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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:

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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.

share|improve this question
    
Related answer and question. –  Américo Tavares May 1 '12 at 15:16
    
For those who do not know what Apéry wrote after the formula $(A)$ it can be found here. –  Américo Tavares Jun 8 '12 at 15:35
    
Translation: "We write $\equiv$ 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 $x$." –  Generic Human Jun 8 '12 at 16:29
    
@GenericHuman Thanks for the translation. –  Américo Tavares Jun 8 '12 at 16:30

3 Answers 3

up vote 7 down vote accepted
+100

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.

share|improve this answer

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?

share|improve this answer
1  
Wouldn't it be $$\frac{(1+a_{1}x)f(x)}{x}=c_{1}+\frac{c_{2}}{1+a_{2}x}x+\frac{c_{3}}{\left( 1+a_{2}x\right) \left( 1+a_{3}x\right) }x^{2}+\ldots \; ?$$ –  Américo Tavares Mar 23 '12 at 22:48
1  
@AméricoTavares About your first comment: sure, but I wanted to pick off the coefficients one by one. That's why I used some other names ($b_j$). The $\equiv$ is also clear by just noting that $$ \frac{x^k}{(1+a_1x)\cdots(1+a_kx)} = x^k + o(x^k) $$ which shows that you can construct the coefficients of $f$ one by one since you leave lower order terms unchanged. –  WimC Mar 24 '12 at 7:10
1  
@AméricoTavares The coefficients $b_j$ are not that important. In fact I could have simply written $c_1 + o(1)$. The essential part is to understand that this step by step process will uniquely generate the coefficients $c_k$, whatever their exact values. I sketched another (less explicit) way to look at this in an earlier comment. –  WimC Mar 24 '12 at 8:31
1  
For $f(x)=1/x^2$ the coefficients are $0, 1, a_1+a_2, a_1a_2+a_1a_3+a_2a_3, \dotsc$. That is, $c_k$ is the coefficient of $x^2$ in the polynomial $\prod_{j=1}^k(x+a_j)$. More generally, for $f(x)=1/x^m$ it turns out that $c_k$ is the coefficient of $x^m$ in that polynomial. –  WimC Apr 14 '12 at 9:37
1  
It appears that a factor $k!$ is missing in the first one. The numerator should be $(-1)^k k!$ (for $k = 0, 1, \dotsc$). –  WimC Apr 14 '12 at 17:47

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.

share|improve this answer
    
For those who do not know what Apéry wrote after the formula $(A)$ it can be found here. –  Américo Tavares Jun 8 '12 at 15:32

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