Estimating the growth of the Taylor coefficients given the growth of the function at the boundary

Question

Let $f(z)=\sum a_nz^n$ be a Taylor series that converges for $|z|<1$ and satisfies $$ |f(z)|\le \frac{1}{(1-|z|)^{k}} $$ for some fixed $k>0$.

Question: What can I deduce about the growth of the Taylor coefficients $a_n$?

Partial result: By judiciously selecting the location of the contour in the formula $a_n=\oint z^{-n}f(z)\tfrac{dz}{2\pi i z}$, namely, by performing the integration over the contour $|z|=\tfrac{n}{n+k}$ [which is the minimum of $|z|^{-n}(1-|z|)^{-k}$], I can get the "trivial bound" $|a_n|< c\cdot n^k$.

But I suspect that this is not sharp.
In particular, the growth of the Taylor coefficients of $(1-z)^{-k}$ is only $n^{k-1}$. Not $n^k$.

More precise formulation of the question:
What is the optimal $k'>0$ such that $$ |f(z)|\le \frac{1}{(1-|z|)^{k}}\quad\Rightarrow\quad |a_n|< c\cdot n^{k'} $$ for all $f(z)=\sum a_nz^n$.

From the above arguments, I know that $k-1\le k'\le k$.

Answer 1

The optimal exponent is $k$. Such examples are given by sparse power series. This is actually trivial in the case $k=0$ (which was not included in the OP). Then we can simply take $f(z)=\sum j^{-2} z^{N(j)}$, say. This is obviously bounded, and the coefficients $a_n$ will not satisfy $|a_n|\lesssim n^{-\epsilon}$ for any $\epsilon>0$ if $N(j)$ increases fast enough.

For positive $k$, we can similarly consider something like $$ f(z)=\sum n^{-2} \left(n^n\right)^k z^{n^n} . $$ Using calculus to find the maximum, we see that $$ x^k(1-\delta)^x \le C\delta^{-k} . $$ Thus $f$ satisfies the desired bound, but the coefficients do not satisfy $|a_n|\lesssim n^{k-\epsilon}$ for any $\epsilon>0$.