# Inequality with two sums

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 $$x \geq 2$$?..

• 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. – Clayton Feb 6 at 1:45
• 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. – Macavity Feb 6 at 7:05
• 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... – Yauhen Yakimenka Feb 6 at 9:36