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

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 linear function

un=4(2n1)=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 Mathematica that,

ζ(3)=6v1+131+13v2+231+23v3+

where the vn are now given by the cubic function

vn=4(2n1)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:

65u1v1u2v2u3v3

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=akbkbk1 for k>1, where we used μk=1bk. Applying this transformation to Tito’s CF yields the CF

321+w21+w31+

where w2k=k34(2k1)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 wk14, 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+w2w2w31+w3+w4w4w51+w5+w6

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

3254β1δ1β2δ2

where

βk=k34(2k1)3k34(2k+1)3=k616(2k1)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

6516β14δ116β24δ2

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

65u1v1u2v2u3v3

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)=3281λ1η1λ2η2λ3

where

ηk=4×(4k4+8k3+k23k)3=4×103,4×1263,λk=8×(68k645k4+12k21)=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

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