A problem on power series Let $P(T) = \frac { 1 + T + T^2 + \cdots + T^m}{ 1 - T^2( 1 + T + T^2 + \cdots + T^n)} = \sum _{i = 0}^\infty \beta_n T^n$ be a formal power series expansion. This kind of series arose while I was reading that Betti numbers of a module over a local ring has polynomial growth. I am not able to prove that there exists a number $\alpha$ such that $\beta_n \leq \alpha^n$ $\forall n \geq 1$.
 A: Write
$\begin{align}
P(T) &= \frac { 1 + T + T^2 + \cdots + T^m}{ 1 - T^2( 1 + T + T^2 + \cdots + T^n)}\\
&= \frac{(1-T^{m+1})/(1-T)}{1-T^2(1-T^{n+1})/(1-T)}\\
&= \frac{1-T^{m+1}}{(1-T)-T^2(1-T^{n+1})}\\
&= \frac{1-T^{m+1}}{1-T-T^2+T^{n+3}}\\
\end{align}
$
The radius of convergence of the denominator
is the root of
$r(T) =\frac{1-T-T^2+T^{n+3}}{1-T}
= 1-T^2\frac{1-T^{n+1}}{1-T}
$.
$r(0) = 1$.
Since $1-T^n \ge 1-T$,
$r(T) \le 1-T^2$.
$r(1) = -n < 0$,
so $r$ has a root between 0 and 1.
Since $\frac{1-T^{n+1}}{1-T} = 1+T+...+T^{n} \le n+1$,
$r(T) \ge 1-(n+1)T^2$,
so the root of $r$
is at least $1/\sqrt{n+1}$.
This shows that a power series gotten from this
will have reasonable radius of convergence,
so the coefficients can be worked with.
The next step is the usual one of writing
$P(T) = \frac{1-T^{m+1}}{1-T-T^2+T^{n+3}}
=\sum_{i=0}^{\infty} a_i T^i
$
in the form
$\begin{align}
1-T^{m+1} &= (1-T-T^2+T^{n+3})\sum_{i=0}^{\infty} a_i T^i \\
&= \sum_{i=0}^{\infty} a_i T^i
-\sum_{i=0}^{\infty} a_i T^{i+1}
-\sum_{i=0}^{\infty} a_i T^{i+2}
+\sum_{i=0}^{\infty} a_i T^{i+n+3}\\
&= \sum_{i=0}^{\infty} a_i T^i
-\sum_{i=0}^{\infty} a_{i-1} T^i
-\sum_{i=0}^{\infty} a_{i-2} T^i
+\sum_{i=0}^{\infty} a_{i-n-3} T^i\\
&= \sum_{i=0}^{\infty} (a_i-a_{i-1}-a_{i-2}+a_{i-n-3}) T^i
\end{align}
$
This, with appropriate boundary conditions,
gives a recurrence relation for the $a_i$.
For $i=0$, $a_0=1$
(since $a_j=0$ for $j < 0$).
For $i=m+1$,
$-1 = a_i-a_{i-1}-a_{i-2}+a_{i-n-3}$.
For all other $i$,
$a_i-a_{i-1}-a_{i-2}+a_{i-n-3} = 0$.
I'll leave it at that.
Note: I may have made some errors in the
exponents of $T$ in the fraction
making up $P(T)$,
but I am sure the basic idea is correct.
