1
$\begingroup$

Background: In the context of divergent summation I'm analyzing the matrix of eulerian numbers for a regular matrix-summation method. Beginning indexes at zero (r for "row", c for "column") the entries of M are

$m_{r,c}=\frac{eul(r,c)}{r!} $ and $ eul(r,c)=\displaystyle\sum_{k=0}^{c}(-1)^k \binom{r+1}{k} (c+1-k)^r $. (see for instance wikipedia)

The top-left segment of this (infinite, triangular) matrix is
$ \small{ \begin{array} {rrrrr} 1 & . & . & . & . & . \\ 1 & 0 & . & . & . & . \\ 1/2! & 1/2! & 0 & . & . & . \\ 1/3! & 4/3! & 1/3! & 0 & . & . \\ 1/4! & 11/4! & 11/4! & 1/4! & 0 & . \\ 1/5! & 26/5! & 66/5! & 26/5! & 1/5! & 0 \end{array} } $

The idea is, to sum a sequence $ \{a_k\}_{k=0..\infty} $ using the double sum
$ s = \sum_{r=0}^{\infty} a_r = \sum_{r=0}^{\infty} ( a_r \sum_{c=0}^r m_{r,c} ) = \sum_{c=0}^{\infty} ( \sum_{r=0}^{\infty} a_r m_{r,c} )= \sum_{c=0}^{\infty} b_c $

I was studying that summation with various sequences $ {a_k}$ but I wanted to optimize the computation. For instance, if $ \{a_k\}_{k=0..\infty} = q^k $ and thus define a geometric series with quotient q the $ b_c $ are finite compositions of exponentialseries of $q $ and of powers of $q$: $b_c = e^{q(1+c)}+\sum_{k=1}^c (-1)^k \frac{z^k+kz^{k-1}}{k!}e^z $ where for sum-term $ z=(c+1-k)q $ is simply a shortcut.

Writing the formula for the partial sums $ ps(q,c)= \sum_{k=0}^c b_k $ this boils down to the following series-transformation:

$ ps(x,c) = e^{x(1+c)} \sum_{k=0}^c \frac{(c+1-k)^k}{k!}(-x e^{-x})^k $ and we have in the limit $ \lim_{c\to \infty} ps(x,c) = \frac{1}{1-x} $

After arriving at the term $-x e^{-x} $ I've a vague impression I should have seen this transformation; but even if: I cannot remember. On the other hand - this summation-procedure is powerful, so this transformation is possibly interesting in more general use.

Question: Does someone know this transformation and/or can provide a source where I can read more about it?

0 Answers 0