MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Note: I asked this question in math.stackexchange but did not receive an answer

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

$$ \begin{array} {lll} s &=& \sum_{r=0}^{\infty} a_r &=& \sum_{r=0}^{\infty} ( a_r \sum_{c=0}^r m_{r,c} ) \\\ &=& \sum_{c=0}^{\infty} ( \sum_{r=0}^{\infty} a_r m_{r,c} ) &=& \sum_{c=0}^{\infty} b_c \end{array}$$

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?

Here is an image showing the range of summability for the geometric series $- \infty \lt x \lt 1$ by the convergence of partial sums up to some index k: enter image description here

share|cite|improve this question
Not an answer, but in the context of Eulerian numbers and divergent series, you might be interested in – Stopple Mar 10 '11 at 16:56
@Stopple: Hm, nice! I'll read it later in more depth. I'd done some heuristics concerning the Eulerian numbers but mostly lacking proofs. Perhaps you like as well. – Gottfried Helms Mar 10 '11 at 17:29
It looks like it might be related to Borel summation. – Fabian Mar 10 '11 at 19:17
You could try looking in Hardy's final book, Divergent Series - although this is rather old fashioned. – Zen Harper Mar 11 '11 at 7:13
@Zen: thanks, seems I've to take a day in the lib. I hoped someone around could have known an immediate link/reference. I've read Hardy's book some years ago, but I think I've seen that "inverse Lambert-W"-term elsewhere in some article later; I think it is not in the book of K.Knopp, but possibly in one of its follow-up articles. sigh – Gottfried Helms Mar 14 '11 at 10:24
up vote 0 down vote accepted

After 3 years without a hint I think I've surely mis-recalled that I had come across that summation (I didn't find it in G.H.Hardy and also not on K.Knopp and not in my downloaded references), so besides its similarity to the Borel-summation it may be original and I've to proceed on my own for more whereabouts of the procedure.

I'll "accept" this (my own) answer just to "close the case" here on MO. Thanks all for their comments above so far.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.