Harmonic Trace

We note that $\begin{aligned} \left(\begin{array}{ccccc|c} & & & & & 0 \\ & & & & & \vdots \\ & & B(n)^{-1} & & & \vdots \\ & & & & & 0 \\ & & & & & 0 \\\hline 0 & 0 & \cdots & 0 & -(n+1) & n+1 \end{array}\right) B(n+1) & = \; \left(\begin{array}{ccccc|c} & & & & & 0 \\ & & & & & \vdots \\ & & B(n)^{-1} & & & \vdots \\ & & & & & 0 \\ & & & & & 0 \\\hline 0 & 0 & \cdots & 0 & -(n+1) & n+1 \end{array}\right) \left(\begin{array}{ccccc|c} & & & & & H_1 \\ & & & & & H_2 \\ & & B(n) & & & \vdots \\ & & & & & H_{n-1} \\ & & & & & H_n \\\hline H_1 & H_2 & \cdots & H_{n-1} & H_n & H_{n+1} \end{array}\right) \\[4ex] & = \; \left(\begin{array}{ccccc|c} & & & & & 0 \\ & & & & & \vdots \\ & & I_n & & & \vdots \\ & & & & & 0 \\ & & & & & 1 \\\hline 0 & 0 & \cdots & 0 & 0 & 1 \end{array}\right) \end{aligned}$ and hence $\begin{aligned} B(n+1)^{-1} & = \; \left(\begin{array}{ccccc|c} & & & & & 0 \\ & & & & & \vdots \\ & & I_n & & & \vdots \\ & & & & & 0 \\ & & & & & -1 \\\hline 0 & 0 & \cdots & 0 & 0 & 1 \end{array}\right) \left(\begin{array}{ccccc|c} & & & & & 0 \\ & & & & & \vdots \\ & & B(n)^{-1} & & & \vdots \\ & & & & & 0 \\ & & & & & 0 \\\hline 0 & 0 & \cdots & 0 & -(n+1) & n+1 \end{array}\right) \\[4ex] & = \; \left(\begin{array}{ccccc|c} & & & & & 0 \\ & & & & & \vdots \\ & & B(n)^{-1} + (n+1)K(n) & & & \vdots \\ & & & & & 0 \\ & & & & & -(n+1) \\\hline 0 & 0 & \cdots & 0 & -(n+1) & n+1 \end{array}\right) \end{aligned}$ where $K(n)$ is the $n \times n$ matrix all of whose components are $0$ , except for $1$ in the bottom-right corner. Thus $\mathrm{Tr}\big[B(n+1)^{-1}\big] \; = \; \mathrm{Tr}\big[B(n)^{-1} +(n+1)K(n)\big] + n + 1 \; = \; \mathrm{Tr}\big[B(n)^{-1}\big] + 2(n+1)$ and hence we deduce that $\mathrm{Tr}\big[B(n)^{-1}\big] \; = \; 1 + 2\sum_{r=1}^{n-1}(r+1) \; = \; n^2 + n - 1 \hspace{2cm} n \in \mathbb{N}$ so that $\sum_{n=1}^\infty \frac{1}{4\,\mathrm{Tr}\big[B(n)^{-1}\big] + 5} \; = \; \sum_{n=1}^\infty \frac{1}{(2n+1)^2} \; = \; \tfrac18\pi^2 - 1$ making the answer $8 + 2 + 1 = \boxed{11}$ .