Find the inverse of the following lower triangular Toeplitz matrix
$$\mathbf{A}_{M\times M}=\left[\begin{array}{ccccc} 1\\ -a_{1} & 1\\ -a_{2} & -a_{1} & 1\\ \vdots & & & \ddots\\ -a_{M-1} & \cdots & -a_{2} & -a_{1} & 1 \end{array}\right], $$
where $$a_i=\frac{c\Gamma(i-c)}{\Gamma(i+1)\Gamma(1-c)}$$ and $c\in(0,1)$ is a real constant.
I hope to find the closed-form expression of $\Vert \mathbf{A} ^{-1} \Vert_1$, where $\Vert\cdot\Vert_1$ represents the $L_1$ induced norm. (Actually, in this case, it is the summation of the first column of $\mathbf{A}^{-1}$.) And if possible, I hope to find how the matrix size $M$ affects the value of $\Vert \mathbf{A} ^{-1} \Vert_1$, i.e., $$ f(M)=\Vert \mathbf{A}_{M\times M} ^{-1} \Vert_1. $$ Even a tight approximation or bounds are acceptable.
There are some properties of $\{a_i\}$, for example, $\sum_{i=1}^{\infty}a_i=1$. But by now I haven't had ideas how to solve it. I appreciate if anyone could help me.
You can try to start by using a "closed" form formula for the inverse of $A$. Set $A=I-N$, where $N$ is strict lower triangular ($a$'s with $+$ signs), then $$ A^{-1}=(I-N)^{-1}=\sum_{i=0}^{M-1}N^i. $$ So if $\|A^{-1}\|_1=\|A^{-1}e_1\|_1$ (which is true provided that $a$'s are non-negative as $A^{-1}$ is Toeplitz as well), then you would need to evaluate the terms $N^{-1}e_1$.
Another possible hint: Partition $A$ as $$ A=\begin{bmatrix}\tilde{A}&\\a^T&1\end{bmatrix}, $$ where $\tilde{A}$ is the leading principal $(M-1)\times(M-1)$ sub-matrix of $A$ and $a^T=[-a_{M-1},\ldots,-a_1]$. The inverse can be written explicitly as $$ A^{-1}=\begin{bmatrix}\tilde{A}^{-1} & 0 \\ -a^T\tilde{A}^{-1} & 1\end{bmatrix}, $$ so $$ A^{-1}e_1 = \begin{bmatrix}\tilde{A}^{-1}e_1 \\ -a^T\tilde{A}^{-1}e_1\end{bmatrix}. $$ Hence if $x_i$ is the $i$th component of $A^{-1}e_1$ (with $x_1=1$), there is a recursion for $x_M$ in the form $$ x_M=\sum_{i=1}^{M-1}a_{M-i}x_i. $$