# Prove that ${}_2F_1(0,b;c;z)=1$

I do not know how I could prove that ${}_2F_1(0,\beta;\gamma;t)=1$ because when I apply the definition I get $0$, namely..

$$\sum_{n=0}^{\infty}\frac{(0)_n(\beta)_n}{n!(\gamma)_n}t^n=0$$

someone help?

This is also true for $_2F_1(a,0,c;z)$. The reason is that
$$_2F_1(a,b,c;z) = 1 + \frac{a b}{c} z + \frac{a (1+a) b (1+b)}{c (1+c)} z^2 + \frac{a (1+a) (2+a) b (1+b) (2+b)}{c (1+c)(2+c)} z^3 + \cdots$$
Because all coefficients of $z^n$ for $n \ge 1$ have a factor of $a$ in them, they all disappear when $a=0$.
• You missed a $b$ in the lhs. I fixed it hoping that you will not worry ! – Claude Leibovici Jan 24 '16 at 5:15