Ordinary problem

A Dirichlet series which involves divisor function is $\displaystyle \sum_{n=1}^\infty \frac{\sigma_a (n) }{n^s} = \zeta(s) \zeta(s-a)$ for $s > 1$ , $s > a + 1$ . Therefore, $\begin{aligned} S & = \sum_{n=1}^\infty \frac{\sigma_4 (n)}{n^8} = \zeta(8) \zeta(4) \end{aligned}$ , where $\zeta(x)$ is the Riemann zeta function.

The relation between Bernoulli numbers $B_n^*$ and Riemann zeta function is $B_n^* = \dfrac{2(2n)!}{(2\pi)^{2n}} \zeta(2n)$ , then we have:

$\begin{aligned} \zeta (2n) & = \frac{(2\pi)^{2n}}{2(2n)!}B_n^* \\ \Rightarrow \zeta(4) & = \frac{2^4 \pi^4}{2\cdot{} 4!} \cdot{} \color{#3D99F6}{\frac{1}{30}} = \frac{\pi^4}{90} \quad \quad \quad \small \color{#3D99F6}{B_2^* = \frac{1}{30}} \\ \Rightarrow \zeta(8) & = \frac{2^8\pi^4}{2\cdot{} 8!} \cdot{} \color{#3D99F6}{\frac{1}{30}} = \frac{\pi^4}{9450} \quad \quad \small \color{#3D99F6}{B_4^* = \frac{1}{30}} \end{aligned}$

$\Rightarrow S = \dfrac{\pi^8} {9450} \cdot{} \dfrac{\pi^4}{90} = \dfrac{\pi^{12}}{850500}$

$\Rightarrow \dfrac{\pi^{12}}{S} = \boxed{850500}$

By Dirichlet convolution, the divisor function of the ath powers is the convolution of n^a and the identity. Since a=4, our convolution is n^4 and n^8. Hence the sum is zeta(4)zeta(8), which as like Chew-Seong Cheong said is pi^12/850500. Hence the answer is 850500.