derivation of generating function for the reciprocal central binomial coefficients

According to the article, the ordinary generating function for $\\dbinom{2n}{n}^{-1}$ is

\\frac{4\\left(\\sqrt{4-x}+\\sqrt{x}\\arcsin\\left(\\frac{\\sqrt{x}}{2}\\right)\\right)}%{(4-x)^{3/2}}

To see this, let $C_{n}=\\binom{2n}{n}^{-1}$ , and $C(x)=\\sum_{n\\geq 0}C_{n}x^{n}$ its ordinary generating function. Then

	$\\displaystyle C_{n+1}$	$\\displaystyle=\\dbinom{2n+2}{n+1}^{-1}=\\frac{(n+1)!(n+1)!}{(2n+2)!}$
		$\\displaystyle=\\frac{(n+1)(n+1)}{(2n+2)(2n+1)}\\cdot\\frac{n!n!}{(2n)!}$
		$\\displaystyle=\\frac{n+1}{2(2n+1)}\\cdot C_{n}$

Thus

(4n+2)C_{n+1}=(n+1)C_{n}

so that

\\sum_{n\\geq 0}(4n+2)C_{n+1}x^{n}=\\sum_{n\\geq 0}(n+1)C_{n}x^{n}

A little algebra gives

4\\sum_{n\\geq 0}(n+1)C_{n+1}x^{n}-2\\sum_{n\\geq 0}C_{n+1}x^{n}=\\sum_{n\\geq 0}nC_%{n}x^{n}+\\sum_{n\\geq 0}C_{n}x^{n}

so that

4C^{\\prime}(x)-\\frac{2}{x}(C(x)-1)=xC^{\\prime}(x)+C(x)

and, collecting terms,

(4x-x^{2})C^{\\prime}(x)=(x+2)C(x)-2

We now have a first-order linear ODE to solve. Put it in the form

C^{\\prime}(x)+\\frac{-x-2}{x(4-x)}C(x)=\\frac{-2}{4x-x^{2}}

and we must now integrate the coefficient of $C(x)$ . Expand by partial fractions and integrate to get

\\int\\frac{-x-2}{x(4-x)}dx=\\ln\\left(\\frac{(4-x)^{3/2}}{\\sqrt{x}}\\right)

Thus the solution to the equation is

	$\\displaystyle C(x)$	$\\displaystyle=\\frac{\\sqrt{x}}{(4-x)^{3/2}}\\left(k+\\int\\frac{(4-x)^{3/2}}{\\sqrt%{x}}\\cdot\\frac{-2}{x(4-x)}dx\\right)$
		$\\displaystyle=\\frac{k\\sqrt{x}}{(4-x)^{3/2}}-\\frac{2\\sqrt{x}}{(4-x)^{3/2}}\\int%\\frac{\\sqrt{4-x}}{x^{3/2}}dx$
		$\\displaystyle=\\frac{k\\sqrt{x}}{(4-x)^{3/2}}-\\frac{2\\sqrt{x}}{(4-x)^{3/2}}\\left%(\\frac{-2(4-x)}{\\sqrt{x(4-x)}}-\\arcsin\\left(\\frac{x}{2}-1\\right)\\right)$
		$\\displaystyle=\\frac{4}{4-x}+\\frac{\\sqrt{x}}{(4-x)^{3/2}}\\left(k+2\\arcsin\\left(%\\frac{x}{2}-1\\right)\\right)$

To determine the constant $k$ , note that we should have $C^{\\prime}(x)\\big{\\lvert}_{x=0}=\\frac{1}{2}$ ; looking at $\\lim_{x\\to 0}C^{\\prime}(x)$ we see that for $k=\\pi$ this equation holds. Thus

C(x)=\\frac{4}{4-x}+\\frac{\\sqrt{x}}{(4-x)^{3/2}}\\left(\\pi+2\\arcsin\\left(\\frac{x%}{2}-1\\right)\\right)

We show below that the following is an identity:

\\sqrt{\\frac{z+1}{2}}=\\sin\\left(\\frac{\\pi}{4}+\\frac{1}{2}\\arcsin(z)\\right)

Assuming that result, substitute $\\frac{x}{2}-1$ for $z$ and simplify to get

\\frac{\\sqrt{x}}{2}=\\sin\\left(\\frac{\\pi}{4}+\\frac{1}{2}\\arcsin\\left(\\frac{x}{2}%-1\\right)\\right)

so that

4\\arcsin\\left(\\frac{\\sqrt{x}}{2}\\right)=\\pi+2\\arcsin\\left(\\frac{x}{2}-1\\right)

and then

	$\\displaystyle C(x)$	$\\displaystyle=\\frac{4}{4-x}+\\frac{\\sqrt{x}}{(4-x)^{3/2}}\\left(4\\arcsin\\left(%\\frac{\\sqrt{x}}{2}\\right)\\right)$
		$\\displaystyle=\\frac{4\\left(\\sqrt{4-x}+\\sqrt{x}\\arcsin\\left(\\frac{\\sqrt{x}}{2}%\\right)\\right)}{(4-x)^{3/2}}$

as desired.

Finally, to prove the identity, first expand the right-hand using the formula for $\\sin(a+b)$ , and then apply the half-angle formulas:

	$\\displaystyle\\sin\\left(\\frac{\\pi}{4}+\\frac{1}{2}\\arcsin(z)\\right)$	$\\displaystyle=\\frac{\\sqrt{2}}{2}\\left(\\cos\\left(\\frac{1}{2}\\arcsin(z)\\right)+%\\sin\\left(\\frac{1}{2}\\arcsin(z)\\right)\\right)$
		$\\displaystyle=\\frac{\\sqrt{2}}{2}\\left(\\sqrt{\\frac{1+\\cos(\\arcsin(z))}{2}}+%\\sqrt{\\frac{1-\\cos(\\arcsin(z))}{2}}\\right)$
		$\\displaystyle=\\frac{\\sqrt{2}}{2}\\left(\\sqrt{\\frac{1+\\sqrt{1-z^{2}}}{2}}+\\sqrt{%\\frac{1-\\sqrt{1-z^{2}}}{2}}\\right)$
		$\\displaystyle=\\frac{1}{2}\\left(\\sqrt{1+\\sqrt{1-z^{2}}}+\\sqrt{1-\\sqrt{1-z^{2}}}\\right)$

Now square this expression to get

\\frac{1}{4}\\left(2+2\\sqrt{1-1+z^{2}}\\right)=\\frac{\\lvert z\\rvert+1}{2}

Thus the identity holds for $0\\leq z\\leq 1$ ; an almost identical computation using $-z$ in of $z$ shows that it also holds for $-1\\leq z\\leq 0$ .