Find the radius of convergence of
$$F(\alpha,\beta,\gamma,z)=1+\sum_\limits{n=1}^\infty\frac{\alpha(\alpha+1)\cdots(\alpha+n-1)\beta(\beta+1)\cdots(\beta+n-1)}{n!\gamma(\gamma+1)\cdots(\gamma+n-1)}z^n$$
Here $\alpha,\beta\in\mathbb{C}$ and $\gamma \neq 0,-1,-2,\cdots$.
I set
$$c_n=\frac{\alpha(\alpha+1)\cdots(\alpha+n-1)\beta(\beta+1)\cdots(\beta+n-1)}{\gamma(\gamma+1)\cdots(\gamma+n-1)},\qquad a_n=\frac{c_n}{n!}$$
and
$$\color{red}{c_n\sim\left(\frac{\alpha\,\beta}{\gamma}\right)^n}, \qquad n\to\infty$$
Thus, the radius of convergence $R$ can be found using Cauchy-Hadamard's formula
$$R=\frac{1}{\lim\sup|a_n|^{1/n}}=\left|\frac{\gamma}{\alpha\,\beta}\right|\cdot\lim_\limits{n\to\infty}\sqrt[n]{|n!|}=\left|\frac{\gamma}{\alpha\,\beta}\right|\cdot\infty=\infty.$$
This would conclude that $F(\alpha,\beta,\gamma,z)$ is an entire function, but I do not know if these steps are correct.
I realized that the part in red is not true. What would be a better approximation?
Best Answer
It is not correct that $c_n\sim\left(\frac{\alpha\beta}{\gamma}\right)^n$. To get a correct asymptotic, you will need Stirling's approximation. As $x\to\infty$, we have $\Gamma(x)\sim x^{x-1}e^{-x}\sqrt{2\pi x}$. Using $\alpha(\alpha+1)\ldots(\alpha+n-1)=\Gamma(\alpha+n)/\Gamma(\alpha)$, we can compute $$ a_n\sim\frac{\Gamma(\gamma)}{\Gamma(\alpha)\Gamma(\beta)} n^{\alpha+\beta-\gamma}\asymp n^{\alpha+\beta-\gamma}, $$ from which it can be computed that $\lim_{n\to\infty}|a_n|^{1/n}=1$.
A different approach to determine the radius of convergence would be to use the ratio test. The radius of convergence is given by the limit $$ \lim_{n\to\infty}\frac{a_{n}}{a_{n+1}}=\lim_{n\to\infty}\frac{n(\gamma+n)}{(\alpha+n)(\beta+n)}=1, $$ provided that the limit exists.