# 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 $$f\left( x\right) \equiv \sum_{k\geq 1}\frac{c_{k}}{\left( x+a_{1}\right) \left( x+a_{2}\right) \ldots \left( x+a_{k}\right) }.\tag{A}$$ 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.

• 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

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.

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?

• 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
• @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
• @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
• 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
• 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.

• 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