(This is related to a recent post.)

In Mathworld's Figure Eight knot, it mentioned several formulas for the hyperbolic volume $V = 2\kappa$ where $\kappa$ is Gieseking's constant $\kappa \approx 1.01494$. Mathworld gave the BBP-type formulas,

$$\kappa = \frac{\sqrt3}{3^2} \small\sum_{k=0}^\infty \left(\frac{-1}{3^3}\right)^k \left(\frac{3^2}{(6k+1)^2}-\frac{3^2}{(6k+2)^2}-\frac{3\times4}{(6k+3)^2}-\frac3{(6k+4)^2}+\frac1{(6k+5)^2}\right)$$

$$\kappa\, = \frac{\sqrt3}{3^5} \small\sum_{k=0}^\infty \left(\frac1{3^6}\right)^k \left(\frac{3^5}{(12k+1)^2}-\frac{3^5}{(12k+2)^2}-\frac{3^4\times4}{(12k+3)^2}-\dots-\frac1{(12k+11)^2}\right)$$

$$\kappa\, =\; \frac{\sqrt3}{3^{11}}\small\sum_{k=0}^\infty \left(\frac1{3^{12}}\right)^k \left(\frac{3^{11}}{(24k+1)^2}-\frac{3^{11}}{(24k+2)^2}-\frac{3^{10}\times4}{(24k+3)^2}-\dots-\frac1{(24k+23)^2}\right)$$

This looks like the beginning of a pattern, missing the alternating formula for $\frac1{18k+a}$. After a lot of experimentation as the exponents and negative signs were tricky, it seems the general formula is,

$$\kappa = \frac{\sqrt3\,}{3^{3n-1}}\sum_{k=0}^\infty \left(\frac{(-1)^n}{3^{3n}}\right)^k\Big(P_1(k)+P_2(k)\Big)\tag1$$


$$P_1(k)=\sum_{p=1}^\color{blue}5 (-1)^{\alpha} \sum_{q=0}^{n-1} \frac{3^{\beta}}{(6nk+p+6q)^2}\,(-1)^{q+1}$$

$$P_2(k) = \sum_{r=0}^{n-1} \frac{3^{\gamma}}{(6nk+6r+3)^2}\,(-1)^{r+1}$$

and exponents,

$$\alpha = (p+6)\lfloor (p+6)/2 \rfloor\\ \beta =\lfloor (6n-p)/2 \rfloor-3q\\ \gamma = 3n-3r-1$$

with floor function $\lfloor x\rfloor$.

Note that the Mathworld formulas has $4$ as numerators. But since $4-1=3$, then separating $P_1(k)$ and $P_2(k)$ allows numerators that are purely powers of $3^n$.

I've tested it with $1<n<20$ and it works fine.

Q: How do we prove $(1)$ in fact is true for all integer $n\ge1$?


1 Answer 1


Considering that $\operatorname{Cl}_2$ is a series about $\sin$, I think it should be written in the following form:

$$ \begin{aligned} P_1(k)&=(-1)^{k+1}\sum_{n=0}^∞ i^{\left(1+(-1)^k\right) n} \sum_{m=0}^{3k+3}\frac{\cos(πm)\sin \left(\frac{2πm}{3}\right)}{(3nk+3n+3k-m+3)^2}\\ P_2(k)&=\left(-\frac{1}{27}\right)^k \left(\frac{2}{3}\right)^2\underset{n=0}{\overset{\infty }{\sum }}\frac{i^{\left(1+(-1)^k\right) n} }{27^{kn+n}} \underset{m=0}{\overset{6 k+6}{\sum }}\frac{3^{m/2} \cos \left(\frac{m π }{3}\right) \sin \left(\frac{m π }{6}\right)}{(6nk+6n+6k-m+6)^2} \end{aligned} $$

$$ κ=P_1(k)=P_2(k)=\operatorname{Cl}_2\left(\frac{π}{3}\right), ∀k∈\mathbb{N}\\ $$

You can use this code for numerical validation:

    I^((1 + (-1)^k )*n)*Sum[
    (Cos[m*Pi] Sin[(2m*Pi)/3])/
    (3 + 3k - m + 3n + 3k*n)^2, 
    {m, 0, 3 + 3k}], 
    {n, 0, Infinity}
Clausen6[k_] := (-27^(-1))^k*(2/3)^2NSum[
    I^((1 + (-1)^k)*n)*3^(-3n - 3k*n)* Sum[
    (6 + 6k - m + 6n + 6k*n)^2, 
    {m, 0, 6 + 6*k}], 
    {n, 0, Infinity}

Here is a detailed proof of $P_2$.

$P_1$ can be proved in the same way, with the only difference being the upper limit of the integral.


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