Find the sum of this infinite series 
Sum the series $$\frac{1^2}{2^2}+\frac{1^2\cdot3^2}{2^2\cdot4^2}+\frac{1^2\cdot3^2\cdot5^2}{2^2\cdot4^2\cdot6^2}+\ldots$$

Someone please help me in finding sum of this infinite series. I have never found the sum of this kind of series.
 A: The function $${}_2 F_1(1/2,1/2;1;z) = 1 + \sum_{n=1}^{\infty} \frac{((1/2)(3/2)...(1/2 + n - 1))^2}{n!^2} z^{n} = 1 + \sum_{n=1}^{\infty} \Big( \frac{1 \cdot 3 \cdot ... \cdot (2n-1)}{2 \cdot 4 \cdot ... \cdot (2n)} \Big)^2 z^{n}$$
can be expressed as an elliptic integral $$\frac{2}{\pi} \int_0^{\pi/2} \frac{1}{\sqrt{1 - z \sin^2(\theta)}} \, \mathrm{d}\theta$$ whenever $|z| < 1.$ 
From the integral representation it is easy to see that diverges as $z$ approaches $1$. For fun, the alternating series with $z = -1$ can be evaluated as $$1 -\frac{1^2}{2^2} + \frac{1^2 \cdot 3^2}{2^2 \cdot 4^2} \pm ... = \frac{\Gamma(1/4)^2}{(2\pi)^{3/2}} \approx 0.8346$$
A: One may observe that the general term of this series rewrites
$$
a_n:=\frac{1^2\cdots(2n-1)^2}{2^2\cdot4^2\cdots(2n)^2}=\frac{[(2n)!]^2}{2^{4n}\cdot(n!)^4}
$$ thus, as $n \to \infty$, it satisfies
$$
\left|\frac{a_n}{a_{n+1}}\right| = 1+ \frac{1}{n} + \frac{C_n}{n^2},\quad |C_n|<1,
$$ and the series diverges by Gauss' test which is a refined ratio test.
