Suppose $$f(z)=\sum_{n=0}^\infty a_nz^n$$ extends meromorphically to the whole complex plane. Show that if $$\frac{r^n a_n}{n^k}\to c\neq 0\quad \mathrm{as}\quad n\to\infty\quad$$ for some integer $k\geq 0$ and some $r>0$, then $f(z)$ has a pole of order $k+1$ at $z=r$.
I tried finding $\lim_{z\to r}(z-r)f(z)$ but didn't succeed. I also tried to use the root test for $(z-r)f(z)$ but there isn't enough information. Now I don't know where I should start with. Any help is very much appreciated.
Cauchy's integral formula implies that the Taylor series of $f$ at $z=0$ converges for $|z|<r$ where $r$ is the modulus of the first pole. With $s_1,\ldots,s_K$ the finitely many poles on $|z|=r$ of order $e_1,\ldots,e_k$ we get that the Taylor series of $f-\sum_{k\le K}\sum_{j\le e_k} \frac{c_{k,j}}{(z-s_k)^j}$ converges for $|z|< r+\epsilon$. That is to say $$f(z)=\sum_{n\ge 0} a_n z^n, \qquad a_n = \sum_{k\le K} \sum_{j\le e_k} C_{k,j} n^{j-1} s_k^{-n} + O( (r+\epsilon)^{-n})$$ And $a_n\sim C_{e_1,1} s_1^{-n} n^{e_1-1}$ iff $e_1$ is larger than all the other $e_k$