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Wadim Zudilin
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Positivity of sequences via generating series

There are different ways of showing that a given sequence $a_0,a_1,a_2,\dots$ of integers, say, is nonnegative. For example, one can show that $a_n$ count something, or express $a_n$ as a (multiple) sum of obviously positive numbers. Another recipe is manipulating with the corresponding generating series $A(x)=\sum_{n=0}^\infty a_nx^n$ and showing that $A(x)\ge0$ (this is the notation for a series which has all coefficients nonnegative, and this extends to formal power series in as many variables as needed).

An example of criterion in this direction is $$ (*) \qquad A(xy)\ge0 \iff \frac1{(1-x)(1-y)}A\biggl(\frac{xy}{(1-x)(1-y)}\biggr)\ge0 $$ (the multiple $1/(1-x)(1-y)$ is introduced for cosmetic purposes only and, of course, both $A(x)\ge0$ and $A(xy)\ge0$ are by definition equivalent to the nonnegativity of the sequence $a_n$). The latter can be verified by comparing the corresponding coefficients in the power series expansion $$ \frac1{(1-x)(1-y)}A\biggl(\frac{xy}{(1-x)(1-y)}\biggr) =\sum_{n=0}^\infty a_n\sum_{k,m=0}^\infty\binom{n+k}n\binom{n+m}nx^{n+k}y^{n+m}. $$

On the other hand, the one-dimensional version of $( * )$, $$ A(x)\ge0 \iff \frac1{1-x}A\biggl(\frac{x}{1-x}\biggr)\ge0, $$ is simply false.

My question is whether it is possible to find two nontrivial rational functions $p(x)\in\mathbb Q[[x]]$ and $r(x)\in x\mathbb Q[[x]]$ in one variable $x$ such that $$ A(x)\ge0 \iff p(x)A\bigl(r(x)\bigr)\ge0. $$ Although I am not supposed to put several problems in one question, I would also ask about a more direct proof of $( * )$ and about general ways of constructing such $p$ and $r$ in more than one variable.

Motivation. Basically I am interested in proving nonnegativity of certain $q$-series sequences $a_0(q),a_1(q),a_2(q),\dots$ by manipulating with the corresponding generating series $A_q(x)=\sum_{n=0}^\infty a_n(q)x^n$. Some of them can be "guessed" from non-$q$-versions, for example there is a neat $q$-analogue of the criterion $( * )$.

Wadim Zudilin
  • 13.5k
  • 4
  • 58
  • 102