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We know that $$\sum_{n=0}^\infty C_nx^n=\frac{1-\sqrt{1-4x}}{2x}$$. But because $C_n=\frac1{n+1}\binom{2n}n=\frac{(2n)!}{n!(n+1)!}$, that sum is also $$\sum_{n=0}^\infty\frac{(2n)!}{(n+1)!}\frac{x^n}{n!}$$ $$=\sum_{n=0}^\infty\frac{(\frac12)_n(1)_n}{(2)_n}\frac{(4x)^n}{n!}$$ $$={}_2F_1(\frac12,1;2;4x)$$, so we get the identity $$_2F_1(\frac12,1;2;4x)=\frac{1-\sqrt{1-4x}}{2x}$$ My two questions are:
Is this correct?
Can you proof this identity without the use of generating functions or are there well-known hypergeometric series identities from which this follows?

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  • $\begingroup$ Check out my edit to see how you can prevent the left-hand subscript on $_2F_1$ from being attached to the preceding symbol. $\endgroup$ – joriki Jul 18 '18 at 11:59
  • $\begingroup$ It's correct for $x=\dfrac14$ $\endgroup$ – Nosrati Jul 18 '18 at 12:04
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This is correct: you can use one of the contiguous relations, in particular $$ (c-1)(F(c-)-F) = \frac{(c-b)F(b-)+(b-c+az)F}{1-z}, $$ which here becomes $$ F(1/2,1;1;4x)-F(1/2,1;2;4x) = \frac{F(1/2,0;2;4x)+(-1+2x)F(1/2,1;2;4x)}{1-4x}. $$ $F(a,0;c;z)=1$, and similarly it is easy to check that $F(a,b;b;z)=(1-z)^{-a}$, so we have $$ \frac{1}{\sqrt{1-4x}} = \frac{1}{1-4x} + \left( 1 + \frac{-1+2x}{1-4x} \right) F(1/2,1;2;4x) \\ \sqrt{1-4x} -1 = (-2x)F(1/2,1;2;4x), $$ and rearranging gives the result. The same strategy works for any hypergeometric functions with some parameters integers.

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