Lemma 1

The element $1 - T$ of $\mathbf{Z}((T))$ is invertible. For every integer $r > 0$ one has

In fact, $1 - T$ is invertible in the ring $\mathbf{Z}[[T]]$, with inverse $\sum_{m \geq 0} T^m$; one therefore has

and the announced formula follows from E, III, p. 44, prop. 15.

Let $Q(T) \in \mathbf{Z}[T, T^{-1}]$, let $r$ be an integer $> 0$, and let $F = (1 - T)^{-r} Q \in \mathbf{Z}((T))$. Put

Then, by lemma 1, one has

Let $n_1$ be the least upper bound in $\overline{\mathbf{R}}$ of the set of integers $i \in \mathbf{Z}$ such that $a_i \neq 0$. For every integer $n \geq n_1$, one has $\alpha_n = \tilde{\alpha}(n)$, where $\tilde{\alpha}$ is the polynomial of $\mathbf{Q}[X]$ defined by

If one sets $c = Q(1) = \sum_{i \in \mathbf{Z}} a_i$, one has $\tilde{\alpha}(X) = c X^{r-1}/(r-1)! + \theta(X)$, where $\theta$ is a polynomial of degree $\leq r-2$. Consequently, one has

where the rational number $\rho_n$ tends to a limit as $n$ increases indefinitely. Hence one deduces the relation

If $F = \sum_{n \in \mathbf{Z}} a_n T^n$ and $G = \sum_{n \in \mathbf{Z}} b_n T^n$ are two elements of $\mathbf{Z}((T))$, one denotes by “ $F \leq G$ ” the relation “ $a_n \leq b_n$ for all $n \in \mathbf{Z}$ ”. This is an order relation compatible with the ring structure of $\mathbf{Z}((T))$ (A, VI, p. 18, Def. 1). One has $(1-T)^{-1} \geq 1$. If $Q \in \mathbf{Z}[T, T^{-1}]$ is $\geq 0$, then the integer $Q(1)$ is positive.