As a background, Ramanujan also gave a continued fraction for ζ(3) as
where the sequence of un, starting with n=1, is given by the linear function
This has rather slow convergence. Using an approach similar to Apéry’s of finding a faster converging version, I found via Mathematica that,
where the vn are now given by the cubic function
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.
Here’s a nice little Mathematica routine for evaluating Tito’s continued fraction with precision
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.
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:
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
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
with some sequence μk,k>0 looks like this:
Now, given a CF
one can transform this into a CF of the form
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
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
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
its even part is the CF
Thus, the even part of the previously transformed CF is given by
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
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
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:
The formula is somewhat more complicated, and converges at the same rate as the even part.