TITLE: Analytical proof for the convergence of a sequence
QUESTION [8 upvotes]: Consider the following sequence
$\Xi_N=N\sum\limits_{i=0}^{N-1} {N-1 \choose i} (-1)^{(i+1)} \log\left(i+1\right)$.
I numerically compute the asymptotic behavior of sequence and it turns out that the sequence approaches to a non-zero value as N goes to infinity. Now, I want to analytically prove that this sequence converges to a non-zero value as N goes to infinity.
Also, it can be proved that the sequence has another form as follows
$\Xi_N=\sum\limits_{i=1}^{N} {N \choose i} (-1)^{(i)} i \log\left(i\right)$.
Moreover, Using
$\int_{0}^{1} \sum_{m=1}^{i} \frac{1}{x+m} dx=\log(i+1)$
Then
$\Xi_N=N\sum_{m=1}^{N-1}{N-1 \choose m-1} (-1)^{m-1}\int_{0}^{1} \frac{1}{x+m} dx $
Could you give me some advice?
Thanks

REPLY [3 votes]: By Frullani's theorem and the binomial theorem:
$$\Xi_{N+1} = \int_{0}^{+\infty}\sum_{k=0}^{N}\binom{N}{k}(-1)^{k+1}\left(e^{-x}-e^{-(k+1)x}\right)\frac{dx}{x}=\int_{0}^{+\infty}\frac{(1-e^{-x})^N}{x e^{x}}\,dx$$
hence your limit is simply $\color{red}{\large 0}$ by the dominated convergence theorem, since
$$ f_N(x) = (1-e^{-x})^N $$
is bounded between $0$ and $1$ and behaves like $x^N$ in a right neighbourhood of the origin.