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 $1 \leq m < n$ 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.
 A: We can extend the sum to $j=0$ since the summand is $0$ for $j=0$. We have
$$
\begin{align}
\sum_{j=0}^m(-1)^j\binom mjj^l
&=\left.\sum_{j=0}^m\left(q\frac{\partial}{\partial q}\right)^lq^j\binom mj\right|_{q=-1}
\\
&=\left.\left(q\frac{\partial}{\partial q}\right)^l(1+q)^m\right|_{q=-1}
\\
&=(-1)^mm!\left\{{l\atop m}\right\}\;,
\end{align}
$$
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 $1/n^m$:
$$
\begin{align}
\sum_{j=1}^m (-1)^j\binom mj\log(1 - j/n)
&=
-\sum_{j=0}^m (-1)^j\binom mj\sum_{l=1}^\infty\frac1l\left(\frac jn\right)^l
\\
&=
(-1)^{m+1}m!\sum_{l=m}^\infty\left\{{l\atop m}\right\}\frac{n^{-l}}l\;.
\end{align}
$$
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.
A: Here is an asymptotic expression for the above sum; thanks for all the help. This took a little time to figure out.
$$
\begin{aligned}
\sum_{j=1}^{m} (-1)^{j} \binom mj {\log(1 - j/n)} &= \sum_{j=1}^m (-1)^j \binom mj (\log(n-j) - \log n) \\
&= \sum_{j=1}^m (-1)^j \binom mj \log(n-m + m-j) - \log n \sum_{j=1}^m (-1)^j \binom mj \\
& = \sum_{m-j = 0}^{m-1} \left((-1)^{m - (m-j)} {m \choose m-j} \log((n-m) + (m-j))\right) + \log n\\
& = \sum_{k = 0}^{m-1} \left((-1)^{m-k} \binom mk \log((n-m) + k)\right) + (-1)^m (-1)^m\log n \\
& = (-1)^m \sum_{k = 0}^{m} \left((-1)^{k} \binom mk \log((n-m) + k)\right)
\end{aligned}
$$
Letting $r=n-m$, we now estimate $\sum_{k=0}^{m} (-1)^{k} \binom mk \log(r+k)$.
$$
\begin{aligned}
\sum_{k=0}^{m} (-1)^{k} \binom mk \log(r+k) &= \sum_{k=0}^{m} (-1)^{k} \binom mk \int_1^{r+k} \frac{1}{x}\, dx \\
&= \sum_{k=0}^{m} (-1)^{k} \binom mk \int_0^1\,\, \sum_{j=1}^{r+k-1} \frac{1}{x+j}\, dx \\
&= \int_0^1 \,\, \sum_{j=1}^{r+m-1} \frac{1}{x+j} \sum_{k = \text{max} \{j+1-r,0\}}^m (-1)^k \binom mk \, dx \\
&= \int_0^1 \,\, \sum_{j=1}^{r+m-1} \frac{1}{x+j} \cdot (-1)^{j-r+1} {m-1\choose j-r}\, dx \\
&= \int_0^1 \,\, \sum_{j=r}^{r+m-1} \frac{1}{x+j} \cdot (-1)^{j-r+1} {m-1\choose j-r}\, dx \\
&= \int_0^1 \,\, \sum_{i=0}^{m-1} \frac{(-1)^{i+1} {m-1\choose i}}{x+r+i} \, dx, \qquad i = j-r \\
&= \int_0^1 \,\, \frac{-\Gamma(m+1) \Gamma(r+x)}{\Gamma(m+r+x+1)}\, dx
%&= -\Gamma(m+1) \int_0^1 \,\, \frac{\Gamma(n-m+x)}{\Gamma(n+x+1)}\, dx
\end{aligned}
$$
We have the following estimate for the ratio of Gamma functions: $\Gamma(N)/\Gamma(N+x) = N^{-x}(1 + O(1/N))$. Therefore, $\displaystyle{\Gamma(r+x)/\Gamma(r+x+m+1) = (r+x)^{-m-1} (1 + O(1/r))}$. Plugging this back into the previous equation and integrating, we get
\begin{equation}
\sum_{k=0}^{m} (-1)^{k} \binom mk \log(r+k) = \frac{-\Gamma(m)}{m} \left( \frac{1}{r^m} - \frac{1}{(r+1)^m} \right) ( 1 + O(1/r) )
\end{equation}
where $r=n-m$. Putting all this together we have:
\begin{equation}
\sum_{j=1}^{m} (-1)^{j} \binom mj \log(1 - j/n) = (-1)^{m+1} \frac{\Gamma(m)}{m} \left( \frac{1}{(n-m)^m} - \frac{1}{(n-m+1)^m} \right) ( 1 + O(1/(n-m)) )
\end{equation}
A: 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
$$ -n!\,\sum_{m=1}^{\infty}{\left(\frac{b}{a}\right)}^{m}\frac{1}{m} \left(\frac{1}{n!}\sum\limits_{k=1}^{n} (-1)^{m-k} {n \choose k} k^m \right) =  -n!\,\sum_{m=1}^{\infty}\left\{\begin{matrix} m \\ n \end{matrix}\right\} {\left(\frac{b}{a}\right)}^{m}\frac{1}{m} $$
$$ = -n!\,\sum_{m=1}^{\infty}\left\{\begin{matrix} m \\ n \end{matrix}\right\} {\left(\frac{b}{a}\right)}^{m}\frac{1}{m} \,,$$
$$ \Rightarrow \sum\limits_{k=1}^{n} (-1)^k {n \choose k} \ln(a+bk) = -\ln(a) -n!\,\sum_{m=1}^{\infty}\left\{\begin{matrix} m \\ n \end{matrix}\right\} {\left(\frac{b}{a}\right)}^{m}\frac{1}{m} \,. $$
