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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.

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    In your sum, the term with $j=m$ is going to give $\log0$ if $m=n$. Maybe this is why that other formula doesn't work for you --- your sum inherently involves an undefined term.2012-10-12
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    Sorry, I should have said that $m. The problem still persists, though.2012-10-12
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    @user44441: There's an edit link under the question. Please fix the question itself; people shouldn't have to read through the comments to understand the question. Note that the result in the other question is asymptotic for large $m$, so Gerry's point is still relevant, since the asymptotic expansion doesn't know whether you're excluding $m=n$.2012-10-12
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    You can get binomial coefficients using `\binom mj` or `{m\choose j}`.2012-10-12
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    Ok, thanks. My first post, so please excuse the mistakes.2012-10-12
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    Similar problem with identical title now posted to MO: http://mathoverflow.net/questions/109975/alternating-sum-of-binomial-coefficients-times-logarithm2012-10-18

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