Convergence of a complex power series Let $a,b,c \in \mathbb C$ with $c \in \mathbb N$. Then I have to calculate the radius of convergence of the following power series:
$$
1+ \frac{ab}{c \cdot 1!} z + \frac{a (a+1)b(b+1)}{c(c+1)2!} z^2+ \frac{a(a+1)(a+2)b(b+1)(b+2)}{c(c+1)(c+2)3!}z^3 + \cdots
$$
Using the ratio-test I get that
$$
\left | \frac {a_{n+1}}{a_n} \right | = \left | \frac{(a+n)(b+n)}{(c+n)(n+1)}z \right |
$$
How can I proceed ? This is an exam-question and a given hint says to consider whether $a$ or $b$ are in $\mathbb N$ or not. I have no idea how to use the hint.
Thanks in advance.
 A: It would seem to me that
$$
\lim_{n\to\infty} \left| \frac{(a+b)(b+n)}{(c+n)(n+1)}z \right| = 0
$$
Regardless of $z$ because you're dividing a polynomial of degree $1$ by a polynomial of degree $2$. So, if you went through the ratio test correctly, the radius of convergence should just be $\infty$.

EDIT:
As per your comment, we now have
$$
\lim_{n\to\infty} \left| \frac{(a+n)(b+n)}{(c+n)(n+1)}z \right| = 
\lim_{n\to\infty} \left| \frac{n^2+O(n)}{n^2+O(n)}\right| \cdot\left|z\right|=\left|z\right|
$$
Which means that although the ratio test fails when $\left|z\right|=1$, our radius of convergence is just $1$ (since the series converges for $|z|<1$).
I have no idea how to use the hint either; it seems you may have outsmarted the exam.

SECOND EDIT: I now understand the exam hint!  The ratio test assumes $(a+n)$ and $(b+n)$ are always non-zero.  In the event that $a$ or $b$ is a negative integer, the radius of convergence changes to $\infty$ since we will always have finitely many non-zero terms.
