A Trace of Maximum

Note that the components in right-hand column and the bottom row of $C(n)$ are all equal to $n^{-1}$ . Then we have $\begin{aligned} \left(\begin{array}{ccccc|c} & & & & & 0 \\ & & & & & \vdots \\ & & C(n)^{-1} & & & \vdots \\ & & & & & 0 \\ & & & & & 0 \\\hline 0 & 0 & \cdots & 0 & -n(n+1) & (n+1)^2 \end{array}\right)C(n+1) & = \; \left(\begin{array}{ccccc|c} & & & & & 0 \\ & & & & & \vdots \\ & & C(n)^{-1} & & & \vdots \\ & & & & & 0 \\ & & & & & 0 \\\hline 0 & 0 & \cdots & 0 & -n(n+1) & (n+1)^2 \end{array}\right) \left(\begin{array}{ccccc|c} & & & & & \frac{1}{n+1} \\ & & & & & \vdots \\ & & C(n) & & & \vdots \\ & & & & & \frac{1}{n+1} \\ & & & & & \frac{1}{n+1} \\\hline \frac{1}{n+1} & \frac{1}{n+1} & \cdots & \frac{1}{n+1} & \frac{1}{n+1} & \frac{1}{n+1} \end{array}\right) \\ & = \; \left(\begin{array}{ccccc|c} & & & & & 0 \\ & & & & & \vdots \\ & & I_n & & & \vdots \\ & & & & & 0 \\ & & & & & \frac{n}{n+1} \\\hline 0 & 0 & \cdots & 0 & 0 & 1 \end{array}\right) \end{aligned}$ so that $\begin{aligned} C(n+1)^{-1} & = \; \left(\begin{array}{ccccc|c} & & & & & 0 \\ & & & & & \vdots \\ & & I_n & & & \vdots \\ & & & & & 0 \\ & & & & & -\frac{n}{n+1} \\\hline 0 & \cdots & \cdots & 0 & 0 & 1 \end{array}\right) \left(\begin{array}{ccccc|c} & & & & & 0 \\ & & & & & \vdots \\ & & C(n)^{-1} & & & \vdots \\ & & & & & 0 \\ & & & & & 0 \\\hline 0 & 0 & \cdots & 0 & -n(n+1) & (n+1)^2 \end{array}\right) \\ & = \; \left(\begin{array}{ccccc|c} & & & & & 0 \\ & & & & & \vdots \\ & & C(n)^{-1} + n^2J(n) & & & \vdots \\ & & & & & 0 \\ & & & & & -n(n+1) \\\hline 0 & 0 & \cdots & 0 & -n(n+1) & (n+1)^2 \end{array}\right) \end{aligned}$ where $J(n)$ is the $n \times n$ matrix with $1$ in the bottom-right corner, and $0$ everywhere else. Thus $\mathrm{Tr}[C(n+1)^{-1}] \; = \; \mathrm{Tr}[C(n)^{-1} + n^2J(n)] + (n+1)^2 \; = \; \mathrm{Tr}[C(n)^{-1}] + n^2 + (n+1)^2$ and hence $\mathrm{Tr}[C(n)^{-1}] = \sum_{r=1}^n r^2 + \sum_{r=1}^{n-1}r^2 = \tfrac13n(2n^2+1) \hspace{2cm} n \ge 1$ Finally, then, $\sum_{n=2}^\infty \frac{1}{\mathrm{Tr}[C(n)^{-1}] - n} \; = \; \sum_{n=2}^\infty \frac{3}{2(n-1)n(n+1)} \; = \frac34\sum_{n=2}^\infty\left(\frac{1}{(n-1)n} - \frac{1}{n(n+1)}\right) \; = \; \frac38$ making the answer $3 + 8 = \boxed{11}$ .