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.
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) $.
On of methods for finding an estimating or finding bound for a finite sum is using the following formula
$\sum_{n=a}^b f(n) \sim \int_a^b f(x)\,dx + \frac{f(a) + f(b)}{2} + \sum_{k=1}^\infty \,\frac{B_{2k}}{(2k)!}\left(f^{(2k - 1)'}(b) - f^{(2k - 1)'}(a)\right)$ 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.
Moreover, the sharp bounds of Bernoulli numbers has been computed ,(see here ) So you can also try to find a bound for your sum.
