A binomial sum

It's not too hard to show that \(\displaystyle \binom{-\frac{1}{2}}{n} = \left( -\dfrac{1}{4} \right)^n \binom{2n}{n}\) and hence, \(\displaystyle \frac{1}{\sqrt{1-4x}} = \sum_{n=0}^\infty \binom{-\frac{1}{2}}{n}(-4x)^n = \sum_{n=0}^\infty \binom{2n}{n}x^n\).

Some tinkering around on WolframAlpha revealed that \(\displaystyle \sum_{n=0}^\infty \binom{3n}{n}x^n = \frac{2\cos(\frac{1}{3}\sin^{-1}(\frac{3\sqrt{3}}{2}\sqrt{x}))}{\sqrt{4-27x}}\). By a rather natural substitution \(x = \frac{4}{27} \sin^2 \theta\), we get

\[\sum_{n=0}^\infty \binom{3n}{n}x^n = \boxed{\dfrac{\cos \frac{\theta}{3}}{\cos \theta}}.\]

Sadly, WA revealed nothing about \(\displaystyle \sum_{n=0}^\infty \binom{4n}{n}x^n\). A cool expression for \(\displaystyle \sum_{n=0}^\infty \binom{4n}{2n}x^n\) is easily obtained with the substitution \(x = \frac{1}{16}\sin^2 \theta\), however:

\[\begin{align} \sum_{n=0}^\infty \binom{4n}{2n}x^n &= \frac{1}{2} \left[ \sum_{n=0}^\infty \binom{2n}{n}(\sqrt{x})^n + \sum_{n=0}^\infty \binom{2n}{n}(-\sqrt{x})^n \right] \\ &= \frac{1}{2} \left[ \frac{1}{\sqrt{1-4\sqrt{x}}} + \frac{1}{\sqrt{1+4\sqrt{x}}} \right] \\ &= \frac{1}{2} \left[ \frac{\sqrt{1-4\sqrt{x}}+\sqrt{1+4\sqrt{x}}}{\sqrt{1-16x}} \right] \\ &= \frac{1}{2} \left[ \frac{\sqrt{1-\sin \theta}+\sqrt{1+\sin \theta}}{\cos \theta} \right] \\ &= \frac{1}{2} \left[ \frac{\sqrt{2 + 2\cos \theta}}{\cos \theta} \right] \\ &= \boxed{\dfrac{\cos \frac{\theta}{2}}{\cos \theta}}. \end{align}\]

Let's define the sum \(\displaystyle F_{a,b}(x) = \sum_{n=0}^\infty \binom{an}{bn}x^n\). For convenience let's also define \(K_{a,b} = \dfrac{b^b(a-b)^{a-b}}{a^a}\). So \(K_{2,1} = \dfrac{1}{4}\), \(K_{3,1} = \dfrac{4}{27}\), and \(K_{4,2} = \dfrac{1}{16}\). It seems that the factor associated with the \(\sin^2 \theta\) comes from the series' radius of convergence, ie

\[\begin{align} \binom{an}{bn} &= \frac{(an)!}{(bn)!((a-b)n)!} \\ &\sim \frac{\text{Insert Stirling's approximation}}{\text{Insert more Stirling's approximations}} \\ &\sim Cn^{-1/2} \left( \frac{a^a}{b^b(a-b)^{a-b}} \right)^n \\ &\sim Cn^{-1/2} K_{a,b}^{-n}. \end{align}\]

Conjecture. \(\displaystyle F_{a,b}(K_{a,b}\sin^2 \theta) = \frac{\text{Some expression}}{\cos \theta}\).

Note by Jake Lai
5 days, 4 hours ago

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