## Zudilin’s continued fraction for Zeta(4)

Euler proved the following general continued fraction formula,

$\frac{1}{u_1}+\frac{1}{u_2}+\frac{1}{u_3} = \cfrac{1}{u_1 - \cfrac{u_1^2}{u_1+u_2 - \cfrac{u_2^2}{u_2+u_3 - \cfrac{u_3^2}{u_3+u_4 - \ddots}}}}$

which automatically gives a representation for the Riemann zeta function $\zeta(s)$.  However, the convergence is rather slow.  Apery found a much faster version for $\zeta(3)$ and proved its irrationality.  The status for other odd s, however, remains open.  While it has already been proved irrational for all even s, Wadim Zudilin nonetheless found an interesting faster version for $\zeta(4)$.

First, consider the following known binomial sums,

$\sum_{n=1}^\infty \frac{1}{n\,\binom{2n}n} = \frac{1}{3\sqrt{3}}\,\pi$

$\sum_{n=1}^\infty \frac{1}{n^2\,\binom{2n}n} = \frac{1}{3}\,\zeta(2)$

$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^3\,\binom{2n}n} = \frac{2}{5}\,\zeta(3)$

$\sum_{n=1}^\infty \frac{1}{n^4\,\binom{2n}n} = \frac{17}{36}\,\zeta(4)$

This nice pattern stops at s = 4.  (There are sums for other s, but they do not have this succinct form.)  It is suggestive then that $\zeta(4)$ belongs to the same family and may share certain similarities.

Zudilim found that,

$\zeta(4) = \cfrac{13}{p_0 + \cfrac{1^8\, q_1}{p_1+ \cfrac{2^8\,q_2}{p_2+\cfrac{3^8\,q_3}{p_3 + \ddots}}}}$

where, starting with n = 0,

$p_n = 3(2n+1)(3n^2+3n+1)(15n^2+15n+4) = 12, 2142, 26790, 142968,\dots$

$q_n = 3(3n-1)(3n+1) = -3, 24, 105, 240, 429\dots$

(Note that the cfrac does not use $q_0$.)  Like Apery’s accelerated version for $\zeta(2)$, the partial denominators use addition, in contrast to Euler’s form which has subtraction.  Also, like its siblings, it obeys a recurrence relation,

$(n+1)^5U_{n+1}=p_n U_n+n^3q_nU_{n-1}$

Starting with $U_0=1, U_1=12$, one gets the sequence,

$U_n = 1, 12, 804, 88680, \dots$

which is not yet in the OEIS.  In Theorem 4, Zudilin mentions they are positive rationals, but some Mathematica experimentation will show that apparently all the $U_i$ are integral.  (Proof anyone?)  Furthermore, they have the limit,

$\lim_{n \to \infty} \frac{U_{n+1}}{U_n} = (3+2\sqrt{3})^3$

For more details, refer to Zudilin’s An Apery-like difference equation for Catalan’s constant.  Now if only someone will do something more about $\zeta(5)$