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 $k-1\le k'\le k$.

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 $k-1\le k'\le k$.