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

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: