Continued fraction for Apéry's constant conjectured by The Ramanujan Machine

Solution 1:

Finally! Found this result buried in the literature... In a French article Sur l’accélération de la convergence de certaines fractions continues by Christian Batut and Michel Olivier from 1980, a method for accelerating continued fractions for series of form $\sum_{n=1}^{\infty}\frac{1}{f(n)}$ is described. In examples we then find accelerated version of well-known $\sum_{n=1}^{\infty}\frac{1}{(2n-1)^3}=\frac{7}{8} \zeta(3)$ which gives the desired continued fraction (the accelaration process and meaning of sequences $a_{n,k},b_{n,k},r_{n,k},R_{n,k},d_{n,k}$ is described earlier in the article):

3.2.1 - Pour la série $S=\sum_{n=1}^{\infty}\frac{1}{f(n)}$, on obtient: \begin{align} a_{n+1,k}&=f(n)+f(n+1)+2ak(k+1)(2n+1)\\ b_{n+1,k}&=-f(n)^2\\ r_{n,k}&=-an^3+2a(k+1)n^2-[2a(k+1)^2+b]n+(k+1)[(k+1)^2a+b]\\ R_{n,k}&=an^3+2a(k+1)n^2+[2a(k+1)^2+b]n+(k+1)[(k+1)^2a+b]\\ d_{n,k}&=-(k+1)^2[(k+1)^2a+b]^2 \end{align} Notons que: $r_{n+1,k-1}=-r_{k,n}$, d'oú il résulte que le tableau des $(\frac{p_{n,k}}{q_{n,k}})_{n \geq 0, k \geq 0}$ est symétrique.

3.2.2 - Cas particuliers

En remplacant dans le cas précédent, $n$ par $n + \frac{b}{a}$ et en multipliant par $a^3$ on obtient les suites $(r_{n,k}),(a_{n,k}),(b_{n,k})$ et $(R_{n,k})$ correspondant au polynome $f(n)=(an+b)^3$.

$\vdots$

3.2.4 - $S=\sum_{n=1}^{\infty}\frac{1}{(2n-1)^3}$

Dans (3.2.2) on choisit $a=2, b=-1$. On construit ainsi la fraction continue: $$ \frac{7}{8} \zeta(3)=\frac{1}{p(0)-\frac{1^6}{p(1)-\frac{2^6}{p(2)-\ddots}}} $$ ou $p(n)=6n^3+9n^2+5n+1$.

Here $6n^3+9n^2+5n+1=(2n+1)(3n^2+3n+1)$.