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\newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$\begin{align}
&\color{#88f}{\large\sum_{k = 0}^{D}\pars{-1}^{k}{n \choose k}}
=\sum_{k = 0}^{D}\pars{-1}^{k}\ \overbrace{\oint_{0\ <\ \verts{z}\ =\ a\ <\ 1}
{\pars{1 + z}^{n} \over z^{k + 1}}\,{\dd z \over 2\pi\ic}}
^{\ds{=\ {n \choose k}}}
\\[5mm]& \
=\oint_{0\ <\ \verts{z}\ =\ a\ <\ 1}{\pars{1 + z}^{n} \over z}
\sum_{k = 0}^{D}\pars{-\,{1 \over z}}^{k}\,{\dd z \over 2\pi\ic}
\\[5mm] = &\ \oint_{0\ <\ \verts{z}\ =\ a\ <\ 1}{\pars{1 + z}^{n} \over z}
{\pars{-1/z}^{D} + z \over 1 + z}\,{\dd z \over 2\pi\ic}
\\[5mm]&=\pars{-1}^{D}\ \underbrace{\oint_{\verts{z}\ =\ a\ <\ 1}
{\pars{1 + z}^{n - 1} \over z^{D + 1}}\,{\dd z \over 2\pi\ic}}
_{\ds{=\ {n - 1 \choose D}}}\ +\
\underbrace{\oint_{0\ <\ \verts{z}\ =\ a\ <\ 1}\pars{1 + z}^{n - 1}
\,{\dd z \over 2\pi\ic}}_{\ds{=\ 0}}
\\[5mm]&\ =
\bbox[10px,border:1px groove navy]{\pars{-1}^{D}{n - 1 \choose D}}
\\ &
\end{align}
$$S_k = \sum_{n\geq k}\frac{2k(2n-1)!}{(n-k)!(n+k)!}\cdot\frac{1}{4^n}=\sum_{n\geq k}\frac{(2n-1)!\left[(n+k)-(n-k)\right]}{(n-k)!(n+k)!}\cdot\frac{1}{4^n} $$
equals
$$ \sum_{n\geq k}\left[\frac{(2n-1)!}{(n-k)!(n+k-1)!}-\frac{(2n-1)!}{(n-1-k)!(n+k)!}\right]\frac{1}{4^n}$$
or
$$ \frac{1}{ 4^k}\sum_{m\geq 0}\left[\frac{(2m+2k-1)!}{m!(m+2k-1)!}-\frac{(2m+2k-1)!}{(m-1)!(m+2k)!}\right]\frac{1}{4^{m}}$$
or
$$ \frac{1}{4^k}\sum_{m\geq 0}\frac{1}{4^m}[x^m]\left[(1+x)^{2m+2k}\frac{1-x}{1+x}\right]$$
or
$$ \frac{1}{4^k}\sum_{m\geq 0}[x^m]\left[\left(\frac{1+x}{2}\right)^{2m}(1-x)(1+x)^{2k-1}\right]$$
or
$$ \frac{1}{4^k}\sum_{m\geq 0}\operatorname*{Res}_{x=0}\left[\left(\frac{1+x^2}{2x}\right)^{2m}\frac{(1-x^2)(1+x^2)^{2k-1}}{x}\right]$$
which finally boils down to something simple to compute (i.e. the opposite of the residue at infinity of a rational function), by switching $\text{Res}$ and $\sum_{m\geq 0}$. I am pretty sure I messed up some transformations but the approach in itself should be fine.
Best Answer
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align} &\bbox[10px,#ffd]{\sum_{\ell = r}^{\infty}\pi_{\ell r}\pars{p}} = \sum_{\ell = r}^{\infty}{\ell \choose r}p^{r}\pars{1 - p}^{\ell - r} = p^{r}\sum_{\ell = r}^{\infty}\overbrace{\ell \choose \ell - r} ^{\ds{=\ {\ell \choose r}}}\ \pars{1 - p}^{\ell - r} \\[5mm] = &\ p^{r}\sum_{\ell = r}^{\infty} \overbrace{{-r - 1 \choose \ell - r}\pars{-1}^{\ell - r}} ^{\ds{=\ {\ell \choose \ell - r}}}\ \pars{1 - p}^{\ell - r} \\[5mm] \stackrel{\ell\ -\ r\ \mapsto\ \ell}{=}\,\,\,& p^{r}\sum_{\ell = 0}^{\infty} {-r - 1 \choose \ell}\pars{-1}^{\ell}\pars{1 - p}^{\ell} \\[5mm] = &\ p^{r}\sum_{\ell = 0}^{\infty}{-r - 1 \choose \ell}\pars{p - 1}^{\ell} = p^{r}\,\bracks{1 + \pars{p - 1}}^{-r - 1} = \bbx{1 \over p} \end{align}