A binomial sum

It's not too hard to show that (12n)=(14)n(2nn)\displaystyle \binom{-\frac{1}{2}}{n} = \left( -\dfrac{1}{4} \right)^n \binom{2n}{n} and hence, 114x=n=0(12n)(4x)n=n=0(2nn)xn\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 n=0(3nn)xn=2cos(13sin1(332x))427x\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=427sin2θx = \frac{4}{27} \sin^2 \theta, we get

n=0(3nn)xn=cosθ3cosθ.\sum_{n=0}^\infty \binom{3n}{n}x^n = \boxed{\dfrac{\cos \frac{\theta}{3}}{\cos \theta}}.

Sadly, WA revealed nothing about n=0(4nn)xn\displaystyle \sum_{n=0}^\infty \binom{4n}{n}x^n. A cool expression for n=0(4n2n)xn\displaystyle \sum_{n=0}^\infty \binom{4n}{2n}x^n is easily obtained with the substitution x=116sin2θx = \frac{1}{16}\sin^2 \theta, however:

n=0(4n2n)xn=12[n=0(2nn)(x)n+n=0(2nn)(x)n]=12[114x+11+4x]=12[14x+1+4x116x]=12[1sinθ+1+sinθcosθ]=12[2+2cosθcosθ]=cosθ2cosθ.\begin{aligned} \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{aligned}

Let's define the sum Fa,b(x)=n=0(anbn)xn\displaystyle F_{a,b}(x) = \sum_{n=0}^\infty \binom{an}{bn}x^n. For convenience let's also define Ka,b=bb(ab)abaaK_{a,b} = \dfrac{b^b(a-b)^{a-b}}{a^a}. So K2,1=14K_{2,1} = \dfrac{1}{4}, K3,1=427K_{3,1} = \dfrac{4}{27}, and K4,2=116K_{4,2} = \dfrac{1}{16}. It seems that the factor associated with the sin2θ\sin^2 \theta comes from the series' radius of convergence, ie

(anbn)=(an)!(bn)!((ab)n)!Insert Stirling’s approximationInsert more Stirling’s approximationsCn1/2(aabb(ab)ab)nCn1/2Ka,bn.\begin{aligned} \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{aligned}

Conjecture. Fa,b(Ka,bsin2θ)=Some expressioncosθ\displaystyle F_{a,b}(K_{a,b}\sin^2 \theta) = \frac{\text{Some expression}}{\cos \theta}.

Note by Jake Lai
1 year, 2 months ago

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