Evaluation of infinite series of ratios (with denominators given as finite products) in terms of rational numbers and $\pi\sqrt3$ How can we prove whether
$$\sum_{k=0}^\infty \frac{1}{\prod_{i=1}^{6n+2}(3k+i)}=q_1+q_2\sqrt{3}\pi$$
for all natural $n$ with rational $q_1$ and $q_2$?

Some series related to rational approximations to $\sqrt{3}\pi$ are:
$$\sum_{k=0}^\infty \frac{1}{(3k+1)(3k+2)}=\sum_{k=0}^\infty \frac{1}{\prod_{i=1}^2(3k+i)}=\frac{\pi}{3\sqrt{3}}$$
$$\sum_{k=0}^\infty \frac{1}{\prod_{i=1}^8(3k+i)}=\frac{329-60\sqrt{3}\pi}{100800}$$
$$\sum_{k=0}^\infty \frac{1}{\prod_{i=1}^{14}(3k+i)}=\frac{748440\sqrt{3}\pi-4071899}{57537672192000}$$
 A: Let us consider the series $\frac{1}{1-x^3} = \sum_{k=0}^\infty x^{3k}$.  If you take the double integral with base 0, then you get $\sum_{k=0}^\infty \frac{x^{3k+2}}{(3k+1)(3k+2)}$, and this function's value at $x=1$ is the first sum.  Likewise, if you take the eighth iterated integral, you get $\sum_{k=0}^\infty \frac{x^{3k+8}}{\prod_{i=1}^8(3k+i)}$, and this function's value at $x=1$ is the second sum; and so on.
However, there is an alternate formula for the iterated integral: namely, the $n$th iterated integral of $f(x)$ is equal to $\frac{1}{(n-1)!} \int_0^x (x-t)^{n-1} f(t)\,dt$.  Therefore, the sums in question are equal to $\frac{1}{(6k+1)!} \int_0^1 \frac{(1-t)^{6k+1}}{1-t^3}dt = \frac{1}{(6k+1)!} \int_0^1 \frac{(1-t)^{6k}}{1+t+t^2} dt$.
On the other hand, it is straightforward to check that $(1-t)^{6} \equiv -27 \pmod{1+t+t^2}$.  (One shortcut: first show that $(1-t)^{6} \equiv -18 + 9t + 9t^2 \pmod{1-t^3}$.)  Therefore, $(1-t)^{6k} = p(t) (1+t+t^2) + (-27)^k$ for some $p \in \mathbb{Z}[t]$, and $\frac{1}{(6k+1)!} \int_0^1 \frac{(1-t)^{6k}}{1+t+t^2} = \frac{1}{(6k+1)!} \int_0^1 \left(p(t) + \frac{(-27)^k}{1+t+t^2}\right) dt$.  From this expression, it should be easy to see why this number is in $\mathbb{Q} + \mathbb{Q} \cdot (\pi \sqrt{3})$.
A: Probably not very useful.
$$a_n=\prod_{i=1}^{6n+2}(3k+i)=(3 k+1)_{6 n+2}$$ where appears Pochhammer symbols.
$$\sum_{k=0}^\infty \frac{1}{a_n}=\frac{\, _4F_3\left(\frac{1}{3},\frac{2}{3},1,1;2 n+1,2 n+\frac{4}{3},2
   n+\frac{5}{3};1\right)}{(1)_{6 n+2}}$$ where appears in numerator the generalized hypergeometric function.
From here, it would be necessary to prove that, because of its arguments, the numerator itself write as $\alpha_n + \beta_n \pi \sqrt 3$, $\alpha_n$ and $\beta_n$ being rational numbers. 
