Apery’s Proof of Irrationality of Zeta(3)

We want to prove the irrationality of ${\zeta(3)}$ . We will use the following remarkable formula to achieve that.

$\displaystyle \zeta(3)= \sum_{n-1}^{\infty} \frac{1}{n^{3}}=\frac{5}{2} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{3}\left(\begin{array}{l} 2 n \\ n \end{array}\right)}$

In fact, we will prove that for any rational ${p/q}$ , we have

$\displaystyle \left|\zeta(3)-\frac{p}{q}\right|>\frac{1}{q^{(\theta+e)}}, \quad \theta=13.417820 \ldots$

The strategy to prove irrationality is simple. If you can approximate the number to well by rationals, then the number has to be irrational. Precisely if we have ${|\alpha -\frac{p}{q}| < \frac{1}{q^{1+\delta}}}$ for infinitely many rationals ${\frac{p}{q} }$ , then ${\alpha}$ has to be irrational. This follows from the estimate, if ${\frac{a}{b} \neq \frac{p}{q}.}$ ,

$\displaystyle |\frac{a}{b} - -\frac{p}{q}| =\frac{aq-bp}{bq} \ge \frac{1}{bq}.$

The question is how do we find the approximations or give an argument that shows the existence of good approximations. It’s mysterious!

We begin with the proof of the formula

$\displaystyle \zeta(3)= \sum_{n-1}^{\infty} \frac{1}{n^{3}}=\frac{5}{2} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{3}\left(\begin{array}{l} 2 n \\ n \end{array}\right)}$

Telescoping

$\displaystyle \frac{a_{1} a_{2} \ldots a_{k-1}}{\left(x+a_{1}\right)\left(x+a_{2}\right) \ldots\left(x+a_{k}\right)}=\frac{a_{1} a_{2} \ldots a_{k-1}}{x\left(x+a_{1}\right)\left(x+a_{2}\right) \ldots\left(x+a_{k-1}\right)}-\frac{a_{1} a_{2} \ldots a_{k}}{\left(x+a_{1}\right)\left(x+a_{2}\right) \ldots\left(x+a_{k}\right)}$

we get

$\displaystyle \sum_{k=1}^{K} \frac{a_{1} a_{2} \ldots a_{k-1}}{\left(x+a_{1}\right) \ldots\left(x+a_{k}\right)}=\frac{1}{x}-\frac{a_{1} a_{2} \ldots a_{K}}{x\left(x+a_{1}\right) \ldots\left(x+a_{K}\right)}$

Take ${x=n^2}$ and ${a_k -k^2}$ to get

$\displaystyle \sum_{k=1}^{K} \frac{-1^{2} \cdot-2^{2} \ldots-(k-1)^{2}}{\left(n^{2}-1\right) \ldots\left(n^{2}-k^{2}\right)}=\frac{1}{n^{2}}-\frac{(-1)^{n-1} 2}{n^{2}\left(\begin{array}{c} 2 n \\ n \end{array}\right)}$

Now with

$\displaystyle \epsilon_{n, k}=\frac{1}{2} \frac{k !^{2}(n-k) !}{k^{3}(n+k) !}$ ,

we get

$\displaystyle (-1)^{k} n\left(\epsilon_{n, k}-\epsilon_{n-1, k}\right)=\frac{(-1)^{k-1}(k-1) !^{2}}{\left(n^{2}-1^{2}\right) \ldots\left(n^{2}-k^{2}\right)}.$

and

$\displaystyle \sum_{n=1}^{N} \sum_{k=1}^{n-1}(-1)^{k}\left(\epsilon_{n, k}-\epsilon_{n-1, k}\right)$

$\displaystyle =\sum_{n=1}^{N} \frac{1}{n^{3}}-2 \sum_{n=1}^{N} \frac{(-1)^{n-1}}{n^{3}\left(\begin{array}{c} 2 n \\ n \end{array}\right)}$

But this is also equal to

$\displaystyle \sum_{k=1}^{N}(-1)^{k}\left(\epsilon_{N, k}-\epsilon_{k, k}\right)=$

$\displaystyle \sum_{k=1}^{N} \frac{(-1)^{k}}{2 k^{3}\left(\begin{array}{c} N+k \\ k \end{array}\right)\left(\begin{array}{c} N \\ k \end{array}\right)}+\frac{1}{2} \sum_{k=1}^{N} \frac{(-1)^{k-1}}{k^{3}\left(\begin{array}{c} 2 k \\ k \end{array}\right)}$

The first term vanishes in the limit ${N \rightarrow \infty}$ and we get

$\displaystyle \sum_{k=1}^{\infty} \frac{1}{n^{3}}=\frac{5}{2} \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k^{3}\left(\begin{array}{c} 2 k \\ k \end{array}\right)}.$

OK, how do we use this formula for ${\zeta(3)}$ to get good approximations?

The sequence

$\displaystyle c_{n, k}=\sum_{m=1}^{n} \frac{1}{m^{3}}+\sum_{m=1}^{k} \frac{(-1)^{m-1}}{2 m^{3}\left(\begin{array}{c} n \\ m \end{array}\right)\left(\begin{array}{c} n+m \\ m \end{array}\right)}$

approaches to ${\zeta(3)}$ uniformly in ${k}$ .

But this as a rational approximation doesn’t work for our purposes. It can be shown that the denominator of ${c_{n,k}}$ is quite large (like ${e^n}$ ) compared to distance between the ${c_{n,k}}$ and ${\zeta(3)}$ which is like ${\frac{1}{n^3}}$ .

$\displaystyle a_{n}=\sum_{k=0}^{n}\left(\begin{array}{c} n \\ k \end{array}\right)^{2}\left(\begin{array}{c} n+k \\ k \end{array}\right)^{2} c_{n, k}$

$\displaystyle b_{n}=\sum_{k=0}^{n}\left(\begin{array}{c} n \\ k \end{array}\right)^{2}\left(\begin{array}{c} n+k \\ k \end{array}\right)^{2}$

Both the sequences ${a_n, b_n}$ satisfy the recurrence relation

$\displaystyle n^{3} u_{n}+(n-1)^{3} u_{n-2}=\left(34 n^{3}-51 n^{2}+27 n-5\right) u_{n-1}, \quad n > 1 .$

and we can deduce

$\displaystyle a_{n} b_{n-1}-a_{n-1} b_{n}=\frac{6}{n^{3}}$

We now define the integer sequences

$\displaystyle p_n = 2[1, 2, 3, \cdots n]^3 a_n, \quad q_n =2[1, 2, 3, \cdots n]^3 b_n$

Before proving the recurrence and the relation, we show how we can use these sequences to get very good approximations to ${\zeta(3)}$

First note that we have ${\frac{a_n}{b_n} \rightarrow \zeta(3)}$ and

$\displaystyle \zeta(3)-\frac{a_{n}}{b_{n}}=\sum_{k=n+1}^{\infty} \frac{6}{k^{3} b_{k} b_{k-1}}.$

Also from

$\displaystyle n^{3} b_{n}+(n-1)^{3} b_{n-2}=\left(34 n^{3}-51 n^{2}+27 n-5\right) b_{n-1}$

we get

$\displaystyle b_{n}-\left(34-51 n^{-1}+27 n^{-2}-5 n^{-3}\right) b_{n-1}+\left(1-3 n^{-1}+3 n^{-2}-n^{-3}\right) b_{n-2}=0$

And hence ${b_n}$ is bounded by the ${\alpha^n}$ where ${\alpha}$ is the bigger root of ${X^2-34X+1=0.}$

In fact,

$\displaystyle \frac{(1+\sqrt{2})^{2}}{(2 \pi \sqrt{2})^{3 / 2}} \frac{(1+\sqrt{2})^{4 n}}{n^{3 / 2}}\left(1-\frac{48-15 \sqrt{2}}{64 n}+O\left(n^{-2}\right)\right)$

Therefore

$\displaystyle \left|\zeta(3)-\frac{p_{n}}{q_{n}} \right| \le \frac{C}{b_n^2} \le \frac{1}{\alpha^{2n}} \le \frac{1}{q_n^{1+\delta}}.$

Proof sketch of recurrence relations:

$\displaystyle B_{n, k}=4(2 n+1)\left(k(2 k+1)-(2 n+1)^{2}\right)\left(\begin{array}{l} n \\ k \end{array}\right)^{2}\left(\begin{array}{c} n+k \\ k \end{array}\right)^{2}$

$\displaystyle \begin{aligned} B_{n, k}-B_{n, k-1} &=(n+1)^{3}\left(\begin{array}{c} n+1 \\ k \end{array}\right)^{2}{{n+1+k} \choose k}^{2}-\\ &-\left(34 n^{3}+51 n^{2}+27 n+5\right)\left(\begin{array}{c} n \\ k \end{array}\right)^{2}\left(\begin{array}{c} n+k \\ k \end{array}\right)^{2}+\\ &+n^{3}\left(\begin{array}{c} n-1 \\ k \end{array}\right)^{2}\left(\begin{array}{c} n-1+k \\ k \end{array}\right)^{2} \end{aligned}$

$\displaystyle \implies (n+1)^{3} b_{n+1}- \left(34 n^{3}+51 n^{2}+27 n+5\right)b_{n}-n^{3} b_{n-1}=0 .$

$\displaystyle (n+1)^{3} b_{n+1, k} c_{n+1, k}-\left(34 n^{3}+51 n^{2}+27 n+5\right) b_{n, k} c_{n, k}+n^{3} b_{n-1, k} c_{n-1, k}$

$\displaystyle =\left(B_{n, k}-B_{n, k-1}\right) c_{n, k}+ +(n+1)^{3} b_{n+1, k}\left(c_{n+1, k}-c_{n, k}\right) -n^{3} b_{n-1, k}\left(c_{n, k}-c_{n-1, k}\right)$

We can see this implies

$\displaystyle \implies (n+1)^{3} a_{n+1}- \left(34 n^{3}+51 n^{2}+27 n+5\right)a_{n}-n^{3} a_{n-1}=0 .$

So both the sequences satisfy the recurence. Now it’s easy to show that

$\displaystyle a_{n} b_{n-1}-a_{n-1} b_{n}=\frac{6}{n^{3}}$

We can deal with ${\zeta(2)}$ similarly.

$\displaystyle \zeta(2)=\sum_{n=1}^{\infty} \frac{1}{n^{2}}=\frac{\pi^{2}}{6}=3 \sum_{n=1}^{\infty} \frac{1}{n^{2}\left(\begin{array}{l} 2 n \\ n \end{array}\right)}$

The quantities to consider in the case are

$\displaystyle \begin{aligned} &B_{n, k}=\left(k^{2}+3(2 n+1) k-11 n^{2}-9 n-2\right)\left(\begin{array}{c}. n \\ k \end{array}\right)^{2}\left(\begin{array}{c} n+k \\ k \end{array}\right), \\ &A_{n, k}=B_{n, k} c_{n, k}+3(-1)^{n+k-1} \frac{(n-1) !}{(k-1) !}. \end{aligned}$

$\displaystyle c_{n, k}-c_{n-1, k}=2(-1)^{n+k-1} \frac{k !^{2}(n-k-1) !}{n(n+k) !}.$

$\displaystyle b_{n}=\frac{\left(\frac{1}{2}(1+\sqrt{5})\right)^{4}}{2 \pi \sqrt{5+2 \sqrt{5}}} \frac{\left(\frac{1}{2}(1+\sqrt{5})\right)^{5 n}}{n}\left(1+O\left(n^{-1}\right)\right).$

Questions:

How do we think about and come up with these approximations? The constructions and the double sequences involved all seem to work like magic!

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