Apery-like formulas for zeta(2n)

3 Jun

Apery-like formulas for zeta(2n)

Posted June 3, 2012 by tpiezas in equations, mathematics. Tagged: binomial, Riemann zeta function. Leave a Comment

It is well-known that,

$\begin{aligned}\zeta(2) &= 3\sum_{k=1}^\infty \frac{1}{k^2\,\binom{2k}k}\end{aligned}$

D. Bailey, J. Borwein, D. Bradley gave a generalization. First define,

$\begin{aligned}&A(a_0) = \sum_{k=1}^\infty \frac{1}{k^{a_0}\,\binom{2k}k}\\ &A(a_0, a_1, a_2,\dots) = \sum_{k=1}^\infty \frac{1}{k^{a_0}\,\binom{2k}k} \sum_{p=1}^{k-1}\frac{1}{p^{a_1}} \sum_{q=1}^{k-1} \frac{1}{q^{a_2}}\dots\end{aligned}$

Obviously,

$\zeta(2) = 3A(2)$

However, a little experiment with Mathematica’s LatticeReduce command will show there are two solutions for $\zeta(4)$ ,

$\begin{aligned} &a\big(\zeta(4)-3A(4)+9A(2,2)\big) =0\\ &b\big(5\zeta(4)-10A(4)-6A(2,2)\big) = 0\end{aligned}$

where {a, b} are scaling variables. Adding the two together,

$(a-5b)\zeta(4)-(3a-10b)A(4)+3(3a+2b)A(2,2) = 0$

hence there are an infinite number of solutions. For appropriately chosen {a,b}, we can also eliminate one term. Thus,

$\begin{aligned} \zeta(4) &= \frac{36}{17}A(4)\\ &=\frac{108}{5}A(2,2)\end{aligned}$

For $\zeta(6)$ , there are now three solutions. Given the five terms,

$\zeta(6),\, A(6),\, A(4,2),\, A(2,4),\, A(2,2,2)$

then the coefficients such that their sum is equal to zero are,

$\begin{aligned} &\text{1st sol:}\; (5, -9, 1, -15, 3)\\ &\text{2nd sol:}\; (2, -7, 23, 9, 15)\\ &\text{3rd sol:}\;\, (10,-17,-3,6,-36)\end{aligned}$

Using the same approach above, we can eliminate two of the terms. One solution has an interesting number pop up,

$163\zeta(6) = 288A(6)+432A(2,4)$

though the appearance of 163 is probably only a coincidence. See Bailey, Borwein, Bradley’s paper for more details.

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