Vandermonde infinite matrix inverse

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I am searching for an inverse of a certain infinite matrix, Vandermonde one.

I have been searching in bibliography and some well known examples exist in literature:

  • Pascal Matrix Inverse -> Alternating Pascal Matrix
  • Stirling 2nd Kind Inverse -> Stirling 1st Kind

In my case I am searching for the inverse Vandermonde infinite matrix:

$$V=lim_{n=\infty}{V_n} =lim_{n=\infty}{(i^j)_{i=1...n,j=1...n}}$$

General formula for it exists as it can be stated as the product of 3 other invertible (det <> 0 and finite elements):

$$V_n = S_n · D_n · P_n^T$$

where:

  • S_n = Stirling 2nd kind matrix
  • D_n = Diagonal matrix with $d_{ii} = i!$
  • P_n = Pascal matrix (elements are binomial numbers)

Following properties on inverting matrix product would allow to invert it, but it leads into a non finite elements matrix. So I suspect it could exist a similar "renormalization procedure" as in Ramanujan sums, but not fluent on it to see how to apply this concept to it.

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In the dot-product of the inverse factors of the Vandermonde-matrix occur series of the type $1+1/2+1/3+ \cdots$ which -if cohenrently interpreted with the whole inversion assumption- would mean the infinite expression for $\log(0)$ (and similar for its powers) which is/are not regularizable and remain singularities.
So there is no inverse. However, I've seen that in many situations where the need of this inversion formally occurs, partial evaluations left of the final singular dotproduct and right of it might be evalauatable and the dot-product of that partial evaluation is then regularizable. I have an example in the essay "pascal-matrix tetrated" (on page 3 using the label $W$ for the matrix) and perhaps in some older erticle more such explicite (and may be better to the point) examples - I'll have to look into my web-files. (I'll add more references later as they might be helpful).
Just as a hint, after your observation of the relation between the Stirling-numbers -matrices $S2 = S1^{-1}$ and that of the Pascalmatrix $P$ and the Vandermondematrix $VZ$ that they can be expressed as $VZ = S2 \cdot F \cdot P^\tau$ where $P^\tau$ means the transposed, the upper-triangular pascalmatrix and $F$ the diagonalmatrix of the factorials (the F taken from "Factorial-function") : in this the inverse of the Vandermondematrix would occur by inversion of the matrix-factors be $VZ^{-1}= (P^\tau)^{-1} \cdot F^{-1} \not * S1$ where the symbol $\not *$ shall mean that the evaluation of the dot-product cannot be done (due to unremovable singularities) and must be postponed in any formula where it occurs. Now I had cases, where some extended formula in fact was doable (mainly in context of tetration via Carlemanmatrices) - by example with some more matrixfactors $X$ and $A$ in an actual matrix-problem $$\begin{array}{rll} X \; \;& \overset?= A \cdot VZ^{-1} \qquad \text{ is this possible?} \\ X \; \;&= A \cdot \left( (P^\tau)^{-1} \cdot F^{-1} \not * S1 \right) &\text{this cannot be evaluated} \\ X \; \; &=( A \cdot (P^\tau)^{-1} \cdot F^{-1}) \not * S1 &\text{so evaluate parenthese first} \\ &Y = A \cdot (P^\tau)^{-1} \cdot F^{-1} & \text{ might become lower triangular} \\ X \; \;&= Y \cdot S1 \qquad \text { might then be evaluatable!} \end{array} $$


Remarks
A more detailed study with a matrix much related to your vandermonde-matrix in a similar way lead to the interesting finding, that all dot-products vanish and the inverse would come out to be the null-matrix... see this and this

see also an earlier answer of mine here on MSE and another earlier answer of mine
A very basic and rough exploration of your problem at all was one of my first amateurish approaches with infinite matrices and I had put my observations together in this text . This is very informal: I even did not know all proper nomenclatura for this numbertheoretical approach and it does not yet contain any of the examples in which I later overcame the problem of the non-invertibility which I've found in context of the "tetration", and thus should be completely rewritten - but perhaps it gives company and impulses to your own current approaches....