What a summation

it is easy to show that $\int\int \cdots \int_{0 < x_1 < x_2 < \cdots < x_n} e^{-(a_1x_n + a_2x_{n-1} + \cdots + a_{n-1}x_2 + a_nx_1)}\,dx_1\,dx_2\,\cdots\,dx_n \; = \; \frac{1}{a_1(a_1+a_2)\cdots(a_1+a_2+\cdots a_n)}$ for all $a_1,a_2,...,a_n > 0$ . Thus $\begin{aligned} A_n & = \; \sum_{k_1,k_2,...,k_n=1}^\infty \frac{1}{k_1^2(k_1^2+k_2^2) \cdots (k_1^2+k_2^2 +\cdots k_n^2)} \\ & = \; \sum_{k_1,k_2,...,k_n=1}^\infty \int\int \cdots \int_{0 < x_1 < x_2 < \cdots < x_n} e^{-(k_1^2x_n + k_2^2x_{n-1} + \cdots + k_{n-1}^2x_2 + k_n^2x_1)}\,dx_1\,dx_2\,\cdots\,dx_n \\ & = \; \int\int \cdots \int_{0 < x_1 < x_2 < \cdots < x_n} F(x_1)F(x_2)\cdots F(x_n)\,dx_1\,dx_2\,\cdots\,dx_n \end{aligned}$ for any $n \in \mathbb{N}$ , where $F(x) \; = \; \sum_{k=1}^\infty e^{-k^2x}$ By symmetry it is clear that $A_n \; = \; \frac{1}{n!} \int_0^\infty\int_0^\infty\cdots \int_0^\infty F(x_1)F(x_2)\cdots F(x_n)\,dx_1\,dx_2\,\cdots\,dx_n \; = \; \frac{1}{n!}\left(\int_0^\infty F(x)\,dx\right)^n \; = \; \frac{1}{n!}\left(\sum_{k=1}^\infty k^{-2}\right)^2 \; = \; \frac{1}{n!}\left(\tfrac16\pi^2\right)^n$ and hence $\sum_{n=1}^\infty A_n \; = \; e^{\frac16\pi^2} - 1$ making the answer $2+6 = \boxed{8}$ .