A problem on power series

Question

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 $\beta_n \leq \alpha^n$ $\forall n \geq 1$.

Answer 1

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.