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Is the simplification from line 1 to 2 correct? BTW, $(a)_{k}$ is the usual Pochhammer symbol.

$\rho=\frac{b_0}{1-q}\frac{x^a}{a}\sum_{k=0}^\infty(-1)^k\frac{(b-1)_k}{k!}\frac{(a)_k}{(a+1)_k} x^k$

$=\frac{b_0}{1-q}\frac{x^a}{a}\sum_{k=0}^\infty\frac{(1-b)_k(a)_k}{(a+1)_k}\frac{x^k}{k!}$

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  • $\begingroup$ It seems correct, since $(-1)^k (c)_k = (-c)_k$. $\endgroup$ – Rigel Jan 26 at 8:17

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