Let $S_p=\sum_{k=0}^{\infty} {-p \choose k} (1+k)^{-p}$, how to show that $S_1=S_2=\ln 2$

Question

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.

Answer 1

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 $$S_2=\sum_{k=0}^{\infty} (-1)^k (k+1) \frac{x^k}{(k+1)^2}=\ln 2.$$

Answer 2

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 $$\sum_{k=0}^\infty\frac{(-1)^k}{k+1}.$$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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} $$ \left\{\begin{array}{l} \ds{S_{\color{red}{1}} = {\pars{-1}^{\ \color{red}{1} + 1} \over \Gamma\pars{\color{red}{1}}} \int_{0}^{1}\ln^{\color{red}{1} - 1}\pars{x} \pars{1 + x}^{-\color{red}{1}}\,\dd x = \int_{0}^{1}{\dd x \over 1 + x} = {\large\ln\pars{2}}} \\[5mm] \ds{S_{\color{red}{2}} = {\pars{-1}^{\ \color{red}{2} + 1} \over \Gamma\pars{\color{red}{2}}} \int_{0}^{1}\ln^{\color{red}{2} - 1}\pars{x} \pars{1 + x}^{-\color{red}{2}}\,\dd x \\[2mm] \ds{\phantom{S_{2}\,} = -\int_{0}^{1}{\ln\pars{x} \over \pars{1 + x}^{2}}\,\dd x = {\large\ln\pars{2}} }} \end{array}\right. $$