Continued fractions for Zeta(2) and Zeta(3)

It seems there is a nice “pattern” between the continued fractions for the Riemann zeta function,

\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} = \frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+\dots

at s = 2, and s = 3.  First though, a little introduction.

The origins of this function go back to 1644 when, at the tender age of 18, the Italian mathematician Pietro Mengoli (1626-1686) first proposed what would be later known as the Basel Problem, namely to determine the exact value of the sum of the reciprocals of the squares.  Euler would later find that, for n a positive integer, then \zeta(2n) is a rational multiple of \pi^{2n}, with the first case s = 2 as,

\zeta(2) = \frac{\pi^2}{6}

which obviously is irrational.  However, the status of odd s was not as easy to resolve.  It was only in 1979 that Apery proved that \zeta(3) is irrational, which henceforth was called Apery’s constant.

I. Zeta(2)

One of its infinite number of continued fractions can be given as,

m\,\zeta(2) = \cfrac{1}{u_1+\cfrac{1^4}{u_2+\cfrac{2^4}{u_3+\cfrac{3^4}{u_4+\ddots}}}}

where m = \frac{1}{2} and the u_n, starting with n = 1, are generated by,

u_n = 2n-1 = 1, 3, 5, 7,\dots

or simply the odd numbers.  The convergence is slow, but Apery found it can accelerated by using a quadratic function,

u_n = 11n^2-11n+3 = 3, 25, 69, 135,\dots

but now m = \frac{1}{5}.

II. Zeta(3)

Likewise, this also has an infinite number of continued fractions (see Ramanujan’s versions here), but one important form is,

m\,\zeta(3) = \cfrac{1}{v_1-\cfrac{1^6}{v_2-\cfrac{2^6}{v_3-\cfrac{3^6}{v_4-\ddots}}}}

where m = 1 and the v_n, again starting with n = 1, are,

v_n = (n-1)^3+n^3 = (2n-1)(n^2-n+1) = 1, 9, 35, 91, \dots

Apery again found an accelerated version,

v_n = 34n^3-51n^2+27n-5 = (2n-1)(17n^2-17n+5) = 5, 117, 535, 1463, \dots

where now m = \frac{1}{6}, and established that its rate of convergence was such that \zeta(3) could not be a ratio of two integers.

III. Connection between Zeta(2) and Zeta(3)

Define the two sequences,

B_n = \sum_{k=0}^n {\binom n k}^2 {\binom {n+k}k} = 1, 3, 19, 147, 1251,\dots

A_n = \sum_{k=0}^n {\binom n k}^2 {\binom {n+k}k}^2 = 1, 5, 73, 1445, 33001, \dots

then it was established that these Apery numbers have the recurrence relations,

n^2 B_n = (11n^2-11n+3)B_{n-1}+(n-1)^2B_{n-2}

n^3 A_n = (34n^3-51n^2+27n-5)A_{n-1}-(n-1)^3A_{n-2}

Interesting that the same polynomials pop up, isnt’ it?  Furthermore, their limiting ratios have the common form,


thus for b = 1, 2,

\lim_{n \to \infty} \frac{B_{n+1}}{B_n} = \left(\frac{1+\sqrt{5}}{2}\right)^5 = 11.0901\dots

\lim_{n \to \infty} \frac{A_{n+1}}{A_n} = \left(1+\sqrt{2}\right)^4 = 33.9705\dots

hence the golden ratio and the silver ratio surprisingly turn up in the continued fractions for \zeta(2) and \zeta(3), respectively.  It is easy to check other sequences,

C_n = \sum_{k=0}^n {\binom n k}^p {\binom {n+k}k}^q

for some small {p, q}, but there are no other recurrence relations similar in form to the two above.

IV. Zeta(5)

There is an orderly non-simple continued fraction for all \zeta(s), with the next odd s as,

m \zeta(5) = \cfrac{1}{w_1-\cfrac{1^{10}}{w_2-\cfrac{2^{10}}{w_3-\cfrac{3^{10}}{w_4-\ddots}}}}

where m = 1 and the w_n are,

w_n = (n-1)^5+n^5 = (2n-1)(n^4-2n^3+4n^2-3n+1) = 1, 33, 275, 1267, \dots

Unfortunately, no one has yet found an accelerated version where the w_n are generated by a 5th degree (or higher) polynomial, and m is rational.

Can you find one?

For further reading, refer to Alfred van der Poorten’s excellent, A Proof That Euler Missed.

One response to this post.

  1. […] About « Continued fractions for Zeta(2) and Zeta(3) […]


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