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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.

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The following Mathematica code computes 2*len terms of cfrac. Apery2[len_] := Module[{res, vfun = 4 (2 # - 1)^3 &}, res = vfun[len + 1]; Do[res = vfun[k] + Divide[k^3, 1 + Divide[k^3, res]], {k, len, 1, -1}]; Divide[6, res]]. With this code I checked Apery2[330000] == N[Zeta[3], 10^6] gives True. Computation took about 6 minutes. –  Sasha May 4 '12 at 19:40
    
Thanks, Sasha! I used a rather crude Mathematica code of looking for $v_n = an^3+bn^2+cn+d$ with unknown $(a,b,c,d)$. With a limited number of terms, its difference from $\zeta(3)$ was $10^{-50}$, several magnitudes MUCH less than the next best candidate which was $10^{-7}$. Hence I knew this polynomial had to be it. But a rigorous proof is still needed. –  Tito Piezas III May 4 '12 at 20:07
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1 Answer

up vote 20 down vote accepted

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:

$$\zeta(3)=\frac32-\cfrac{81}{272-\cfrac{\eta_1}{\lambda_1-\cfrac{\eta_2}{\lambda_2-\cdots}}}$$

where

$$\begin{align*} \eta_k&=4k^3(4k^3+8k^2+k-3)^3\\ \lambda_k&=8(68k^6+408k^5+975k^4+1180k^3+762k^2+252k+34) \end{align*}$$

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

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5  
Since $v_{k}=\left( 17k^{2}+17k+5\right) (2k+1)=34k^{3}+51k^{2}+27k+5$, this even part is exactly Apéry's continued fraction expansion for $\zeta (3)$ $$\begin{equation*} \zeta (3)=\dfrac{6}{5+\overset{\infty }{\underset{n=1}{\mathbb{K}}}\dfrac{-n^{6}}{34n^{3}+51n^{2}+27n+5}}, \end{equation*}$$ which can be found in section 7 of this van der Poorten's paper. –  Américo Tavares May 5 '12 at 12:51
2  
@Américo: I guess that completes the proof of convergence to $\zeta(3)$. As I said, it was derived from the CF Tito had; both the original CF and the even part converge, so they should converge to the same value. –  J. M. May 5 '12 at 12:54
    
Right! Perhaps you might add the derivation of this even part. –  Américo Tavares May 5 '12 at 12:56
1  
Thanks, J.M. Yes, it certainly was useful. I didn't realize one could derive Apery's cfrac from it! :-) –  Tito Piezas III May 5 '12 at 15:01
2  
Thanks Américo! I couldn't have made it complete without your comment... –  J. M. May 5 '12 at 15:51
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