Alternating sum of binomial coefficients times logarithm Would like to compute the following sum: $\sum_{j=1}^m (-1)^j \begin{pmatrix} m\\ j\end{pmatrix} \log(1 - j/n)$, where $1 \leq m &lt; n$ are fixed integers. A similar sum has already been computed on Stackexchange at Limit of alternating sum with binomial coefficient, where the logarithmic term is $\log(a+bk)$; the answer is in terms of $r = a/b$ and involves the gamma function, $\Gamma(r)$. When I apply this to my situation, I get (as part of the answer) $a=1, b=-1/n$ and therefore, I get the term $\Gamma(r) = \Gamma(-n)$ which is not defined, since the gamma function has a pole there. Please help with the correct expression for the sum in my case.
We can extend the sum to $j=0$ since the summand is $0$ for $j=0$. We have

$$
\begin{align}
\sum_{j=0}^m(-1)^j\binom mjj^l
&amp;=\left.\sum_{j=0}^m\left(q\frac{\partial}{\partial q}\right)^lq^j\binom mj\right|_{q=-1}
\\
&amp;=\left.\left(q\frac{\partial}{\partial q}\right)^l(1+q)^m\right|_{q=-1}
\\
&amp;=(-1)^mm!\left\{{l\atop m}\right\}\;,
\end{align}
$$

where $\left\{{l\atop m}\right\}$ is a Stirling number of the second kind.

This vanishes for $0\le l\lt m$, so expanding the logarithm yields a power series in $1/n$ with leading term $1/n^m$:

$$
\begin{align}
\sum_{j=1}^m (-1)^j\binom mj\log(1 - j/n)
&amp;=
-\sum_{j=0}^m (-1)^j\binom mj\sum_{l=1}^\infty\frac1l\left(\frac jn\right)^l
\\
&amp;=
(-1)^{m+1}m!\sum_{l=m}^\infty\left\{{l\atop m}\right\}\frac{n^{-l}}l\;.
\end{align}
$$

Formally, we can also write

$$
\sum_{j=0}^m(-1)^j\binom mj\log(1 - j/n)=\log\left(1-\frac qn\frac{\partial}{\partial q}\right)(1+q)^m\;,
$$

but I'm not sure whether that's of any use.
Yesterday, I posted my answer by mistake on the other post while I was supposed to post it here. Anyways, your case is a special case of the case, we are going to deal with.

Recalling the identity of Stirling numbers of the second kind in terms of the binomial, 

$$\left\{\begin{matrix} n \\ m \end{matrix}\right\} = \frac{1}{m!}\sum_{j=0}^m (-1)^{m-j}{m \choose j } j^n\,. $$

Writing the sum as 
$$ \sum\limits_{k=1}^{n} (-1)^k {n \choose k} \ln(a) + \sum\limits_{k=1}^{n} (-1)^k {n \choose k} \ln(1+\frac{b k}{a})$$

$$ -\ln(a) + \sum\limits_{k=1}^{n} (-1)^k {n \choose k} \ln(1+\frac{b k}{a}) \,.$$

Expanding the $\ln(1+\frac{b k}{a})$ in terms of its Taylor series and interchanging the order of summation gives
$$ -n!\,\sum_{m=1}^{\infty}{\left(\frac{b}{a}\right)}^{m}\frac{1}{m} \left(\frac{1}{n!}\sum\limits_{k=1}^{n} (-1)^{m-k} {n \choose k} k^m \right) =  -n!\,\sum_{m=1}^{\infty}\left\{\begin{matrix} m \\ n \end{matrix}\right\} {\left(\frac{b}{a}\right)}^{m}\frac{1}{m} $$

$$ = -n!\,\sum_{m=1}^{\infty}\left\{\begin{matrix} m \\ n \end{matrix}\right\} {\left(\frac{b}{a}\right)}^{m}\frac{1}{m} \,,$$

$$ \Rightarrow \sum\limits_{k=1}^{n} (-1)^k {n \choose k} \ln(a+bk) = -\ln(a) -n!\,\sum_{m=1}^{\infty}\left\{\begin{matrix} m \\ n \end{matrix}\right\} {\left(\frac{b}{a}\right)}^{m}\frac{1}{m} \,. $$