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



Estimating the growth of the Taylor coefficients given the growth of the function at the boundary
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 $\epsilon>0$.


