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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.

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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.

Ben Grossmann
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    Oh typo. It must be $(a+n)(b+n)$ over $(c+n)(n+1)$ –  Jun 27 '13 at 17:48
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    Unless $a$ or $b$ is zero or a negative integer. In that case the power series has only finitely many non-zero terms and the radius of convergence is infinite. - And btw., this is the http://en.wikipedia.org/wiki/Hypergeometric_function. – Martin R Jun 27 '13 at 17:55
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    Ok Martin. I got it :) And otherwise if $|z| < 1$. Thank you, guys. –  Jun 27 '13 at 17:58
  • I accept this as answer, even though Martin made the core of the question clear to me. –  Jun 27 '13 at 18:01
  • Ah I guess Martin beat me to the punch there. – Ben Grossmann Jun 27 '13 at 18:26