## Generating Function for the Reciprocals of the Central Binomial Coefficients

In this post we generalize the result from the last post to find the generating function for the reciprocals of the central binomial coefficients.  As we did with that one, we start with the beta integral expression for $1/\binom{2n}{n}$: $\displaystyle \frac{1}{\binom{2n}{n}} = (2n+1) \int_0^1 y^n (1-y)^n \, dy = \int_0^1 (2n+1) (y(1-y))^n \, dy$.

Now, multiply by $\frac{x^n}{2n+1}$ (for now, it’s easier not to deal with the extra factor of $2n+1$ on the right) and sum up: $\displaystyle \sum_{n=0}^{\infty} \frac{x^n}{\binom{2n}{n} (2n+1)} = \sum_{n=0}^{\infty} \int_0^1 (xy(1-y))^n \, dy = \int_0^1 \left( \sum_{n=0}^{\infty} (xy(1-y))^n \right) \, dy = \int_0^1 \frac{1}{1-xy(1-y)} \, dy,$

where we use the geometric series formula in the last step.  Now, apply the substitution $t = \sqrt{x} (2y-1)$.  It’s not obvious at this point that this will be helpful, but it will.  The denominator of the integral becomes $\displaystyle 1-xy(1-y) = 1-x \left(\frac{t + \sqrt{x}}{2\sqrt{x}}\right) \left( \frac{\sqrt{x}-t}{2 \sqrt{x}}\right) = 1- \frac{x-t^2}{4} = \frac{4 - x+ t^2}{4}.$

This gives us $\displaystyle \int_0^1 \frac{1}{1-xy(1-y)} \, dy = \int_{-\sqrt{x}}^{\sqrt{x}} \frac{4}{2 \sqrt{x} (t^2 + 4-x)} \, dt = \frac{2}{\sqrt{x}} \int_{-\sqrt{x}}^{\sqrt{x}} \frac{1}{t^2 + 4-x} \, dt \\ = \left.\frac{2}{\sqrt{x (4-x)}} \arctan \left( \frac{t}{\sqrt{4-x}} \right) \right|_{-\sqrt{x}}^{\sqrt{x}} \\ = \frac{2}{\sqrt{4x-x^2}} \left( \arctan \left( \frac{\sqrt{x}}{\sqrt{4-x}} \right) - \arctan \left( \frac{-\sqrt{x}}{\sqrt{4-x}} \right) \right)\\ = \frac{4}{\sqrt{4x-x^2}} \arctan \left( \frac{\sqrt{x}}{\sqrt{4-x}} \right) \\ = \frac{4}{\sqrt{4x-x^2}} \arcsin \left( \frac{\sqrt{x}}{2} \right),$

where in the second-to-last step we use the fact that arctangent is an odd function.

In sum, we now have $\displaystyle \sum_{n=0}^{\infty} \frac{x^n}{\binom{2n}{n} (2n+1)} = \frac{4}{\sqrt{4x-x^2}} \arcsin \left( \frac{\sqrt{x}}{2} \right).$

We have some more work to do to get the left side where we want it.  Replace $x$ with $x^2$ to get $\displaystyle \sum_{n=0}^{\infty} \frac{x^{2n}}{\binom{2n}{n} (2n+1)} = \frac{4}{\sqrt{4x^2-x^4}} \arcsin \left( \frac{x}{2} \right) = \frac{4}{x\sqrt{4-x^2}} \arcsin \left( \frac{x}{2} \right) .$

(Technically, we get $|x|$ in place of $x$ on the right, but since arcsine is odd the expression simplifies to what we have here.)  Now, multiply both sides by $x$ and differentiate.  After some simplification, we get the following: $\displaystyle \sum_{n=0}^{\infty} \frac{x^{2n}}{\binom{2n}{n}} = \frac{1}{1-\left(\frac{x}{2}\right)^2} + \frac{4x \arcsin \left(\frac{x}{2}\right)}{(4-x^2)^{3/2}}.$

To finish, we replace $x$ with $\sqrt{x}$: $\displaystyle \sum_{n=0}^{\infty} \frac{x^n}{\binom{2n}{n}} = \frac{1}{1-\frac{x}{4}} + \frac{4 \sqrt{x}\arcsin \left(\frac{\sqrt{x}}{2}\right)}{(4-x)^{3/2}}.$

(To verify with the result from the previous post, we need to investigate this result for convergence.  Convergence matters only with the geometric series evaluation with common ratio $xy(1-y)$.  Since $y$ must be between 0 and 1, the max value of $y(1-y)$ is 1/4, and so the values of $x$ for which the series converges are $-4 < x < 4$.  Thus we can safely substitute 1 for $x$ in the formula we just derived, obtaining $\frac{4}{3} + \frac{2 \pi \sqrt{3}}{27}$, as we found in the last post.)