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

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.

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

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

If some background is of interest: here are the questions on MSE
http://math.stackexchange.com/questions/16228 // question of some user which motivated me to look at an example
http://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

[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:

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.

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]

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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.
Very slick. I think you mean $k s_1(k,k-R) = s_1(k+1, k-R) - s_1(k, k-R-1)$ (you have the opposite sign). – David Speyer Aug 19 '14 at 15:19
It depends on whether you use signed or unsigned Stirling numbers (the signed number $s_1(n,k)$ is off by a factor $(-1)^{n-k}$). In case of signed Stirling numbers my equation is correct, I think. :) – Fabian Gundlach Aug 19 '14 at 15:28
Very nice, ideed. I suspected some separation into two sums in the manner as you did it with the difference of pairs of $s_1()$ but couldn't get the key entry. I'll have to go through your solution step by step and check the relevant identities with the $s_1$, although I think that this answers my question and it's ok to accept it now. Thank you very much, this has now been an open problem for 3 years... – Gottfried Helms Aug 19 '14 at 19:11