How to find a closed form expression for the following summation? How to find a closed form expression for the following summation?
$$ \sum_{m\geq 0} \frac{m r^m \Gamma(m+c)}{\Gamma(m+1)} $$
 A: Since $\Gamma(m+c)=\int_{0}^{+\infty}t^{m+c-1}e^{-t}\,dt$, for any $r<1$ the given series can be written as
$$ \int_{0}^{+\infty}\sum_{m\geq 0}\frac{m r^m t^{m+c-1}}{m!} e^{-t}\,dt =\int_{0}^{+\infty} r t^c e^{(r-1)t}\,dt=\color{red}{\frac{r\,\Gamma(c+1)}{(1-r)^{c+1}}}.$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\sum_{m \geq 0}{mr^{m}\Gamma\pars{m + c} \over \Gamma\pars{m + 1}} & =
\pars{c - 1}!\,r\,\totald{}{r}
\sum_{m \geq 0}{\pars{m + c - 1}! \over m!\pars{c - 1}!}\,r^{m} =
\pars{c - 1}!\,r\,\totald{}{r}
\sum_{m \geq 0}{m + c - 1 \choose m}\,r^{m}
\\[5mm] & =
\pars{c - 1}!\,r\,\totald{}{r}
\sum_{m \geq 0}{-c \choose m}\pars{-1}^{m}\,r^{m} =
\pars{c - 1}!\,r\,\totald{}{r}\pars{1 - r}^{-c}
\\[5mm] & =
\pars{c - 1}!\,r\,\bracks{-c\pars{1 - r}^{-c - 1}\pars{-1}} =
\bbx{c!\,{r \over \pars{1 - r}^{c + 1}}}
\end{align}
