Q: Proving that $\lim \limits_{b \rightarrow \infty} F(1,b,1;\frac{z}{b})=e^z$ I am trying to prove that $\lim \limits_{b \rightarrow \infty} F(1,b,1;\frac{z}{b})=e^z$ without using dominated convergence theorem. Here $F$
 is the hypergeometric function. 
I have been  able to show that if the limit exists it has to be equal to $e^x$ but I have not been able to show that the limit exists. I am struck as to why we can exchange the sum and limit here. But again I was able to show that the limit exists when $x>0$. Since as $b$ increases $ F(1,b,1;\frac{x}{b})$ decreases and it is greater than $0$ so limit exists.

A: From the series definition of the hypergeometric function it follows that
$$_2F_1(1,b;1;z)=(1-z)^{-b}.$$
Does this solve your problem?
