ScalingStacks

Proof. [024G]

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Proof.

Integrating

y−2(1−yn)−1/2=∑k=0∞(12)kk!yn​k−2,y^{-2}(1-y^{n})^{-1/2}=\sum_{k=0}^{\infty}\frac{(\frac{1}{2})_{k}}{k!}y^{nk-2},

and utilizing the initial conditions to fix the leading coefficients,

g⁡(y)=∑k=0∞(12)kk!​yn​kn​k−1.g(y)=\sum_{k=0}^{\infty}\frac{(\frac{1}{2})_{k}}{k!}\frac{y^{nk}}{nk-1}.

The formula for g⁡(y)g(y) follows from the observation −1n​k−1=(−1n)k(n−1n)k.\frac{-1}{nk-1}=\frac{(-\frac{1}{n})_{k}}{(\frac{n-1}{n})_{k}}. The evaluation of g⁡(1)g(1) appeals to Gauss’ hypergeometric theorem [1, Section 1.3]:

Lemma 5.4.

For a,b,ca,b,c with Re⁡(c−a−b)>0{\rm Re}(c-a-b)>0 we have

F12​[a,b;c;1]=Γ⁡(c)​Γ​(c−a−b)Γ⁡(c−a)​Γ​(c−b).\,{}_{2}F_{1}[a,b;c;1]=\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}.

∎

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