# Apéry's constant ($\zeta(3)$) value

I tried to find some proofs about the Apéry's constant, but I didn't find any intuitive proof. Is this constant given by the "brutal force" summing of $1 + \frac{1}{2^3} + \frac{1}{3^3} + \frac{1}{4^3} + ...$ or there is a easy way to do it?

If there is any sum formula, could you give me the intuition behind this? I really can't just accept formulas, would be great if I could have a explanation.

And about the proof of it being irrational, does someone knows an intuitive link that can explain me this? I only find links that accept things that I don't know. And as I'm asking this only for studying and not for homework, I really wanted to understand as great as possible.

Thanks!

• Did you read A Proof That Euler Missed... Apéry's Proof of the Irrationality of $\zeta(3)$, An Informal Report of Alfred van der Poorten? (link). – Américo Tavares May 5 '13 at 22:28
• What do you understand by «brrutal force summing»? There is no such thing, really :-) – Mariano Suárez-Álvarez May 5 '13 at 23:51
• You want intuition for its irrationality or for the constant itself? Also what kind of intuition? You could always imagine any number of constants as the area enclosed by some curve on a specified interval, sense there are numerous definite integral representations of $\zeta(3)$. – Ethan May 5 '13 at 23:56
• I mean "brutal force" by summing handly or in a computer, the original formula for $\zeta(3)$ – Lucas Zanella May 6 '13 at 0:44
• @lucas, but you simply cannot do that! – Mariano Suárez-Álvarez Dec 30 '16 at 3:54

Article $$ by Alfred van der Poorten is a good starting point, the best I know of.

Convergence speed

I mean "brutal force" by summing handly or in a computer, the original formula for $\zeta(3)$.

It is known that the given series converges very slowly. Apéry's rational approximation $a_n/b_n$ below improves the speed of convergence to $\zeta(3)$, while controlling the increase of the size of $b_n$.

Apéry stated the following equality:

$$\zeta (3)\overset{\mathrm{def}}{=}\sum_{n=1}^{\infty }\frac{1}{n^{3}}=\frac{5}{2}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^{3}\binom{2n}{n}}.\tag{1}$$

In section 3 van der Poorten shows that

$$\sum_{n=1}^{N}\frac{1}{n^{3}}+\sum_{k=1}^{N}\frac{(-1)^{k-1}}{2k^{3}\binom{N}{k}\binom{N+k}{k}}=\frac{5}{2}\sum_{k=1}^{N }\frac{(-1)^{n-1}}{k^{3}\binom{2k}{k}},\tag{2}$$

from which $(1)$ follows by letting $N$ tend to infinity (see this answer of mine). The formula $(2)$ is not very intuitive though.

The second series in $(1)$ is a fast convergent series, faster by far than the defining series for the Apéry's constant $\zeta(3)$. There are even faster convergent series, obtained by techniques of convergence acceleration.

Proof of the irrationality

As far as the irrationality concerns, Apéry constructed two sequences $(a_n),(b_n)$ $^1$ whose ratio $a_n/b_n\to\zeta(3)$ and

1. $2(b_{n}\zeta (3)-a_{n})$ satisfies $\lim\sup \left\vert 2(b_{n}\zeta (3)-a_{n})\right\vert^{1/n}\le(\sqrt{2}-1)^4$.
2. $b_{n}\in \mathbb{Z},2(\operatorname{lcm}(1,2,\ldots ,n))^{3}a_{n}\in \mathbb{Z}$.
3. $\left\vert b_{n}\zeta (3)-a_{n}\right\vert >0$.

This is enough to prove the irrationality of $\zeta (3)$ by contradiction. $$.

Intuition

Given that in van der Poorten's words

Those who listened casually, or who were afflicted with being non-Francophone, appeared to hear only a sequence of unlikely assertion.

it is natural to ask: Where did these ideas come from? My tentative explanation is based on the following fact. A few years later after his proof $$, Roger Apéry derived the rational approximation $a_n/b_n$ to $\zeta(3)$ by transforming the defining series for $\zeta(3)$ into a continued fraction and applying iterated transformations to it, which improved the speed of convergence, and obtained the recurrence relations satisfied by $a_{n},b_{n}$ $$.

--

$^1$ The sequences are $$\begin{equation*} a_{n}=\sum_{k=0}^{n}\binom{n}{k}^{2}\binom{n+k}{k}^{2}c_{n,k}, \qquad b_{n}=\sum_{k=0}^{n}\binom{n}{k}^{2}\binom{n+k}{k}^{2},\end{equation*}$$

where

$$\begin{equation*} c_{n,k}=\sum_{m=1}^{n}\frac{1}{m^{3}}+\sum_{m=1}^{k}\frac{\left( -1\right) ^{m-1}}{2m^{3}\binom{n}{m}\binom{n+m}{m}}\quad k\leq n. \end{equation*}$$

References.

$$ Poorten, Alf., A Proof that Euler Missed…, Apéry’s proof of the irrationality of $\zeta(3)$. An informal report, Math. Intelligencer 1, nº 4, 1978/79, pp. 195-203.

$$ Fischler, Stéfane, Irrationalité de valeurs de zêta (d’ après Apéry, Rivoal, …), Séminaire Bourbaki 2002-2003, exposé nº 910 (nov. 2002), Astérisque 294 (2004), 27-62

$$ Apéry, Roger (1979), Irrationalité de $\zeta2$ et $\zeta3$, Astérisque 61: 11–13

$$ Apéry, Roger (1981) Interpolations de Fractions Continues et Irrationalité de certaines Constantes, Bull. section des sciences du C.T.H.S., n.º3, p.37-53

• +1: Note that your papers are available online here and at Numdam. – Raymond Manzoni May 5 '13 at 23:55
• @RaymondManzoni Thanks for the links. – Américo Tavares May 5 '13 at 23:58
• I wanted exactly to know where does the $\zeta (3)=\sum_{n=1}^{\infty }\frac{1}{n^{3}}=\frac{5}{2}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^{3}\binom{2n}{n}}$ comes from by intuition, and then understand the other parts. – Lucas Zanella May 6 '13 at 0:25
• @LucasZanella According to footnote 4 of Alf van der Poorten's article, formula $(2)$ was well known. – Américo Tavares May 6 '13 at 1:38