# A new continued fraction for Apéry’s constant, ζ(3)\zeta(3)?

As a background, Ramanujan also gave a continued fraction for $\zeta(3)$ as

$\zeta(3) = 1+\cfrac{1}{u_1+\cfrac{1^3}{1+\cfrac{1^3}{u_2+\cfrac{2^3}{1+\cfrac{2^3}{u_3 + \ddots}}}}}\tag{1}$

where the sequence of $u_n$, starting with $n = 1$, is given by the linear function

$u_n = 4(2n-1) = 4, 12, 20, 28, \dots$

This has rather slow convergence. Using an approach similar to Apéry’s of finding a faster converging version, I found via Mathematica that,

$\zeta(3) = \cfrac{6}{v_1 + \cfrac{1^3}{1 + \cfrac{1^3}{v_2 + \cfrac{2^3}{1 + \cfrac{2^3}{v_3 +\ddots}}}}}\tag{2}$

where the $v_n$ are now given by the cubic function

$v_n = 4(2n-1)^3 = 4, 108, 500, 1372, \dots$

Question: Can anyone prove that (2), with $v_n$ defined by the cubic function, is indeed true?

Postscript: A short description of Apéry’s accelerated continued fractions for $\zeta(2)$ and $\zeta(3)$ is given here.

Here’s a nice little Mathematica routine for evaluating Tito’s continued fraction with precision prec:

prec = 10^4;
y = N[4, prec];
c = y; d = 0; k = 1;
u = 1; v = y;
While[True,
c = 1 + u/c; d = 1/(1 + u d);
h = c*d; y *= h;
v += 96 k^2 + 8;
c = v + u/c; d = 1/(v + u d);
h = c*d; y *= h;
If[Abs[h - 1] <= 10^-prec, Break[]];
u += 3 k (k + 1) + 1;
k++];
6/y


where I used the Lentz-Thompson-Barnett method for the evaluation.

For prec = 10^4, the thing evaluates in 120 seconds (via AbsoluteTiming[]), giving a result that agrees with $\zeta(3)$ to 10,000 digits.

One can consider the even part of Tito’s CF, which converges at twice the rate of the original:

where

Here’s Mathematica code corresponding to this CF:

prec = 10^4;
y = N[5, prec];
c = y; d = 0; k = 1;
While[True,
u = k^6;
v = (2 k + 1) ((17 k + 17) k + 5);
c = v - u/c; d = 1/(v - u d);
h = c*d; y *= h;
If[Abs[h - 1] <= 10^-prec, Break[]];
k++];
6/y


For prec = 10^4, the thing evaluates in 70 seconds (via AbsoluteTiming[]). There may be further ways to accelerate the convergence of the CF, but I have yet to look into them.

## Added, quite a bit later:

As it turns out, the even part I derived is precisely Apéry’s CF for $\zeta(3)$ (thanks Américo!). Conversely put, Tito’s CF is an extension of Apéry’s CF. Here’s how to derive Apéry’s CF from Tito’s CF (while proving convergence along the way).

We start from an equivalence transformation of Tito’s CF. A general equivalence transformation of a CF

with some sequence $\mu_k, k>0$ looks like this:

Now, given a CF

one can transform this into a CF of the form

where $w_1=\dfrac{a_1}{b_1}$ and $w_k=\dfrac{a_k}{b_k b_{k-1}}$ for $k > 1$, where we used $\mu_k=\dfrac1{b_k}$. Applying this transformation to Tito’s CF yields the CF

where $w_{2k}=\dfrac{k^3}{4(2k-1)^3}$ and $w_{2k+1}=\dfrac{k^3}{4(2k+1)^3}$. (You can easily demonstrate that this transformed CF and Tito’s CF have identical convergents.)

At this point, we find that since the $w_k \leq\dfrac14$, we have convergence of the CF by Worpitzky’s theorem.

Now, we move on to extracting the even part of this transformed CF. Recall that if a CF has the sequence of convergents

then the even part is the CF whose convergents are $u_0,u_2,u_4,\dots$ (Analogously, there is the odd part with the sequence of convergents $u_1,u_3,u_5,\dots$)

Now, given a CF of the form

its even part is the CF

Thus, the even part of the previously transformed CF is given by

where

We’re almost there! We only need to perform another equivalence transformation, which I’ll split into two steps to ease understanding. First, the easy one with $\mu_k=4$, which yields the CF

The last step is to cancel out the odd integer denominators of the $\beta_k$ and $\delta_k$; to do this, we take $\mu_k=(2k+1)^3$; this finally yields the CF

where

and this is Apéry’s CF.

For completeness, I present a formula for the odd part of Tito’s CF, after some post-processing with a few equivalence transformations:

where

The formula is somewhat more complicated, and converges at the same rate as the even part.