## Return to Question

5 removed the remark that the formula was corrected

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 $C=0$ 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

editorial remark: I had a typo in the first version of this question in the first formula and in the wolfram-alpha-formula which I corrected here and also in the .pdf-file. That was induced by the seemingly recent changes of evaluation-method at wolfram-alpha, which do no more evaluate the expression to the exact, but only to an approximate value (in the cost-free version); this is different from the behave in december 2011 when I first wrote the .pdf-file.

4 commented edit of formulae

*edit: I had a typo in the first formula or even an evaluation error in the version without the possible type. I've just corrected the typo. Now I've to doubly recheck my results before it makes sense to proceed *

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 $C=0$ 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

editorial remark: I had a typo in the first version of this question in the first formula and in the wolfram-alpha-formula which I corrected here and also in the .pdf-file. That was induced by the seemingly recent changes of evaluation-method at wolfram-alpha, which do no more evaluate the expression to the exact, but only to an approximate value (in the cost-free version); this is different from the behave in december 2011 when I first wrote the .pdf-file.

3 formula error/typo detected

*edit: I had a typo in the first formula or even an evaluation error in the version without the possible type. I've just corrected the typo. Now I've to doubly recheck my results before it makes sense to proceed *

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):

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

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-R,1+k)/kStirlingS1(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 $C=0$ 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

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