Coefficient $\small c_{k} $ at $\small x^k$ in $\small (x-1)(x-1/2)(x-1/3)...(x-1/n) $?

Question

This is a detail question from my current main question , but came out as a standalone problem. Background: I've found a description for the matrix T but to have this practically usable I need a solution to the following problem:

Consider the sequence of products

$\small \qquad \qquad \begin{array} {lll} 1, \\ (x-1),\\ (x-1)(x-1/2),\\ (x-1)(x-1/2)(x-1/3), \\ \ldots \end{array}$

Now my question is to find efficiently the sequence of the coefficients $\small c_{k,p} $ at the same (say the $p$th) powers of $x$ so that I can generate the powerseries $\small f(t,p)=\sum_{k=0}^\infty c_{k,p} t^k $
For instance if p=0 this is simple: $\small f(t,0)= \sum_{k=0}^\infty (-1)^k/k! \cdot t^k$ where the coefficients occur by $1, -1, (-1)(-1/2) , (-1)(-1/2)(-1/3),\ldots$

So: what is some efficient way to find the coefficients for f(t,p) with some fixed $p$? Some compositions of binomial-coefficients, factorials?
I'm also fine with a recursive expression which provides the coefficients sequentially with increasing powers, like $\small c_{k+1,p} = \langle \text{something}\rangle c_{k,p}$ or similar (I have some fully recursive description which needs all earlier coefficients with $p-1,p-2,\ldots$ but I want an expression which is "standalone" for a selected $p$.)

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[added] I should add, that I know some composition using the Stirling numbers first kind and factorials. However, I need the coefficients up to an index up to (sometimes) a couple of hundreds for approximate testing of a family of functions whose formal powerseries are also composed of this coefficients (cofactored according the matrix-product in my main question). As long as I do not know a non-recursive and relatively simple formula for that Stirling numbers, a description of the coefficients in terms of Stirling-numbers is not of much help. (However, this might in turn indicate, that a simpler formula for my coefficients could not be available at all - of what I'm getting aware as I'm writing this now...)

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[update]

For a crosscheck consider the following table.

The columns contain the coefficients of some product, for instance the column with index $c=4$ has $\small (1/24,-5/12,35/24,-25/12,1) $ because of the expansion of $\small (x-1)(x-1/2)(x-1/3)(x-1/4) $ into $\small 1/24-5/12 x+35/24 x^2-25/12 x^3+1 x^4 $

The $p$th row (index begins at $p=0$) shows the sought coefficients for my power series for $\small f(t,p)$:

$\qquad \small \begin{array} {rrrrrrr} 1 & -1 & 1/2 & -1/6 & 1/24 & -1/120 & 1/720 & -1/5040 \\ . & 1 & -3/2 & 1 & -5/12 & 1/8 & -7/240 & 1/180 \\ . & . & 1 & -11/6 & 35/24 & -17/24 & 35/144 & -23/360 \\ . & . & . & 1 & -25/12 & 15/8 & -49/48 & 7/18 \\ . & . & . & . & 1 & -137/60 & 203/90 & -967/720 \\ . & . & . & . & . & 1 & -49/20 & 469/180 \\ . & . & . & . & . & . & 1 & -363/140 \\ . & . & . & . & . & . & . & 1 \end{array} $
for instance $\small f(t,2)= 1 t^2 - 11/6 t^3 + 35/24 t^4 - \ldots + \ldots $

Answer 1

We have $$\prod_{k=1}^n \left(x - \frac{1}{k}\right) = \frac{(-1)^n x^{n+1}}{n!} \prod_{k=0}^n \left(\frac{1}{x} - k\right).$$

Now, let $y = 1/x$. Then $$\prod_{k=0}^n \left(\frac{1}{x} - k\right) = \prod_{k=0}^n (y-k) = y^{\underline{n+1}} = \sum_{k=0}^{n+1} s(n+1,k) y^k,$$ where $s(n,k)$ is a Stirling number of the first kind. (The fact that Stirling numbers of the first kind are used to convert from falling factorial powers to normal powers is one of their basic properties.)

Putting it all together, and rewriting everything in terms of $x$, we have $$\prod_{k=1}^n \left(x - \frac{1}{k}\right) = \frac{(-1)^n}{n!} \sum_{k=0}^{n+1} s(n+1,k) x^{n+1-k}.$$

(This agrees with the table of coefficients you posted.)

Answer 2

Let $p_n(x) = \prod_{k=1}^n \left( x- \frac{1}{k} \right) = \frac{x^n}{n!} \left( 1 - \frac{1}{x} \right)_{(n)}$, where $(a)_{(n)}$ stands for the Pochhammer's symbol.

Then the generating function for these polynomials is easily found by summing the hypergeometric series, $\sum_{n=0}^\infty (a)_{(n)} \frac{t^n}{n!} = (1-t)^{-a} $: $$ \sum_{n=0}^\infty p_n(x) t^n = \left( 1-t x \right)^{1/x-1} = \mathrm{e}^{-t} \left( 1 + x \cdot t \left( 1- \frac{t}{2} \right) \right) + \mathcal{o}(x) $$

Answer 3

Well, if I accept the definition by Stirling numbers, then the top-left of the matrix, which I called T is composed as
$\qquad \small t_{r,c}=(-1)^c \cdot {s1(c+1,1+c-r) \over c! } $
The top left segment explicitely:

$\qquad \small \begin{array} {rrrr} S1(1,1)/0! & -S1(2,2)/1! & S1(3,3)/2! & -S1(4,4)/3! & S1(5,5)/4! \\ . & -S1(2,1)/1! & S1(3,2)/2! & -S1(4,3)/3! & S1(5,4)/4! \\ . & . & S1(3,1)/2! & -S1(4,2)/3! & S1(5,3)/4! \\ . & . & . & -S1(4,1)/3! & S1(5,2)/4! \\ . & . & . & . & S1(5,1)/4! \end{array} $

where the $\small s1(r,c) $ are the (signed) Stirling numbers first kind as defined for instance in Mathematica (the matrix-indexes are always assumed based zero here)

Some more background:
I'm trying to prove that a certain family of series, which occur as dot-products of two infinite matrices, actually evaluate all to zero (showing either convergence or at least Euler-/Abel summability).

The matrices under discussion are

$\qquad \small W = T \cdot G $

where T is the matrix just discussed in this thread, and G was defined in the main thread by

$\qquad g_{r,c}=(-1)^c (1+c)^r \binom{r+1}{c+1} $

With this for the entries in W we have
$$ w_{r,c}= (-1)^c \sum_{k=r}^{\infty} (-1)^k { s1(k+1,1+k-r) \over k!} \cdot (1+c)^k \binom{k+1}{c+1} $$

Mathematica at wolframalpha is able to evaluate the first few rows, (r=0..2) even with an indeterminate c (the Stirling-number-references must then be replaced by more explicite expressions (obvious)), whose entries result all to be zero. Also I tested a couple of more $\small w_{r,c} $ of the top-left of matrix W individually using wolfram alpha - they all evaluated to zero. Try for instance $\small w_{1,1} $ , $\small w_{5,1} $ , $\small w_{5,2} $ I tried a handful more examples. I'll provide more info in the "main thread" later.

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So this clears all my issues here and I'll append also a note of the result in my "main question".

I've one answer to "accept", so I just select Mike's answer that brought me finally on the track.