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I am going through Van der Poorten's "A Proof that Euler Missed...", which outlines Apéry's proof that $\zeta(3)$ is irrational. In section 3. "Some Irrelevant Explanations" (page 197 in the linked PDF), the author proves that:

$$ \zeta(3)=\frac{5}{2} \sum_{k=1}^{\infty}{\frac{ (-1)^{k-1}} {\binom {2k}{k}k^{3}}}. $$

I have trouble understanding a part of his proof. Here is in my own words what I understand already and where I'm stuck.


Step 1.

First we consider the sum

$$ \sum_{k=1}^{K} \frac{a_1a_2\ldots a_{k-1}}{(x+a_1)(x+a_2)\ldots(x+a_k)}, $$

and note that it is equal to

$$ \frac{1}{x}-\frac{a_1a_2\ldots a_{K}}{x(x+a_1)(x+a_2)\ldots(x+a_K)}. $$

This can easily be proved by defining $A_K=\frac{a_1a_2\ldots a_{K}}{x(x+a_1)(x+a_2)\ldots(x+a_K)}$. The identity then becomes:

$$ \sum_{k=1}^{K} (A_{k-1} - A_k) = A_0 - A_K, $$

which is trivially true.


Step 2.

In a second step, we define $x=n^2$ and $a_k=-k^2$, for $k\leq K\leq n-1$ and use the above sum identity to obtain

$$ \sum_{k=1}^{n-1} \frac{(-1)^{k-1}(k-1)!^2}{(n^2-1^2)\ldots(n^2-k^2)}. $$

From the above we know that this must equal $A_0-A_K$, so we have:

$$ \frac{1}{n^2}-\frac{(-1)^{n-1}(n-1)!^2}{n^2(n^2-1^2)\ldots(n^2-(n-1)^2)}, $$

and after simplifying we have the compact version:

$$ \sum_{k=1}^{n-1} \frac{(-1)^{k-1}(k-1)!^2}{(n^2-1^2)\ldots(n^2-k^2)}=\frac{1}{n^2}-\frac{2(-1)^{n-1}}{n^2 \binom{2n}{n}}. $$


Step 3.

We will now try to find an alternative representation of the terms inside the sum, i.e., $\frac{(-1)^{k-1}(k-1)!^2}{(n^2-1^2)\ldots(n^2-k^2)}$. Indeed, by defining

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

we note that the terms in the sum can be written as:

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

Using this observation, the sum can now be written in an alternative form as:

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


The issue.

Van der Poorten then "concludes" that:

$$ \sum_{k=1}^{n-1} (-1)^k n (\epsilon_{n,k}-\epsilon_{n-1,k}) = \frac{1}{n^3}-\frac{2(-1)^{n-1}}{n^3 \binom{2n}{n}}. $$

But this is in contradiction with the identity obtained in Step 2! Indeed, this would mean that the sum

$$ \sum_{k=1}^{n-1} \frac{(-1)^{k-1}(k-1)!^2}{(n^2-1^2)\ldots(n^2-k^2)} $$

is equal to both

  • $\frac{1}{n^2}-\frac{2(-1)^{n-1}}{n^2 \binom{2n}{n}}$ and
  • $\frac{1}{n^3}-\frac{2(-1)^{n-1}}{n^3 \binom{2n}{n}}$.

I am obviously missing something here. What is it?

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No, it's written:

$$ \sum_{k=1}^{n-1} (-1)^k (\epsilon_{n,k}-\epsilon_{n-1,k}) = \frac{1}{n^3}-\frac{2(-1)^{n-1}}{n^3 \binom{2n}{n}} $$

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  • $\begingroup$ But in Step 2, we obtain a conflicting identity... Obvisouly $\frac{1}{n^2}-\frac{2(-1)^{n-1}}{n^2 \binom{2n}{n}} \neq \frac{1}{n^3}-\frac{2(-1)^{n-1}}{n^3 \binom{2n}{n}}.$ $\endgroup$ – Klangen Jun 21 '18 at 14:07
  • $\begingroup$ @PreservedFruit : You have to distinguish between $(-1)^k (\epsilon_{n,k}-\epsilon_{n-1,k}) $ and $(-1)^k n(\epsilon_{n,k}-\epsilon_{n-1,k}) $ . $\endgroup$ – user90369 Jun 21 '18 at 14:14
  • $\begingroup$ Arrgh, you are right, I missed that $\times n$... But then I don't understand why we need Step 2 at all. Why don't we just start the proof directly from Step 3? $\endgroup$ – Klangen Jun 21 '18 at 14:33
  • $\begingroup$ Maybe he wanted to be very specific, I do not know. Anyway, the main thing is, to understand the context. $\endgroup$ – user90369 Jun 21 '18 at 15:04
  • $\begingroup$ OK thank you. I will follow this up in another MO question. $\endgroup$ – Klangen Jun 21 '18 at 15:08
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Creative telescoping is a very nice way for proving such identity, which is also discussed in the first section of my notes. On the other hand, one might also start with the RHS:

$$ \sum_{n\geq 1}\frac{(-1)^{n+1}}{n^3\binom{2n}{n}} = \sum_{n\geq 1}\frac{(-1)^{n+1}B(n,n)}{2n^2}=\int_{0}^{1}\sum_{n\geq 1}\frac{(-1)^{n+1}}{2n^2}x^{n-1}(1-x)^{n-1}\,dx $$ turning the series in the LHS into $$ -\frac{1}{2}\int_{0}^{1}\frac{\text{Li}_2(-x(1-x))}{x(1-x)}\,dx\stackrel{x\mapsto\frac{1+z}{2}}{=}-\int_{-1}^{1}\frac{\text{Li}_2\left(-\frac{1-z^2}{4}\right)}{1-z^2}\,dz=-\int_{0}^{1}\frac{-\text{Li}_2\left(-\frac{x}{4}\right)}{x\sqrt{1-x}}\,dx.$$ By integration by parts, the RHS depends on $$ \int_{0}^{1}\log\left(1+\frac{1-x^2}{4}\right)\log\left(\frac{1-x}{1+x}\right)\,\frac{dx}{x} $$ which by enforcing the substitution $x=\frac{1-z}{1+z}$ is mapped into $$ \int_{0}^{1}\frac{\log(z)\log\left(\frac{1+3z+z^2}{1+z}\right)}{1-z^2}\,dz. $$ The last integral can be computed from the functional relations for $\text{Li}_2$ and $\text{Li}_3$. We have $$ \int_{0}^{1}\frac{\log(z)\log(1+z)}{1-z^2}\,dz = -\frac{\pi^2}{8}\log(2)+\frac{7}{16}\zeta(3)$$ and by Feynman's trick $$ \int_{0}^{1}\frac{\log(z)\log(1+az)}{1-z^2}\,dz = \int_{0}^{a}\frac{\pi^2(3u-1)+24\,\text{Li}_2(-u)}{24(1-u^2)}\,du $$ which ensures a rather amazing cancellation.

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