# 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.

• No, Robert Z, I have checked and rechecked the problem but I can affirmatively say that it isn't the summation that is being squared but the term of summation. – Subhanjan Saha Jul 18 '17 at 6:30
• I, see. Just a curiosity, what is the title of this Under-Graduate Mathematics Book? – Robert Z Jul 18 '17 at 6:36
• "where x is strictly less than unity." what does that mean exactly? – zhw. Jul 18 '17 at 6:43
• @zhw. Taken literally, it means $x<1$. I go so far as to guess is actually means $|x|<1$. – Arthur Jul 18 '17 at 6:46
• Solution $\frac{2 K\left(x^2\right)}{\pi }-\frac{x^2}{4}-1$ where $K$ is the elliptic integral of the first kind. Without the square the solution is much nicer $\frac{\sqrt{1-x}}{1-x}-1-\frac{x}{2}$ – Raffaele Jul 18 '17 at 12:17

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.