of $h(r)=r^{n-1}-r^n$ vanishes, that is for $r=(n-1)/n.$ Therefore we put $r=(n-1)/n$ in (\ref{eq1}) and obtain $$|a_n|\leq \left(\frac n{n–1}\right)^{n–1}=\left(1+\frac1{n-1}\right)^{n-1}\leq e.$$
For the latter inequality observe that
$$\log\left(\left(1+\frac1{n-1}\right)^{n-1}\right)=(n-1)\log\left(1+\frac1{n-1}\right)=\frac1x\log(1+x)$$
with $x=1/(n-1)$ and $\log(1+x)=\int_0^x\frac1{1+t}\leq x$ for positive $x$; hence $\frac1x\log(1+x)\leq1$ for all positive $x.$
