## A new continued fraction for Zeta(3)?

Continuing the discussion from the previous post, Ramanujan also gave a continued fraction for $\zeta(3)$ as, $\zeta(3) = 1+\cfrac{1}{u_1+\cfrac{1^3}{1+\cfrac{1^3}{u_2+\cfrac{2^3}{1+\cfrac{2^3}{u_3 + \ddots}}}}}$

where the $u_n$, starting with n = 1, are given by the linear function, $u_n = 4(2n-1) = 4, 12, 20, 28, \dots$

(Notice the difference from the other version since this one has the cubes twice used as numerators.)  Using a similar approach to Apery’s of finding a faster converging version, I found via Mathematica that, $\zeta(3) = \cfrac{6}{v_1 + \cfrac{1^3}{1 + \cfrac{1^3}{v_2 + \cfrac{2^3}{1 + \cfrac{2^3}{v_3 +\ddots}}}}}$

where the $v_n$ are now given by the cubic function, $v_n = 4(2n-1)^3 = 4, 108, 500, 1372, \dots$

Of course, a more rigorous mathematical proof is needed that indeed the equality holds.

Update:  J.M. from mathstackexchange.com proved that the continued fraction DOES converge to $\zeta(3)$ by connecting it to Apery’s version!  One consequence of his analysis is that Apery’s generating polynomial can be seen as, $(2n-1)(17n^2-17n+5) = n^3 + (n-1)^3 + 4(2n-1)^3$