Convergence of a hypergeometric

[CAF123]

The hypergeometric function, ${}_{2}F_1(a,b,c;z)$ can be written in terms of a power series in $z$ as follows, $${}_{2}F_1(a,b,c;z) = \sum_{n=0}^{\infty} \frac{(a)_n (b)_n}{(c)_n} \frac{z^n}{n!}\,\,\,\,\,\text{provided}\,\,\,\,|z|<1$$

So we may reexpress any hypergeometric function as a power series like this as long as the last argument modulus is less than 1.

My question is, I also have an expansion of a hypergeometric of the form ${}_2 F_1(\alpha,\beta+\epsilon, \gamma- \epsilon, z)$, where $\alpha, \beta, \gamma$ are real numbers, by using the HypExp package on Mathematica (expansion in $\epsilon$) and was wondering if I use this expansion do I also require $|z|<1$? When I write the code in Mathematica, the last argument is replaced by simply $x$ say and Mathematica gives me an expansion regardless of the size of $x$ so I am thinking ${\it this}$ expansion (and not the power series one) is maybe valid independent of $|x|$ but would be nice to confirm.

Thanks!

 

[micromass]

The hypergeometric function, ${}_{2}F_1(a,b,c;z)$ can be written in terms of a power series in $z$ as follows, $${}_{2}F_1(a,b,c;z) = \sum_{n=0}^{\infty} \frac{(a)_n (b)_n}{(c)_n} \frac{z^n}{n!}\,\,\,\,\,\text{provided}\,\,\,\,|z|<1$$

This specific form of the hypergeometric function is valid for $|z|<1$. The hypergeometric function can be extended beyond $|z|<1$ by analytic continuation. So this analytic continuation is likely what mathematica is calculating.