Lemma 2

a) Let $F$ be a nonzero element of $\mathbf{Z}((T))$ such that there exists $r \in \mathbf{Z}$, with $(1-T)^r F \in \mathbf{Z}[T, T^{-1}]$; then $F$ can be written uniquely in the form $F = (1-T)^{-d}.Q$, where $Q \in \mathbf{Z}[T, T^{-1}]$, $Q(1) \neq 0$ and $d \in \mathbf{Z}$. If $F \geq 0$, then one has $Q(1) > 0$ and $d \geq 0$.

b) Let $Q, R$ be in $\mathbf{Z}[T, T^{-1}]$, $d, d'$ in $\mathbf{Z}$ with $Q(1) > 0$. If

then either $d < d'$, or else $d = d'$ and $Q(1) \leq R(1)$.

a) We may write $F = (1-T)^{-r} T^n P(T)$ with $r, n \in \mathbf{Z}$ and $P(T) \in \mathbf{Z}[T]$. By euclidean division, we may write $P(T) = (1-T)^p R(T)$ with $R(T) \in \mathbf{Z}[T]$ and $R(1) \neq 0$. Hence $F = (1-T)^{-(r-p)} Q(T)$, where $Q(T) = T^n R(T) \in \mathbf{Z}[T, T^{-1}]$ and $Q(1) \neq 0$. This proves the existence of $d$ and $Q$. Moreover, if $(1-T)^r Q(T) = (1-T)^s R(T)$ with $r > s$ and $Q, R$ in $\mathbf{Z}[T, T^{-1}]$, we have $R(T) = (1-T)^{r-s} Q(T)$, hence $R(1) = 0$; this proves the uniqueness. Suppose that $F$ is $\geq 0$; if we had $d < 0$, then we should have $F(1) = 0$, which is impossible since $F$ is nonzero and all its coefficients are positive; thus we have $d \geq 0$. If $d = 0$, then $Q = F \geq 0$, hence $Q(1)$ is positive. If $d \geq 1$, then $Q(1)$ is positive by formula (3). This proves a).

b) Suppose $d \geq d'$. Then $(1-T)^{-d} ((1-T)^{d-d'} R - Q) \geq 0$; since $S(T) = (1-T)^{d-d'} R - Q$ belong to $\mathbf{Z}[T, T^{-1}]$, this implies $S(1) \geq 0$ by what precedes. If $d > d'$, we have $S(1) = -Q(1) < 0$, whence a contradiction; if $d = d'$, we have $S(1) = R(1) - Q(1)$, whence $Q(1) \leq R(1)$.