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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

MMH
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  • How can that be a series while $N<\infty$ ? – Theodoros Mpalis Mar 29 '16 at 08:32
  • @TheodorosMpalis Yes, you're right. Let say sequence. – MMH Mar 29 '16 at 08:33
  • Are you sure that this is the expression to work ? – Claude Leibovici Mar 29 '16 at 08:36
  • @ClaudeLeibovici, Yes, 100%, why? – MMH Mar 29 '16 at 08:38
  • That's kinda complex... So many products in one sum – Theodoros Mpalis Mar 29 '16 at 08:39
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    You may pick $\sup{\Xi_{N}}$ for even values of $i$ and for odd ones. – Theodoros Mpalis Mar 29 '16 at 08:44
  • That's a good one task though :D – Theodoros Mpalis Mar 29 '16 at 08:46
  • Are you sure of your numerics? I tried some computations with Mathematica: the sequence seems to converge to somewhere close to $-0.3$ up to $n=45$, afterwards values become way more chaotic. This could be because we reach some sort of computational limit, but still for $n=50$ or $n=60$ the numbers do not seem wild enough to justify this. On the other hand such a sudden loss of regularity is also suspicious. Thus: are you sure of your numerics? – Giovanni De Gaetano Mar 29 '16 at 09:11
  • @GiovanniDeGaetano, yes, I observed that phenomena, and I considered that as the computational limit of "choose". For instance, for N<42, my numerical computation via matlab shows that the sequence is about 0.2513. – MMH Mar 29 '16 at 09:18
  • @GiovanniDeGaetano. This is why I asked the question in a comment. I stay very skeptical. – Claude Leibovici Mar 29 '16 at 10:55
  • Is clear that the modulus of the sequence goes to infinity – Charly Mar 29 '16 at 18:40
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    The post Proof ∑k=1n(nk)(−1)klogk=loglogn+γ+γlogn+O(1log2n) might help you – Charly Mar 29 '16 at 19:42
  • If I'm not mistaken the reference provided by @Charly answers the question in the negative. I think it would be nice that one of you two, Charly and Mahdi, posts it as an answer, so that it can be marked as accepted and removed from the list of unanswered ones. – Giovanni De Gaetano Apr 01 '16 at 10:45
  • @GiovanniDeGaetano Not exactly, the technique used in the end of Part 1 can not be used in this question. – MMH Apr 02 '16 at 07:29
  • @Charly, Thanks for your reference. I have added some lines to the question, but, unfortunately, the technique used in the end of Part 1 can not be longer used in this question. – MMH Apr 02 '16 at 07:31
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    If we set $${\Xi N}(x) = \sum\limits{i = 1}^N {{x^{i - 1}}\left( \begin{array}{l} N\ i \end{array} \right){{\left( { - 1} \right)}^i}i\log i} $$ and $$\Delta (x) = \sum\limits_{i = 1}^N {{x^i}\left( \begin{array}{l} N\ i \end{array} \right){{\left( { - 1} \right)}^i}\log i} $$ – Charly Apr 05 '16 at 16:36
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    then $$\int {{\Xi N}(x)} = \sum\limits{i = 1}^N {\int {\left( {{x^{i - 1}}\left( \begin{array}{l} N\ i \end{array} \right){{\left( { - 1} \right)}^i}i\log i} \right)dx} } = \sum\limits_{i = 1}^N {{x^i}\left( \begin{array}{l} N\ i \end{array} \right){{\left( { - 1} \right)}^i}\log idx} = \Delta (x)$$. I guess that we could use the technique used in the post to calculate $\Delta (x)$ and then derive. Note that we are looking for the an estimate of the value of ${\Xi _N}(1)$. As soon as I get home I'll give it a shoot! – Charly Apr 05 '16 at 16:41
  • @Charly: Hi, I have really got stuck into this problem. Have you made any progress? – MMH May 20 '16 at 10:38
  • Sorry, I have completely forgotten about this problem. – Charly May 25 '16 at 14:18

2 Answers2

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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.

Jack D'Aurizio
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  • Using Frullani's theorem is brilliant! How did you come up with this nice idea? btw THANKS. – MMH Sep 17 '16 at 20:48
  • @Mahdi: it is simply one of my favourites. To have an integral representation for logarithms is pretty useful, just like here :D – Jack D'Aurizio Sep 17 '16 at 20:50
  • Hi Jack, I think you miss a term from $\Xi_{N+1}$. Based on the integral representation, we have $\Xi_{N+1}=\color{red}{(N+1)}\int_{0}^{+\infty}\frac{(1-e^{-x})^N}{x e^{x}},dx$. In this case, what is the appropriate dominating function? – MMH Apr 20 '17 at 21:21
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Jack's answer above seems to have missed one term, the correct form should be: $$\Xi _{N+1}=\left( N+1 \right) \int_0^{\infty}{\frac{\left( 1-e^{-x} \right) ^N}{xe^x}dx}$$ Just as you say.
Notice that $$ for\,\,\forall n\geqslant 1, n\left( 1-e^{-x} \right) ^{n-1}\leqslant e^x\,\,, x\in \left[ 0,\infty \right) $$ then we have:
$$ \forall N\geqslant 0, \frac{\left( N+1 \right) \left( 1-e^{-x} \right) ^N}{xe^x}\leqslant \frac{1}{x}\,\,,x\in \left( 0,\infty \right) $$
then you can use the dominated convergence theorem.

origamist
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