6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Subhanjan Saha Jul 18 '17 at 6:30
  • $\begingroup$ I, see. Just a curiosity, what is the title of this Under-Graduate Mathematics Book? $\endgroup$ – Robert Z Jul 18 '17 at 6:36
  • $\begingroup$ "where x is strictly less than unity." what does that mean exactly? $\endgroup$ – zhw. Jul 18 '17 at 6:43
  • $\begingroup$ @zhw. Taken literally, it means $x<1$. I go so far as to guess is actually means $|x|<1$. $\endgroup$ – Arthur Jul 18 '17 at 6:46
  • 1
    $\begingroup$ 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 $\frac{\sqrt{1-x}}{1-x}-1-\frac{x}{2}$ $\endgroup$ – Raffaele Jul 18 '17 at 12:17
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.