As a background, Ramanujan also gave a continued fraction for ζ(3) as

ζ(3)=1+1u1+131+13u2+231+23u3+⋱

where the sequence of un, starting with n=1, is given by the

linearfunctionun=4(2n−1)=4,12,20,28,…

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

Mathematicathat,ζ(3)=6v1+131+13v2+231+23v3+⋱

where the vn are now given by the

cubicfunctionvn=4(2n−1)3=4,108,500,1372,…

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

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

**Answer**

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 ζ(3) to 10,000 digits.

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

65−u1v1−u2v2−u3v3−⋯

where

uk=k6vk=(17k2+17k+5)(2k+1)

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 ζ(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

b0+a1b1+a2b2+a3b3+⋯

with some sequence μk,k>0 looks like this:

b0+μ1a1μ1b1+μ1μ2a2μ2b2+μ2μ3a3μ3b3+⋯

Now, given a CF

b0+a1b1+a2b2+⋯

one can transform this into a CF of the form

b0+w11+w21+⋯

where w1=a1b1 and wk=akbkbk−1 for k>1, where we used μk=1bk. Applying this transformation to Tito’s CF yields the CF

321+w21+w31+⋯

where w2k=k34(2k−1)3 and w2k+1=k34(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 wk≤14, 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

u0=b0,u1=b0+a1b1,u2=b0+a1b1+a2b2,…

then the even part is the CF whose convergents are u0,u2,u4,… (Analogously, there is the *odd part* with the sequence of convergents u1,u3,u5,…)

Now, given a CF of the form

b0+w11+w21+⋯

its even part is the CF

b0+w11+w2−w2w31+w3+w4−w4w51+w5+w6−⋯

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

3254−β1δ1−β2δ2−⋯

where

βk=k34(2k−1)3k34(2k+1)3=k616(2k−1)3(2k+1)3δk=1+k34(2k+1)3+(k+1)34(2k+1)3=17k2+17k+54(2k+1)2

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 μk=4, which yields the CF

65−16β14δ1−16β24δ2−⋯

The last step is to cancel out the odd integer denominators of the βk and δk; to do this, we take μk=(2k+1)3; this finally yields the CF

65−u1v1−u2v2−u3v3−⋯

where

uk=k6vk=(17k2+17k+5)(2k+1)

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:

ζ(3)=32−81λ1−η1λ2−η2λ3−⋱

where

ηk=4×(4k4+8k3+k2−3k)3=4×103,4×1263,…λk=8×(68k6−45k4+12k2−1)=8×34,8×3679,…

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

**Attribution***Source : Link , Question Author : Tito Piezas III , Answer Author : Tito Piezas III*