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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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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) $. |
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