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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 // question of some user which motivated me to look at an example // my follow-up question dealing with the current problem
and a more worked out treatize on this in a pdf-file

[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.

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

the LD factor

This is the U factor:

The U factor

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

The reciprocal U^-1

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

The reciprocal LD^-1

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

Hypothese M^-1

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

enter image description here

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.

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1 Answer 1

up vote 6 down vote accepted

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.

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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
I see. Okay, that makes sense (and the sign turned out to be a global sign anyway, so it doesn't matter). Neat! – David Speyer Aug 19 '14 at 15:32
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

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