29
$\begingroup$

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:

enter image description here

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:

  1. Is series $(A)$ indeed divergent?
  2. Which is the theorem stating or from which expansion $(\mathrm{A})$ can be derived?
  3. Could you please indicate a reference?

I've posted on MathOverflow a variant of this question.

  • 0
    @GenericHuman Thanks for the translation.2012-06-08

3 Answers 3

9

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.

4

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?

  • 0
    Please excuse me by asking if the expansion for $f(x)$ is a divergent or a convergent series. I thought based on an remark in Apéry's paper it was divergent but I may be wrong.2012-04-27
2

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.

  • 0
    For those who do not know what Apéry wrote after the formula $(A)$ it can be found [here](http://i.stack.imgur.com/sGRHk.jpg).2012-06-08