# Simplifying terms with Pochhammer symbol

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

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

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

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