The coefficients in the inverse of unit lower triangular Toeplitz matrix I have a question about the inverse of lower triangular Toeplitz matrix. 
There is a matrix $Q$ which can be infinite-dimensional.
$$
Q=\left[\begin{array}{ccccc}
1\\
-k_{1} & 1\\
-k_{2} & -k_{1} & 1\\
-k_{3} & -k_{2} & -k_{1} & 1\\
\vdots &  &  & \ddots & \ddots
\end{array}\right] 
$$
And the coefficients $0<k_i<1$ are monotone decreasing with $i$ and convergent ($\sum_{i=1}^{\infty}k_{i}=c_0<1$). Moreover, we assume $\lim_{i\rightarrow\infty}\frac{k_{i+1}}{k_i}=s_0$ i.e., $\{k_i\}$ is geometric series when $i$ is large.
Now, let $T=Q^{-1}$ is the inverse of this Toeplitz matrix. I read other papers and know $T$ is also lower triangular Teoplitz matrix. Assume the coefficients are $t_i$, and denote
$$
T=Q^{-1}=\left[\begin{array}{ccccc}
1\\
t_{1} & 1\\
t_{2} & t_{1} & 1\\
t_{3} & t_{2} & t_{1} & 1\\
\vdots &  &  &  & \ddots
\end{array}\right]
$$
I want to know whether $\{t_i\}$ are convergent, and moreover, will $\frac{t_{i+1}}{t_i}$ converge to a constant when $i$ is large enough ($i\rightarrow\infty$)? I tried some examples, and the results showed that both of them are convergent to a constant. But I don't know how to prove it.
I hope to discuss with you. Thanks in advance.
[Updated]
Considering @mike's idea, now I am thinking whether we can express the question as the coefficient ratio for a power series.
As $Q=L^{0}+\sum_{i=1}^{\infty}\left(-k_{i}L^{i}\right)$, where $L$ is the shift matrix in which $L_{ij}=\delta_{i,j+1}$ in infinite-dimensional ($L^0=I$). And $T$ can be written as $T=L^{0}+\sum_{i=1}^{\infty}\left(t_{i}L^{i}\right)$. 
So the question is whether ${t_i}$ is convergent and whether $\lim_{i\rightarrow\infty}\frac{t_{i+1}}{t_i}$ is a constant ($z_0$). If yes, what's the expression of  $z_0$. 
Is this related to the power-series? I don't have much knowledge about it...
 A: The key here is that $q=1-\sum_{i>0}k_ix^i$ is the power series around $0$ with radius of convergence $\frac1{s_0}$, and that $1+\sum_{i>0}t_ix^i$ is the series for its reciprocal $\frac1q$ (which series can be written $\sum_{n\geq 0}(\sum_{i>0}k_ix^i)^n$, so in particular all $t_i$ are positive). I can't come up right now with a setting in which the statement "$Q$ is the matrix of multiplication by $q$ on the basis of monomials, and $T$ the matrix of multiplication by $\frac1q$" makes sense formally, but morally that's what is going on.
The question of whether the $t_i$ converge is negative in general. Just take $k_1=a>1$ and $k_i=0$ for $i>1$, then $t_i=a^i$ diverges. (OK, you don't allow me to take $k_i=0$, but taking $k_i>0$ will only make the $t_i$ larger.) In fact the informaion about the $k_i$ tells you about the singularities of $f$, but for the $t_i$ you need infomation about the zeroes of $f$. Since by continuity you know that $f$ is nonzero on some neighborhood of $0$, you know that $\frac1q$ has positive radius of convergence and therefore $\sqrt[n]{t_n}$ is bounded; this does not imply though that $\frac{t_{i+1}}{t_i}$ converges. Maybe you can do something with the condition that $\frac{k_{i+1}}{k_i}$ convergerges, but I can't see just how.
