# multiplying a hypergeometric series

We are able to calculate the value of the sum

$$\sum_{k=0}^\infty \frac{(a_1)_k(a_2)_k\dots(a_p)_k}{(b_1)_k(b_2)_k\dots(b_{p-1})_k}\cdot\frac{x^k}{k!}$$, which equals the generalised hypergeometric function

$${}_pF_q\left[\begin{matrix} a_1,a_2,\dots ,a_p \\ b_1,b_2,\dots ,b_{p-1} \end{matrix}\quad;x\right]$$.

Is it possible to calculate (or bound) $$\sum_{k=0}^\infty \frac{(a_1)_k(a_2)_k\dots(a_p)_k}{(b_1)_k(b_2)_k\dots(b_{p-1})_k}\cdot\frac{x^k}{k!}\cdot k$$ ?

Thank you very much for any suggestions!

$$S=\sum_{k\geq0}\frac{(a_1)_k\cdots(a_{p})_k}{(b_1)_k\cdots(b_{q})_k}\frac{kx^k}{k!}$$ $$S=\frac{(a_1)_0\cdots(a_{p})_0}{(b_1)_0\cdots(b_{q})_0}\frac{0x^0}{0!}+\sum_{k\geq1}\frac{(a_1)_k\cdots(a_{p})_k}{(b_1)_k\cdots(b_{q})_k}\frac{kx^k}{k!}$$ $$S=\sum_{k\geq1}\frac{(a_1)_k\cdots(a_{p})_k}{(b_1)_k\cdots(b_{q})_k}\frac{kx^k}{k!}$$ $$S=x\sum_{k\geq1}\frac{(a_1)_k\cdots(a_{p})_k}{(b_1)_k\cdots(b_{q})_k}\frac{x^{k-1}}{(k-1)!}$$ $$S=x\sum_{k\geq0}\frac{(a_1)_{k+1}\cdots(a_{p})_{k+1}}{(b_1)_{k+1}\cdots(b_{q})_{k+1}}\frac{x^{k}}{k!}$$ Note that $$(a)_{n+1}=a\cdot(a+1)_n$$. Thus, $$S=x\sum_{k\geq0}\frac{a_1\cdots a_{p}}{b_1\cdots b_{q}}\frac{(a_1+1)_k\cdots(a_{p}+1)_k}{(b_1+1)_k\cdots(b_{q}+1)_k}\frac{x^{k}}{k!}$$ $$S=x\frac{a_1\cdots a_{p}}{b_1\cdots b_{q}}\sum_{k\geq0}\frac{(a_1+1)_k\cdots(a_{p}+1)_k}{(b_1+1)_k\cdots(b_{q}+1)_k}\frac{x^{k}}{k!}$$ $$S=x\frac{a_1\cdots a_{p}}{b_1\cdots b_{q}}\;_pF_q\big(a_1+1,\dots,a_{p}+1;b_1+1,\dots,b_q+1;x\big)$$ And there you go.