Let $S_p=\sum_{k=0}^{\infty} {-p \choose k} (1+k)^{-p}$, how to show that $S_1=S_2=\ln 2$ If $$S_p=\sum_{k=0}^{\infty} {-p \choose k} (1+k)^{-p}.$$
I need to prove that that $S_1=S_2=\ln 2.$ I have no idea about binomial coefficient with negative index. I woner about them, please help.
 A: Here
$$\binom{-p}k=\frac{(-p)(-p-1)(-p-2)\cdots(-p-k+1)}{k!}=(-1)^k\binom{p+k-1}k.$$
When $p=1$,
$$\binom{-p}k=(-1)^k\binom{k}k=(-1)^k$$
and your series
is $$\sum_{k=0}^\infty\frac{(-1)^k}{k+1}.$$
When $p=2$,
$$\binom{-p}k=(-1)^k\binom{k+1}k=(-1)^k(k+1)$$
and your series
is also $$\sum_{k=0}^\infty\frac{(-1)^k}{k+1}.$$
A: Note that  $${-n \choose k}=(-1)^k {n+k-1 \choose k}$$ $$ \implies {-1 \choose k}=(-1)^k, ~~~ {-2 \choose k}=(-1)^k {k+1 \choose k}=(-1)^k (k+1)$$
Then $$S_1=\sum_{k=0}^{\infty} \frac{(-1)^k}{k+1}=1-1/2+1/3-1/+....=\ln 2.$$
Again
$$S_2=\sum_{k=0}^{\infty} (-1)^k (k+1) \frac{x^k}{(k+1)^2}=\ln 2.$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
S_{p} & \equiv \bbox[5px,#ffd]{\sum_{k = 0}^{\infty}{-p \choose k} \pars{1 + k}^{-p}}
\\ & =
\sum_{k = 0}^{\infty}{-p \choose k}\
\overbrace{\bracks{{\pars{-1}^{\ p + 1} \over \Gamma\pars{p}}\int_{0}^{1}\ln^{p - 1}\pars{x}\,x^{k}
\dd x}}^{\ds{\pars{1 + k}^{-p}}}
\\[5mm] & =
{\pars{-1}^{\ p + 1} \over \Gamma\pars{p}}
\int_{0}^{1}\ln^{p - 1}\pars{x}\
\overbrace{\bracks{\sum_{k = 0}^{\infty}{-p \choose k}x^{k}}}^{\ds{\pars{1 + x}^{-p}}}\dd x
\\[5mm] & =
\bbx{{\pars{-1}^{\ p + 1} \over \Gamma\pars{p}}
\int_{0}^{1}\ln^{p - 1}\pars{x}\pars{1 + x}^{-p}\,\dd x} \\ &
\end{align}
$$
\left\{\begin{array}{l}
\ds{S_{\color{red}{1}} =
{\pars{-1}^{\ \color{red}{1} + 1} \over
\Gamma\pars{\color{red}{1}}}
\int_{0}^{1}\ln^{\color{red}{1} - 1}\pars{x}
\pars{1 + x}^{-\color{red}{1}}\,\dd x =
\int_{0}^{1}{\dd x \over 1 + x} = {\large\ln\pars{2}}}
\\[5mm]
\ds{S_{\color{red}{2}} = 
{\pars{-1}^{\ \color{red}{2} + 1} \over
\Gamma\pars{\color{red}{2}}}
\int_{0}^{1}\ln^{\color{red}{2} - 1}\pars{x}
\pars{1 + x}^{-\color{red}{2}}\,\dd x
\\[2mm] \ds{\phantom{S_{2}\,} =
-\int_{0}^{1}{\ln\pars{x} \over \pars{1 + x}^{2}}\,\dd x = {\large\ln\pars{2}} }}
\end{array}\right.
$$
