We have a fixed integer $k$. For two sets of non-negative numbers, $0 \leq a_i \leq M$, and $0 \leq b_i \leq M$, $i=0,1,\dotsc,k$, it is known that $$ a_0 + \frac{a_1}{n} + \frac{a_2}{n^2} + \dotsb + \frac{a_k}{n^k} \geq b_0 + \frac{b_1}{n} + \frac{b_2}{n^2} + \dotsb + \frac{b_k}{n^k} $$ for all integer $n \geq 2$.

What can we say about $a_0, a_1, \dotsc, a_k$ and $b_0, b_1, \dotsc, b_k$ that satisfy the inequality for any $n \geq 2$?

In other words, I am looking for the set of solutions $\{(a_0, a_1, \dotsc, a_k, b_0, b_1, \dotsc, b_k)\}$ of the infinite system of inequalities: $$ \begin{cases} a_0 + \frac{a_1}{2} + \frac{a_2}{2^2} + \dotsb + \frac{a_k}{2^k} \geq b_0 + \frac{b_1}{2} + \frac{b_2}{2^2} + \dotsb + \frac{b_k}{2^k},\\ a_0 + \frac{a_1}{3} + \frac{a_2}{3^2} + \dotsb + \frac{a_k}{3^k} \geq b_0 + \frac{b_1}{3} + \frac{b_2}{3^2} + \dotsb + \frac{b_k}{3^k},\\ a_0 + \frac{a_1}{4} + \frac{a_2}{4^2} + \dotsb + \frac{a_k}{4^k} \geq b_0 + \frac{b_1}{4} + \frac{b_2}{4^2} + \dotsb + \frac{b_k}{4^k},\\ \dotsc \end{cases} $$ with condition $0 \leq a_i, b_i \leq M$, $i=0,1,\dotsc,k$.

As the first step, I have the following idea. Take limit $n \to \infty$, which gives the necessary $$ a_0 \geq b_0.$$ But, obviously, this is not sufficient.

UPD: If we multiply both parts of the inequality by $n^k$ and denote $P(n) = (a_0-b_0) n^k + (a_1-b_1) n^{k-1} + \dotsb + (a_{k-1} - b_{k-1})n + (a_k - b_k)$, then then problem can be re-phrased as to find all the polynomials with coefficients in the range $[-M,M]$ so that $P(n) \geq 0$ for all $n \geq 2$.

UPD2: Perhaps, a stronger requirement can make the problem easier? What if require that $P(x) \geq 0$ for all real values $x \geq 2$?..

  • 2
    $\begingroup$ Do you have an idea of what kind of results you’re hoping for? I haven’t thought about it for too long, but beyond $a_0\geq b_0$, I’m not sure you are going to get much. $\endgroup$ – Clayton Feb 6 at 1:45
  • $\begingroup$ Not sure what you expect. Take the case of non-negative polynomials in a single variable. We can can conclude on the nonnegativity of the constant term, and perhaps some results like Descartes but not really anything more definitive. $\endgroup$ – Macavity Feb 6 at 7:05
  • $\begingroup$ I am also not really sure what result I hope for :) The more facts/information about the $a_i$, $b_i$ I can get, the better. But it seems the public opinion is that one can't say much about it... $\endgroup$ – Yauhen Yakimenka Feb 6 at 9:36

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