# Alternating sum of binomial coefficients times logarithm

Trying to find a closed form expression for the following sum, or an asymptotic expression in terms of well known functions (like the Gamma function, for instance).

Let $m,n$ be positive integers such that $2 \leq m < n$. Estimate the sum $$\sum_{j=1}^m (-1)^j \binom m j \frac{\log(n-j)}{j}$$

where $\log$ stands for the natural logarithm. Thanks for any help.

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What you've written looks like an assignment question, though I could be wrong. – David Roberts Oct 18 '12 at 6:04
Similar problem posted to m.se with identical title a few days ago: math.stackexchange.com/questions/211449/… – Gerry Myerson Oct 18 '12 at 6:34
I posted a similar problem on stackexchange, but that was a different problem and easier in that I have an answer in that case. The problem on stackexchange posted a few days ago was this: find a closed form expression for: $\sum_{j=1}^m (-1)^j \binom m j \log(1 - j/n)$. Notice the missing $1/j$ term in the sum. The answer is an asmptotic expression involving the $\Gamma$ function. – krishnan.shankar Oct 18 '12 at 14:13

As $n \to \infty$ for fixed $m$, $$\log(n-j) = \log(n) + \log(1-j/n) = \log(n) - j/n + O(1/n^2)$$ Since $\sum_{j=1}^m (-1)^j {m \choose j} \frac{1}{j} = -\Psi(m+1)-\gamma$ and $\sum_{j=1}^m (-1)^j {m \choose j} = -1$, your sum is $-(\Psi(m+1)+\gamma) \log(n) + \dfrac{1}{n} + O\left( \dfrac{1}{n^2} \right)$.
Yes, but fixed $m$ is the easiest case and Krishnan didn't specify that $m$ is fixed. It is a lot more challenging if $m,n$ both go to infinity. Try $n=m+1$. – Brendan McKay Oct 18 '12 at 8:55
$\sum_{n=a}^b f(n) \sim \int_a^b f(x)\,dx + \frac{f(a) + f(b)}{2} + \sum_{k=1}^\infty \,\frac{B_{2k}}{(2k)!}\left(f^{(2k - 1)'}(b) - f^{(2k - 1)'}(a)\right)$ and by taking $f(j)= (-1)^j\binom {m}{j}\frac{log(n-j)}{j}$ you can estimate right hand side of this formula, which are faster than of left hand side. In fact by computing some first terms of this infinite sum, we can obtain a good estimation for left hand side. Also here, $B_k$ are Bernoulli numbers.