Arithmetic Geometric Mean can be represented by a Hypergeometric function:
$$\text{agm}(1,p)=\frac{1}{{_2F_1} \left(\frac{1}{2},\frac{1}{2};1;1-p^2 \right)}$$
$$0<p \leq 1$$
One of the main properties of the AGM is the following identity:
$$\text{agm}(1,p)=\frac{1+p}{2}\text{agm} \left(1,\frac{2\sqrt{p}}{1+p} \right)$$
This allows the infinite product representation of the AGM.
I wanted to know if it's possible to prove this identity by directly using the Hypergeometric series.
For the Hypergeometric function the identity will take the following form:
$${_2F_1} \left(\frac{1}{2},\frac{1}{2};1;1-p^2 \right)=\frac{2}{1+p} {_2F_1} \left(\frac{1}{2},\frac{1}{2};1;\frac{(1-p)^2}{(1+p)^2} \right)$$
In the series form it will be:
$$ \sum_{k=0}^\infty \frac{1}{k!^2} \left(\frac{1}{2}\right)^2_k (1-p^2)^k=\frac{2}{1+p} \sum_{k=0}^\infty \frac{1}{k!^2} \left(\frac{1}{2}\right)^2_k \frac{(1-p)^{2k}}{(1+p)^{2k}}$$
I think the following substitution will simplify things:
$$1-p=2x$$
$$ \sum_{k=0}^\infty \frac{1}{k!^2} \left(\frac{1}{2}\right)^2_k 2^{2k}x^k(1-x)^k=\frac{1}{1-x} \sum_{k=0}^\infty \frac{1}{k!^2} {\left(\frac{1}{2}\right)^2_k} \frac{x^{2k}}{(1-x)^{2k}}$$
I haven't been able to prove this identity from the series.
Comparing terms in this form is useless, since the partial sums of the series are not equal (the second series converges much faster).
The only idea I have is to use the uniqueness of the power series, which requires expanding everything, so there are only powers of $x$ left.
We have:
$$(1-x)^k=\sum_{l=0}^k (-1)^l \left(\begin{matrix} k \\ l \end{matrix} \right) x^l$$
$$\frac{1}{(1-x)^{2k+1}}=(2k)! \sum_{n=0}^\infty (-1)^{n+2k}~ (n+2k)_{2k} ~x^n$$
Here $(n+2k)_{2k}$ actually means falling factorial, not rising factorial, like above. $(n+2k)_{2k}=(n+2k)(n+2k-1)(n+2k-2) \cdots$. I don't know what other notation to use in this case.
Now I'm stuck. I don't know how to get the single power series for $x$ on each side so we can compare them.
Use the function $$f(x) = \frac{1}{\operatorname{agm}(1 + x, 1 - x)}\tag{1}$$ which has the property that $$f(x) = \frac{1}{1 + x}\cdot f\left(\frac{2\sqrt{x}}{1 + x}\right)\tag{2}$$ and use the series expansion $$f(x) = 1 + a_{1}x^{2} + a_{2}x^{4} + \cdots\tag{3}$$ (note that the function $f$ is even) and find coefficients $a_{n}$ by using series expansion $(3)$ in functional equation $(2)$. This is how Gauss derived the formula $$f(x) = {}_{2}F_{1}\left(\frac{1}{2}, \frac{1}{2};1; x^{2}\right)\tag{4}$$ See this post of mine for details. Note also that $$f(x) = \frac{1}{\operatorname{agm}(1, \sqrt{1 - x^{2}})}\tag{5}$$ and hence on putting $1 - x^{2} = p^{2}$ we get $$\operatorname{agm}(1, p) = \dfrac{1}{{}_{2}F_{1}\left(\dfrac{1}{2}, \dfrac{1}{2};1; 1 - p^{2}\right)}\tag{6}$$
BTW the evaluation of coefficients $a_{n}$ for general $n$ is difficult but Gauss solved it completely. See this paper for more details of the calculation of $a_{n}$.
Another route to the relation between hypergeometric series is to use the differential equation satisfied by the function $y(x) = {}_{2}F_{1}(a, b; c; x)$. The differential equation satisfied by $y$ is given as $$x(1 - x)y'' + \{c - (a + b + 1)x\}y' - aby = 0\tag{7}$$ Let's just write $F$ in place of ${}_{2}F_{1}$ and then we see that $y = F(a, b; 2b; x)$ satisfies the equation $$x(1 - x)y'' + \{2b - (a + b + 1)x\}y' - aby = 0$$ Putting $x = 4z/(1 + z)^{2}$ we can see that that $$\frac{dF}{dx} = \frac{(1 + z)^{3}}{4(1 - z)}\cdot\frac{dF}{dz}$$ and $$\frac{d^{2}F}{dx^{2}} = \frac{(1 + z)^{5}}{16(1 - z)^{3}}\left((1 - z^{2})\frac{d^{2}F}{dz^{2}} + (4 - 2z)\frac{dF}{dz}\right)$$ and $$x(1 - x) = \frac{4z(1 - z)^{2}}{(1 + z)^{4}}$$ and thus after some algebraic manipulation we get $$z(1 - z)(1 + z)^{2}\frac{d^{2}F}{dz^{2}} + 2(1 + z)(b - 2az + bz^{2} - z^{2})\frac{dF}{dz} - 4ab(1 - z)F = 0$$ and the function $F(a, b; 2b; 4z/(1 + z)^{2})$ satisfies this equation. If we put $F = (1 + z)^{2a}G$ we get $$z(1 - z^{2})\frac{d^{2}G}{dz^{2}} + 2\{b - (2a - b + 1)z^{2}\}\frac{dG}{dz} - 2az(1 + 2a - 2b)G = 0$$ It is easily seen that $G(-z)$ also satisfies this equation and hence $G$ is an even function and we can put $z^{2} = t$ to get $$t(1 - t)\frac{d^{2}G}{dt^{2}} + \left(b + \frac{1}{2} - \left(2a - b + \frac{3}{2}\right)t\right)\frac{dG}{dt} - a\left(a - b + \frac{1}{2}\right)G = 0\tag{8}$$ Comparing this with equation $(7)$ we see that solution $G$ is given by $$G = F(a, a - b + 1/2; b + 1/2; t) = F(a, a - b + 1/2; b + 1/2; z^{2})\tag{9}$$ However the way we reached equation $(8)$ shows us that its solution is given by $$G = (1 + z)^{-2a}F(a, b; 2b; 4z/(1 + z)^{2})\tag{10}$$ Both the solutions $(9)$ and $(10)$ are analytic in neighborhood of $0$ and they are equal to $1$ at $z = 0$ and therefore they are equal and we get the quadratic transformation of Gauss $$F\left(a, b; 2b; \frac{4z}{(1 + z)^{2}}\right) = (1 + z)^{2a}F\left(a, a - b + \frac{1}{2}; b + \frac{1}{2}; z^{2}\right)\tag{11}$$ Putting $a = 1/2, b = 1/2$ we get $$F\left(\frac{1}{2}, \frac{1}{2}; 1; \frac{4z}{(1 + z)^{2}}\right) = (1 + z)F\left(\frac{1}{2}, \frac{1}{2}; 1; z^{2}\right)\tag{12}$$ and putting $z = (1 - p)/(1 + p)$ so that $p = (1 - z)/(1 + z)$ we get $$F\left(\frac{1}{2}, \frac{1}{2}; 1; 1 - p^{2}\right) = \frac{2}{1 + p}F\left(\frac{1}{2}, \frac{1}{2}; 1; \frac{(1 - p)^{2}}{(1 + p)^{2}}\right)\tag{13}$$ which is the result in question. Using similar technique we can prove another quadratic transformation $$F\left(a, b; a + b + \frac{1}{2}; 4z(1 - z)\right) = F\left(2a, 2b; a + b + \frac{1}{2}; z\right)\tag{14}$$ See this post and the next one for more details.