Inverse of an infinite matrix with factorial entries

Question

Has anyone come across the following matrix? $$ A= \begin{bmatrix} \frac{1}{1!}&\frac{1}{2!}&\frac{1}{3!}&\cdots\\ -\frac{1}{2!}&-\frac{1}{3!}&-\frac{1}{4!}&\cdots\\ \frac{1}{3!}&\frac{1}{4!}&\frac{1}{5!}&\cdots\\ -\frac{1}{4!}&-\frac{1}{5!}&-\frac{1}{6!}&\cdots\\ \vdots&\vdots&\vdots&\ddots \end{bmatrix}. $$

I am particularly interested in finding $A^{-1}$.

Answer 1

The LDU-decomposition gives dot-products of rows & columns of the following matrices, whose decomposition of entries is much obvious and simple even for the infinite case and allow Cesaro or Eulersummation, which seem to give always zero (I've checked this for a handful of leading dot-products). The row-scaled left and column-scaled right matrices of the product look like

$$ \small \begin{array}{rrrrrrrrr|rrrrrrrrr|} & & & & & & & & & & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \\ & & & & & & & & & & 3 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & \\ & &u^{-1} &\cdot _\mathfrak E&(d^{-1}&\cdot l^{-1})& = &\Large 0 & & & 5 & 5 & 1 & 0 & 0 & 0 & 0 & 0 & \\ & & & & & & & & & & 7 & 14 & 7 & 1 & 0 & 0 & 0 & 0 & \\ & & & & & & & & & & 9 & 30 & 27 & 9 & 1 & 0 & 0 & 0 & \\ & & & & & & & & & & 11 & 55 & 77 & 44 & 11 & 1 & 0 & 0 & \\ & & & & & & & & & & 13 & 91 & 182 & 156 & 65 & 13 & 1 & 0 & \\ & & & & & & & & & & 15 & 140 & 378 & 450 & 275 & 90 & 15 & 1 & \\ & & & & & & & & & & ... & ... & ... & ... & ... & ... & ... & .. & \\ \hline\\ 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & ... & & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \\ 0 & 1 & -3 & 6 & -10 & 15 & -21 & 28 & ... & & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \\ 0 & 0 & 1 & -5 & 15 & -35 & 70 & -126 & ... & & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \\ 0 & 0 & 0 & 1 & -7 & 28 & -84 & 210 & ... & & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \\ 0 & 0 & 0 & 0 & 1 & -9 & 45 & -165 & ... & & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \\ 0 & 0 & 0 & 0 & 0 & 1 & -11 & 66 & ... & & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & -13 & ... & & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & ... & & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \\ .. & .. & .. & .. & .. & .. & .. & .. & ... & & ... & ... & ... & ... & ... & ... & ... & .. & | \\ \hline \end{array}$$ where $u,d,l$ are the similarity-scaled versions of $U,D,L$. Because the results of the dotproducts - taken by Euler-summation - are all zero (supposedly), the required inverse similarity-scaling of the result-matrix is irrelevant and doesn't change its character of being a zero-matrix.

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You can check some entries in the first column of the results using the sumalt()-procedure in Pari/GP which is a Cesaro-sum-related version of divergent summation of alternating series.

sumalt(k=1,(-1)^k*1*(2*k-1)) \\first row by first column

   vector(12,k,-(-1)^k*binomial(k+0,2))    \\second row check only
sumalt(k=2,(-1)^k*binomial(k+0,2)*(2*k-1)) \\second row by first column

   vector(12,k,-(-1)^k*binomial(k+1,4))    \\3'rd row check only
sumalt(k=3,-(-1)^k*binomial(k+1,4)*(2*k-1))\\3'rd row by first column

   vector(12,k,-(-1)^k*binomial(k+2,6))    \\4'th row check only
sumalt(k=4,-(-1)^k*binomial(k+2,6)*(2*k-1))\\4'rd row by first column


   vector(12,k,-(-1)^k*binomial(k+3,8))    \\5'th row check only
sumalt(k=5,-(-1)^k*binomial(k+3,8)*(2*k-1))\\5'rd row by first column

   vector(12,k,-(-1)^k*binomial(k+4,10))    \\6'th row check only
sumalt(k=6,-(-1)^k*binomial(k+4,10)*(2*k-1))\\6'rd row by first column