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Doubt Question about proof of irrationality of $\zeta(3)$

Post Closed as "Not suitable for this site" by Vladimir Dotsenko, user6976, Alex M., Tom De Medts, GH from MO
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Doubt about proof of irrationality of $\zeta(3)$

I'm reading this article of Henri Cohen about Apery's proof of the irrationality of $\zeta(3)$ but I don't really get the details of "THEOREME 1". My first doubt is about the relation $a_n \sim A \alpha^n n^{-3/2}$.

I know that if $a_n$ fulfilled the relation $a_n-34a_{n-1}+1=0$ then as its characteristic polynomial is $x^2-34x+1$ and as $\alpha$ is one of its roots, if we denote by $\bar{\alpha}$ the second root, then we would have $a_n=A_1\alpha^n+A_2\bar{\alpha}^n$.

Then, as $0<\bar{\alpha}<1$ we have that $a_n/\alpha^n \longrightarrow A_1$.

However, the relation for $a_n$ is $$a_n-(34-51n^{1}+27n^{-2}-5n^{-3})+(n-1)^3n^{-3}a_{n-2}=0$$ and I don't know how can we rigurously take care of the extra terms.

Furthermore, how does one get the extra $n^{-3/2}$ term?

Secondly, why does that relation implies that $\zeta(3)-a_n/b_n = O(\alpha^{-2n})?$

After that, it stays that it can be shown that from the prime number theorem we have that $\log d_n \sim n$ where $d_n=\text{lcm}(1,2, \cdots, n)$.

I've managed to prove that $$\dfrac{\log d_n}{n} \leq \pi(n) \dfrac{\log n}{n} $$ but I'm not able to prove that $\log d_n/n$ converges to $1$.

Lastly, I don't know how from this last result it is true that for any $\varepsilon >0$ we get $$\zeta(3)-\dfrac{p_n}{q_n}=O(q_n^{-r-\varepsilon})$$

I'm not really good at asymptotic behaivour and big-O notation so I would really appreciate if someone could answer with rigurous and detailed explanations.

Thank you very much.