A power series involving binomial coefficients I've been playing around with infinite sums for quite a while now, but recently, I've come across the following question in an Under-Graduate Mathematics Book that is specifically targeted at problem solving. The problem is as follows,
Evaluate the following:
$$\sum_{r=2}^\infty \Biggl(\binom{2r}{r}{\biggl(\frac{x}{4}\biggr)^r}\Biggr)^2$$
where x is strictly less than unity.
I've thoroughly checked that the sum is, in fact, convergent, however, I am completely stumped as to how I am to evaluate it. I am guessing that the final expression is one involving the variable 'x' since I do not see any way for it to be eliminated somehow. Any kind of hint/solution/explanation to the problem would be highly appreciated.
 A: Quite simple.
First, consider the full series:
$$S(x)=\sum_{n=0}^\infty \binom{2n}{n}^2 \frac{x^{2n}}{4^{2n}}=\sum_{n=0}^\infty \frac{(2n)!^2}{n!^4} \frac{x^{2n}}{4^{2n}}$$
Consider the ratio of general terms:
$$\frac{T_{n+1}}{T_n}=\frac{(2n+2)^2(2n+1)^2}{16(n+1)^4} x^2=\frac{(n+1/2)^2}{(n+1)^2} x^2=\frac{(n+1/2)^2}{(n+1)} \frac{x^2}{n+1}$$
$$T_0=1$$
By definition:
$$S(x)={_2 F_1} \left(\frac{1}{2},\frac{1}{2};1,x^2 \right)$$
Using the integral representation of the hypergeometric function:
$${_2 F_1} \left(\frac{1}{2},\frac{1}{2};1,x^2 \right)=\frac{1}{B \left(\frac{1}{2},\frac{1}{2}\right)} \int_0^1 t^{-1/2} (1-t)^{-1/2} (1-x^2 t)^{-1/2} dt=$$
We know $B \left(\frac{1}{2},\frac{1}{2}\right)=\Gamma \left(\frac{1}{2}\right)^2 = \pi$. Let's introduce a new variable $t=y^2$, then:
$$=\frac{2}{\pi} \int_0^1 \frac{dy}{\sqrt{(1-y^2)(1-x^2 y^2)}}$$
But this is just the complete elliptic integral of the first kind.
So:
$$S(x)=\frac{2}{\pi}  K(x^2)$$
And, subtracting the two first terms, we get for the original series:

$$\sum_{n=2}^\infty \binom{2n}{n}^2 \frac{x^{2n}}{4^{2n}}=\frac{2}{\pi}  K(x^2)-1-\frac{x^2}{4}$$

Just as Raffaele pointed out in the comments.
