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:

$$\cfrac{6}{5-\cfrac{u_1}{v_1-\cfrac{u_2}{v_2-\cfrac{u_3}{v_3-\cdots}}}}$$

where

$$\begin{align*}
u_k&=k^6\\
v_k&=(17k^2+17k+5)(2k+1)
\end{align*}$$

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

$$b_0+\cfrac{a_1}{b_1+\cfrac{a_2}{b_2+\cfrac{a_3}{b_3+\cdots}}}$$

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

$$b_0+\cfrac{\mu_1 a_1}{\mu_1 b_1+\cfrac{\mu_1 \mu_2 a_2}{\mu_2 b_2+\cfrac{\mu_2 \mu_3 a_3}{\mu_3 b_3+\cdots}}}$$

Now, given a CF

$$b_0+\cfrac{a_1}{b_1+\cfrac{a_2}{b_2+\cdots}}$$

one can transform this into a CF of the form

$$b_0+\cfrac{w_1}{1+\cfrac{w_2}{1+\cdots}}$$

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

$$\cfrac{\frac32}{1+\cfrac{w_2}{1+\cfrac{w_3}{1+\cdots}}}$$

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

$$u_0=b_0,u_1=b_0+\cfrac{a_1}{b_1},u_2=b_0+\cfrac{a_1}{b_1+\cfrac{a_2}{b_2}},\dots$$

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

$$b_0+\cfrac{w_1}{1+\cfrac{w_2}{1+\cdots}}$$

its even part is the CF

$$b_0+\cfrac{w_1}{1+w_2-\cfrac{w_2 w_3}{1+w_3+w_4-\cfrac{w_4 w_5}{1+w_5+w_6-\cdots}}}$$

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

$$\cfrac{\frac32}{\frac54-\cfrac{\beta_1}{\delta_1-\cfrac{\beta_2}{\delta_2-\cdots}}}$$

where

$$\begin{align*}
\beta_k&=\frac{k^3}{4(2k-1)^3}\frac{k^3}{4(2k+1)^3}=\frac{k^6}{16(2k-1)^3(2k+1)^3}\\
\delta_k&=1+\frac{k^3}{4(2k+1)^3}+\frac{(k+1)^3}{4(2k+1)^3}=\frac{17k^2+17k+5}{4(2k+1)^2}
\end{align*}$$

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

$$\cfrac{6}{5-\cfrac{16\beta_1}{4\delta_1-\cfrac{16\beta_2}{4\delta_2-\cdots}}}$$

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

$$\cfrac{6}{5-\cfrac{u_1}{v_1-\cfrac{u_2}{v_2-\cfrac{u_3}{v_3-\cdots}}}}$$

where

$$\begin{align*}
u_k&=k^6\\
v_k&=(17k^2+17k+5)(2k+1)
\end{align*}$$

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:

$$\zeta(3)=\frac32-\cfrac{81}{\lambda_1-\cfrac{\eta_1}{\lambda_2-\cfrac{\eta_2}{\lambda_3-\ddots}}}$$

where

$$\begin{align*}
\eta_k&=4\times(4k^4+8k^3+k^2-3k)^3=4\times10^3,\,4\times126^3,\dots\\
\lambda_k&=8\times(68k^6-45k^4+12k^2-1)=8\times34,\,8\times3679,\dots
\end{align*}$$

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