An infinite set of identities using Stirling numbers 1st kind - are they all zero?

Question

I have the following set of series involving the Stirling numbers 1'st kind and binomials, which can be understood as a set of dot-products of row- and column-vectors of two infinite matrices (where R and C indicate rows and columns, beginning at zero):

$$ w_{R,C} =\sum_{k=\max(R,C)}^\infty (-1)^k {s_1(1+k,1+k-R)\over k!} \cdot (-1)^C (1+C)^k \cdot \binom {1+k}{1+C} $$

I've tested this heuristically for several R and C and always approximated zero; also wolfram-alpha can evaluate this explicitely to zero if feeded with

sum (-1)^k * StirlingS1(k+1,1+k-R)/k! * (1+C)^k * binomial(1+k,1+C), for k=max(C,R) to infty

where we replace $C$, $R$ and $\max(C,R)$ with actual values.

However, I've no option to let wolfram-alpha answer this in general.

I've proved this for $C=0,1,2$ and the first few $R$ using exponential generating functions, but again, a general proof is out of reach for me (possibly I'm overlooking something trivial like telescoping...), so I ask for help here.

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The convention for Stirling numbers first kind as in Math'ica, indexes beginning at zero:

$ \small \qquad \qquad \begin{array} {rrrrr} 1 & . & . & . & . & . \\\ 0 & 1 & . & . & . & . \\\ 0 & -1 & 1 & . & . & . \\\ 0 & 2 & -3 & 1 & . & . \\\ 0 & -6 & 11 & -6 & 1 & . \\\ 0 & 24 & -50 & 35 & -10 & 1 \end{array} $

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If some background is of interest: here are the questions on MSE
https://math.stackexchange.com/questions/16228 // question of some user which motivated me to look at an example
https://math.stackexchange.com/questions/89853 // my follow-up question dealing with the current problem
and a more worked out treatize on this in a pdf-file http://go.helms-net.de/math/divers/InverseNullmatrix.pdf

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[update] Hmm, after 1 1/2 years I've looked at the question again and still do not have an idea how to construct a proof for the whole set of identities. To possibly stimulate helpful answers here I'll insert pictures of the matrices - perhaps it helps to get an immediate idea when the patterns are more visible/obvious than in the bare formula above.

This is (the top-left-segment of) the matrix $M$ in question.

This are the L and D factors of the L D U-decomposition. Because it seems convenient to recognize familiar numbers I've documented the product LD = L D

This is the U factor:

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This is the reciprocal of U (call it UI):

This is the reciprocal of LD (call it LDI):

and in the limit for infinite size of the UI and LDI, the product UI * LDI = MI = 0 by hypothese.

Here are the matrices UI and LDI in a near-symbolic display, the coefficients $s1[r,c]$ are the Stirling numbers first kind.

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Reformulating the dotproducts using their exponential generating functions it is not difficult to prove the identities for a couple of examples.
But what is missing is the proof for the full set of dotproducts.
[/update]

Answer 1

Cancelling a few terms, you see that we want to show that $$ \sum_{k=C}^\infty (-1-C)^k {s_1(1+k,1+k-R)(1+k)\over (k-C)!}=0. $$ Let us further simplify this equation via the coordinate transformations $C\leftarrow -C-1$ and $k\leftarrow k+1$. We then want to prove that $$ I_{R,C}:=\sum_{k=-C}^\infty \frac{C^k}{(k+C)!}\cdot ks_1(k,k-R)=0 $$ for all $C\leq-1$ and $R\geq0$. Recall the recurrence relation $$ks_1(k,k-R)=s_1(k,k-R-1)-s_1(k+1,k-R).$$ Using summation by parts, we obtain $$ \begin{aligned} I_{R,C} &=\sum_{k=-C}^\infty \frac{C^k}{(k+C)!}\cdot (s_1(k,k-R-1)-s_1(k+1,k-R))\\ &=B+\sum_{k=-C}^\infty \left(\frac{C^{k+1}}{(k+1+C)!}-\frac{C^k}{(k+C)!}\right)\cdot s_1(k+1,k-R)\\ &=B+\sum_{k=-C}^\infty \frac{C^{k+1}-C^k(k+1+C)}{(k+1+C)!}\cdot s_1(k+1,k-R)\\ &=B-\sum_{k=-C}^\infty \frac{C^k}{(k+1+C)!}\cdot (k+1)s_1(k+1,k-R)\\ &=B-\sum_{k=-C+1}^\infty \frac{C^{k-1}}{(k+C)!}\cdot ks_1(k,k-R-1)\\ &=-\frac{I_{R+1,C}}C \end{aligned} $$ with boundary term $$ \begin{aligned} B&=\lim_{k\rightarrow\infty}\frac{C^k}{(k+C)!}s_1(k,k-R-1)+\frac{C^{-C}}{(-C+C)!}s_1(-C,-C-R-1)\\ &=\frac{C^{-C-1}}{(-C+C)!}\cdot(-C)s_1(-C,-C-R-1). \end{aligned} $$ We can therefore use induction and only have to show $I_{R,C}=0$ for $R=0$. But as Gottfried Helms pointed out, this is easy.