Apéry [Ap]

\[\zeta(3) = \sum_{n=1}^\infty \frac{1}{n^3}\]
is irrational.
The proof revolves around the fact that for any $z\in[0,1]$ there are two useful ways to write $\log z$: one is as the derivative of $z^\sigma$ (with respect to $\sigma$) at $\sigma=0$. The other is as \[-\int_z^{1} \frac{1}{u}\mathrm{d}u=-(1-z)\int_0^1 \frac{1}{1-(1-z)t}\mathrm{d}t.\]

On the other hand, when $r=s$, the series does not telescope, but again by differentiating and setting $\sigma=0$, we get that \[I_{r,r}=\sum_{k=0}^\infty \frac{-2}{(k+r+1)^3}=-2\left( \zeta(3)-\sum_{k=1}^r\frac{1}{k^3}\right).\] In particular, $I_{r,r}=(A+B\zeta(3))/d_r^3$ for some integers $A,B$ (depending only on $r$). This is the way in which $\zeta(3)$ enters our proof. Note, for example, that in particular \[-2\zeta(3)=\int_0^1\int_0^1 \frac{\log xy}{1-xy}\mathrm{d}x\mathrm{d}y,\] which should be compared to the well-known integral representation of $\zeta(2)$ as $\int_0^1\int_0^1\frac{1}{1-xy}\mathrm{d}x\mathrm{d}y$.

Doing $n$ partial integrations with $x$, then another $n$ with $y$, we get that \[J_n=\int_{0}^1\int_{0}^1\int_0^1 \frac{x^n(1-x)^ny^n(1-y)^nz^n(1-z)^n}{(1-(1-xy)z)^{n+1}}\mathrm{d}x\mathrm{d}y\mathrm{d}z.\] Let \[g(x,y,z)=\frac{x(1-x)y(1-y)z(1-z)}{1-(1-xy)z}.\] We want to get an upper bound for this function, which is just a straightforward calculus maximisation problem. Here is one possible approach. To simply matters, let $w=1-(1-xy)z$, so that $z=(1-w)/(1-xy)$, and \[g(x,y,z)= \frac{xy(1-x)(1-y)(1-w)(w-xy)}{(1-xy)^2w}.\] If the value of $xy$ is fixed, then calculus verifies that the maximum value of $(1-x)(1-y)=1-(x+y)+xy$ occurs when $x=y$, since the maximum of $x\mapsto x+u/x$ occurs at $x^2=u$. Therefore $g(x,y,z)$ is maximised when $x=y$, and hence \[g(x,y,z)\leq \frac{x^2(1-x)^2(1-w)(w-x^2)}{(1-x^2)^2w}=\frac{x^2}{(1+x)^2}\left(1-x^2-\frac{x^2}{w}-w\right).\] For any fixed $x\in(0,1)$, we can find the maximum of the expression in $w$ by differentiating, which again tells us that the maximum occurs when $w=x$, and so \[g(x,y,z)\leq \frac{x^2(1-x)^2}{(1+x)^2}.\] Finally, another application of calculus tells us that the maximum of $x(1-x)/(1+x)$ occurs at $x=\sqrt{2}-1$, and so finally we have \[g(x,y,z) \leq (\sqrt{2}-1)^4.\] In particular, for all $0\leq x,y,z\leq 1$, we have $g(x,y,z)\leq 2^{-5}$, and so \[J_n\leq 2^{-5n}\int_{0}^1\int_{0}^1\int_0^1 \frac{1}{1-(1-xy)z}\mathrm{d}x\mathrm{d}y\mathrm{d}z\ll 2^{-5n}.\] (In fact using Step One again it's easy to check that the implicit constant is $2\zeta(3)$, but that is irrelevant here.)

**[Ap]**R. Apéry, Irrationalité de $\zeta(2)$ et $\zeta(3)$, Luminy Conference on Arithmetic, Astérisque No. 61 (1979), 11–13.**[Be]**F. Beukers, A note on the irrationality of $\zeta(2)$ and $\zeta(3)$, Bull. London Math. Soc. 11 (1979), 268-272.**[vdPo]**A. van der Poorten, A proof that Euler missed...Apéry's proof of the irrationality of $\zeta(3)$, Math. Intelligencer 1 (1978/79), 195-203.