Sum involving binomial coefficients I have the following sum
$\sum_{j=1}^K {K \choose j} (-1)^{j+1}/j$. Now I can write this as the integral $\int_{-1}^0 \frac{(1+x)^K - 1}{x} dx$. However, I wonder whether there is a closed form expression for that integral? Thanks.
 A: The closed form of the integral is
$$
\int_{-1}^{0}\frac{(1+x)^k - 1}{x} =   \frac{s(k+1,2)}{k!}
$$
where $s(k+1,2)$ denote the Stirling number of the first kind.
A: $\newcommand{\+}{^{\dagger}}%
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
 \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}%
 \newcommand{\dd}{{\rm d}}%
 \newcommand{\down}{\downarrow}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\fermi}{\,{\rm f}}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\half}{{1 \over 2}}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\ol}[1]{\overline{#1}}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
 \newcommand{\sech}{\,{\rm sech}}%
 \newcommand{\sgn}{\,{\rm sgn}}%
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
\begin{align}
&\color{#0000ff}{\large\int_{-1}^{0}{\pars{1 + x}^{K} - 1 \over x}\,\dd x}
=\int_{0}^{1}{1 - x^{K} \over 1 - x}\,\dd x
=-\int_{x = 0}^{x = 1}\pars{1 - x^{K}}\,\dd\ln\pars{1 - x}
\\[3mm]&=\int_{0}^{1}\ln\pars{1 - x}\pars{-Kx^{K - 1}}\,\dd x
=-K\lim_{\mu \to 0}\partiald{}{\mu}\int_{0}^{1}\pars{1 - x}^{\mu}x^{K - 1}\,\dd x
\\[3mm]&=
-K\lim_{\mu \to 0}\partiald{{\rm B}\pars{1 + \mu,K}}{\mu}
=
-K\lim_{\mu \to 0}\partiald{}{\mu}
\bracks{\Gamma\pars{1 + \mu}\Gamma\pars{K} \over \Gamma\pars{1 + \mu + K}}
\\[3mm]&=-K\,\Gamma\pars{K}\lim_{\mu \to 0}\bracks{%
{\Gamma\pars{1 + \mu}\Psi\pars{1 + \mu} \over \Gamma\pars{1 + \mu + K}}
-
{\Gamma\pars{1 + \mu}\Psi\pars{1 + \mu + K} \over \Gamma\pars{1 + \mu + K}}}
\\[3mm]&=
-\Gamma\pars{1 + K}\bracks{%
{-\gamma \over \Gamma\pars{1 + K}} - {\Psi\pars{1 + K} \over \Gamma\pars{1 + K}}}
=
\color{#0000ff}{\large\gamma + \Psi\pars{1 + K}}
\end{align}

$$
\begin{array}{rcl}
{\rm B}\pars{\alpha,\beta}=\int_{0}^{1}t^{\alpha - 1}\pars{1 - t}^{\beta - 1}\,\dd t:
&& Beta\ \mbox{function}
\\[2mm]
\Gamma\pars{z}:&& Gamma\ \mbox{function}
\\[2mm]
\Psi\pars{z} \equiv \totald{\ln\pars{\Gamma\pars{z}}}{z}:&& Digamma\ \mbox{function}
\\[2mm]\gamma \approx 0.577216:&& Euler-Mascheroni\ constant 
\end{array}
$$

