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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.

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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?

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  • $\begingroup$ Yes it does it thank you $\endgroup$ – happymath Mar 3 '15 at 9:59

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