Asymptotic (divergent) series MOTIVATION. After having read in detail an article by Alf van der Poorten  I read a very short paper by Roger Apéry. I am interested in finding a proof of a series expansion in the latter, which is in not given in it. So I assumed it should be stated or derived from a theorem on the subject.
In Apéry, R., Irrationalité de $\zeta 2$ et $\zeta
3$, Société Mathématique de France, Astérisque 61 (1979)
there is a divergent series expansion for a function I would like to
understand. Here is my translation of the relevant part for this question 

(...) given a real sequence $a_{1},a_{2},\ldots ,a_{k}$, an analytic function $f\left( x\right) $ with respect to the variable
  $\frac{1}{x}$ tending to $0$ with $\frac{1}{x}$ admits a (unique)
  expansion in the form $$f\left( x\right) \equiv \sum_{k\geq
1}\frac{c_{k}}{\left( x+a_{1}\right) \left( x+a_{2}\right) \ldots
 \left( x+a_{k}\right) }.\tag{A}$$

Added copy of the original:

and the translation by Generic Human of the text after the formula:

"(We write ≡ instead of = to take into account the aversions of
  mathematicians who, following Abel, Cauchy and d'Alembert, hold
  divergent series to be an invention of the devil; in fact, we only
  ever use a finite sum of terms, but the number of terms is an
  unbounded function of x.)"

Remark. As far as I understand, based on this last text, the expansion of $f(x)$ in $(\mathrm{A})$ is in general   a divergent series and not a convergent one, but the existing answer [by WimC] seems to indicate the opposite.
The corresponding finite sum appears and is proved in section 3 of Alfred van der Poorten's article A proof that Euler missed ... Apéry's proof of the irrationality of $\zeta (3)$ as 

For all $a_{1}$, $a_{2}$, $\dots$  $$ \sum_{k=1}^{K}\frac{a_{1}a_{2}\cdots a_{k-1}}{(x+a_{1})(x+a_{2})\cdots(x+a_{k})}= \frac{1}{x}-\frac{a_{1}a_{2}\cdots a_{K}}{x(x+a_{1})(x+a_{2})\cdots(x+a_{K})},$$
  $$\tag{A'} $$

Questions:


*

*Is series $(A)$ indeed divergent?

*Which is the theorem stating or from which expansion $(\mathrm{A})$ can be derived? 

*Could you please indicate a reference?



I've posted on MathOverflow a variant of this question.
 A: Writing $g_1(x)=f(1/x)$ gives
$$
g_1(x)\equiv\sum_{k\ge1}\frac{c_kx^k}{(1+a_1x)(1+a_2x)\dots(1+a_kx)}\tag{1}
$$
which vanishes at $x=0$.
Recursively define
$$
g_{n+1}(x)=\frac{(1+a_nx)g_n(x)}{x}-c_n\tag{2}
$$
where
$$
c_n=\lim_{x\to0}\frac{g_n(x)}{x}\tag{3}
$$
Then
$$
g_n(x)\equiv\sum_{k\ge n}\frac{c_kx^{k-n+1}}{(1+a_nx)(1+a_{n+1}x)\dots(1+a_kx)}\tag{4}
$$
is another series like $(1)$ (which vanishes at $x=0$).
The series in $(1)$ may or may not converge, as with the Euler-Maclaurin Sum Series. As with most asymptotic series, we are only interested in the first several terms; the remainder (not the remaining terms) can be bounded by something smaller than the preceding terms. Therefore, convergence is not an issue.
A: This formula might be easier to understand if it is expressed for $x$ (instead of $\tfrac{1}{x}$) near $0$.  Let the sequence $a_1, a_2, \dotsc$ be given.  For an analytic $f$ with $f(0)=0$ it then says that there exist $c_1, c_2, \dotsc$ such that
$$
f(x) \equiv \sum_{k \geq 1}\frac{c_kx^k}{(1+a_1x)\cdots(1+a_kx)}
$$
Now $(1+a_1x)f(x)$ also vanishes at $x=0$ so
$$
\frac{(1+a_1x)f(x)}{x} = c_1 + b_1x + b_2x^2 + \dotsc
$$
which gives you $c_1$.  Repeat the process with
$$
\frac{(1+a_1x)f(x)}{x} - c_1
$$
to find $c_2$, and so on.  I don't have any references though, and browsing through the references you provided I just wonder: how can people get such wonderful ideas?
A: I'm not sure Apéry was really saying the series was divergent: I think he was just explaining the notation and saying that if convergence is not proven, then it is useful to have a notation that doesn't imply convergence.


*

*For non-negative $a_i$ and positive $x$, you can check that if $f(1/t)=\sum_{k\ge 1} b_k t^k$ ($t=1/x$), we have
$$c_k=\sum_{i=1}^k b_i\left([x^{i-1}]\prod_{j=1}^{k-1} x+a_j\right)$$
Since all coefficients are non-negative:
$$\prod_{j=1}^{k-1} 1/t+a_j\ge 1/t^{i-1} [x^{i-1}] \prod_{j=1}^{k-1} x+a_j$$
Thus $$\left|\frac{c_k}{\prod_{j=1}^k 1/t+a_j}\right|\le \sum_{i=1}^k \frac{|b_i| t^i}{1+a_k t}\le \sum_{i=1}^k |b_i| t^i$$
which proves convergence of the series whenever $t$ is less than the radius of convergence of $f(1/t)$.

*For negative $a_i$, pick $a_i=-2^i$ and $f(1/t)=1/t$, the series is:
$$\sum_{k\ge 1} \frac{1/a_k}{\prod_{i=1}^k 1+1/(ta_i)}$$
which is absolutely convergent for $t\ne -1/a_j$ and equivalent around $t\rightarrow -1/a_j$ to
$$\frac 1{\prod_{i=1}^j 1+1/(ta_i)}\sum_{k\ge j} \frac{1/a_k}{\prod_{i=j+1}^k 1-a_j/a_i}$$
which diverges as $t\rightarrow -1/a_j$. So there is an open set around $-1/a_j$ where the series does not converge to $f(1/t)$ and the series cannot converge to $f(1/t)$ in a neighborhood of $+\infty$, not even almost everywhere.
