3
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Consider the matrix $A_3$ with the definition if $n=k$ then $A_3(n,k)=\binom{n-1}{k-1}=1$, else if $n\ge k$ then $A_3(n,k)=\frac{\binom{n-1}{k-1}}{1-x}$ else $A_3(n,k)=0$.
$\binom{n-1}{k-1}$ means the binomial of $(n-1)$ over $(k-1)$, where $n$ is the row index and $k$ is the column index.

This is an infinite matrix starting starting:

$$A_3=\left( \begin{array}{cccccc} 1 & 0 & 0 & 0 & 0 & 0 \cdots \\ \frac{1}{1-x} & 1 & 0 & 0 & 0 & 0 \\ \frac{1}{1-x} & \frac{2}{1-x} & 1 & 0 & 0 & 0 \\ \frac{1}{1-x} & \frac{3}{1-x} & \frac{3}{1-x} & 1 & 0 & 0 \\ \frac{1}{1-x} & \frac{4}{1-x} & \frac{6}{1-x} & \frac{4}{1-x} & 1 & 0 \\ \frac{1}{1-x} & \frac{5}{1-x} & \frac{10}{1-x} & \frac{10}{1-x} & \frac{5}{1-x} & 1 \\ \vdots&&&&&&& \ddots \end{array} \right)$$

Calculate the matrix inverse of matrix $A_3$ to get another matrix $A_4$, and focus only on the expanded form of the first column, call it $b$:

$$b = \left( \begin{array}{c} 1 \\ \frac{1}{x-1} \\ \frac{x}{(x-1)^2}+\frac{1}{(x-1)^2} \\ \frac{x^2}{(x-1)^3}+\frac{4 x}{(x-1)^3}+\frac{1}{(x-1)^3} \\ \frac{x^3}{(x-1)^4}+\frac{11 x^2}{(x-1)^4}+\frac{11 x}{(x-1)^4}+\frac{1}{(x-1)^4} \\ \frac{x^4}{(x-1)^5}+\frac{26 x^3}{(x-1)^5}+\frac{66 x^2}{(x-1)^5}+\frac{26 x}{(x-1)^5}+\frac{1}{(x-1)^5} \\ \frac{x^5}{(x-1)^6}+\frac{57 x^4}{(x-1)^6}+\frac{302 x^3}{(x-1)^6}+\frac{302 x^2}{(x-1)^6}+\frac{57 x}{(x-1)^6}+\frac{1}{(x-1)^6} \\ \frac{x^6}{(x-1)^7}+\frac{120 x^5}{(x-1)^7}+\frac{1191 x^4}{(x-1)^7}+\frac{2416 x^3}{(x-1)^7}+\frac{1191 x^2}{(x-1)^7}+\frac{120 x}{(x-1)^7}+\frac{1}{(x-1)^7} \\ \vdots \end{array} \right)$$

Does the limit $$ \lim_{n\to \infty} (n - 1)\frac{b(n - 1)}{b(n)}$$

converge to $\log(x)$?

I noticed that the coefficients within the vector $b$ is equal to the Eulerian numbers found in the OEIS as sequence http://oeis.org/A008292.

As a Mathematica program this is:

Clear[x]
x = 4
nn = 12;
A1 = Table[
   Table[If[n == k, 1, If[n > k, 1/(1 - x), 0]], {k, 1, nn}], {n, 1, 
    nn}];
MatrixForm[A1]
A2 = Table[
   Table[If[n >= k, Binomial[n - 1, k - 1], 0], {k, 1, nn}], {n, 1, 
    nn}];
MatrixForm[A2]
A3 = A1*A2;
MatrixForm[A3]
A4 = Inverse[A3];
MatrixForm[Expand[A4]]
(nn - 1)*A4[[nn - 1, 1]]/A4[[nn, 1]]
N[%, 20]
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1
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As you observe, your functions $b(n)$ are related to the Eulerian polynomials $A_n(x)$ by $$A_n(x)=\frac{xb(n)}{1-x}$$so you are asking about the asymptotic behaviour of $A_n(x)$ for large $n$.

It's probably easiest to start from the exponential generating function for the Eulerian polynomials in the form$$F(z) := \frac{1}{1-xe^z} = \frac{1}{1-x}+\frac{A_1(x)}{(1-x)^2}z+\frac{A_2(x)}{(1-x)^3}\frac{z^2}{2!}+... $$ We can then use singularity analysis, as described for example in Wilf's "Generatingfunctionology" - available here, to estimate the growth of the coefficents of the function $F(z)$.

The function $F(z)$ is analytic in $z$ in a sufficiently small disk centred at the origin. The singularities of $F(z)$ are simple poles at $z=-\log\ (x) \pm 2k \pi i $. The singularity of smallest modulus is at $z_0=-\log\ (x)$. Let's assume that $x$ lies in the open interval (0,1) so that $z_0$ is positive.

Theorem 2.4.3 on page 49 of Wilf's book serves to give an estimate for the coefficients of F(z)$$\frac{1}{n!} \frac{A_n (x)}{( 1-x)^{n+1}} \approx\ \left(\frac{-1}{\log\ x}\right)^n.$$ More precise estimates can be obtained using the methods of Chapter 5 of Wilf's book.

We thus find $$n-1\frac{b(n-1)}{b(n)}=\frac{(n-1)A_{n-1}(x)}{A_n(x)}\approx\ \frac{-\log x}{1-x},$$ which differs from what you suggest, but seems to check numerically.

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