In this thread, How to derive this series
I have asked about how to derive:
$\dfrac{1}{1-x}=1+\dfrac{x}{1+x}+\dfrac{1\cdot2\cdot x^2}{(1+x)(1+2x)}+\dfrac{1\cdot2\cdot3\cdot x^3}{(1+x)(1+2x)(1+3x)}...$
Somos provided the excellent answer:
$$ S(x) = {}_2F_1(1,1;1+1/x;1) = \dfrac{1}{(1-x)} $$
To confirm it, I use the definition of the ordinary hypergeometric function to confirm it:
${}_2F_1(\alpha,\beta;\gamma;x)=1+\dfrac{\alpha\beta x}{\gamma\cdot1!}+\dfrac{\alpha(\alpha+1)\beta(\beta+1)x^2}{\gamma(\gamma+1)\cdot2!}+\dfrac{\alpha(\alpha+1)(\alpha+2)\beta(\beta+1)(\beta+2)x^3}{\gamma(\gamma+1)(\gamma+2)\cdot3!}...$
Everything is fine and correct until
I try to use Wolfram to confirm this identity but it doesn't give me the answer that I need.
https://www.wolframalpha.com/input/?i=2F1%281%2C1%3B1%2B1%2Fx%3B1%29
It doesn't prove that $${}_2F_1(1,1;1+1/x;1) = \dfrac{1}{1-x}$$
After tinkering with it for a while, I notice that
$${}_2F_1(1,1;1;x)= \dfrac{1}{1-x}$$
https://www.wolframalpha.com/input/?i=2F1%281%2C1%3B1%3Bx%29
My question is how to prove that ${}_2F_1(1,1;1;x)={}_2F_1(1,1;1+1/x;1)$
My second question how do you determine $\alpha, \beta, \gamma$ in ${}_2F_1(\alpha,\beta;\gamma;x)$ to determine the limiting function?
For example:
$$x{}_2F_1(\color{red}{1,1;2;-x})=\ln(x+1)$$
$$x{}_2F_1(\color{red}{\color{red}{\frac{1}{2},\frac{1}{2};\frac{3}{2};x^2})}=\arcsin(x)$$
(Is there a way to prove this without recoursing to gamma function, since I haven't learnt gamma function yet)