Tricky Trace

Let $A(n)$ be the $n\times n$ version of this matrix, and let $F_n(t) = \mathrm{det}\,\big(t I_n - A(n)\big)$ be the characteristic polynomial of the matrix $A(n)$ . Then $F_1(t) = t-1$ and $F_2(t) = t^2 - \tfrac32t - \tfrac12$ . If $n \ge 3$ we can use elementary row operations to show that $\begin{aligned} F_n(t) & = \; \left|\begin{array}{ccccc|cc} & & & & & -1 & -1 \\ & & & & & -\frac12 & -\frac12 \\ & & tI_{n-2} - A(n-2) & & & \vdots & \vdots \\ & & & & & \vdots & \vdots \\ & & & & & -\frac{1}{n-2} & -\frac{1}{n-2} \\ \hline -1 & -\frac12 & \cdots & \cdots & -\frac{1}{n-2} & t - \frac{1}{n-1} & -\frac{1}{n-1} \\ -1 & -\frac12 & \cdots & \cdots & -\frac{1}{n-2} & -\frac{1}{n-1} & t - \frac{1}{n} \end{array} \right| \; = \; \left|\begin{array}{ccccc|cc} & & & & & -1 & -1 \\ & & & & & -\frac12 & -\frac12 \\ & & tI_{n-2} - A(n-2) & & & \vdots & \vdots \\ & & & & & \vdots & \vdots \\ & & & & & -\frac{1}{n-2} & -\frac{1}{n-2} \\ \hline -1 & -\frac12 & \cdots & \cdots & -\frac{1}{n-2} & t - \frac{1}{n-1} & -\frac{1}{n-1} \\ 0 & 0 & \cdots & \cdots & 0 & -t & t + \frac{1}{n(n+1)} \end{array} \right| \\[4ex] & = \; \left(t + \tfrac{1}{n(n-1)}\right)\left|\begin{array}{ccccc|c} & & & & & -1 \\ & & & & & -\frac12 \\ & & tI_{n-2} - A(n-2) & & & \vdots \\ & & & & & \vdots \\ & & & & & -\frac{1}{n-2} \\ \hline -1 & -\frac12 & \cdots & \cdots & -\frac{1}{n-2} & t - \frac{1}{n-1} \end{array} \right| + t\left|\begin{array}{ccccc|c} & & & & & -1 \\ & & & & & -\frac12 \\ & & tI_{n-2} - A(n-2) & & & \vdots \\ & & & & & \vdots \\ & & & & & -\frac{1}{n-2} \\ \hline -1 & -\frac12 & \cdots & \cdots & -\frac{1}{n-2} & -\frac{1}{n-1} \end{array} \right| \\ & = \; \left(t + \tfrac{1}{n(n-1)}\right)F_{n-1}(t) + t\mathrm{det}\,B_{n-1}(t) \end{aligned}$ where $B_{n-1}(t)$ is the $(n-1) \times (n-1)$ matrix that only differs from $tI_{n-1} - A(n-1)$ by having $-\tfrac{1}{n-1}$ , instead of $t - \tfrac{1}{n-1}$ , as the bottom right element. In particular, then $\mathrm{det}\,B_{n-1}(0) = \mathrm{det}(-A(n-1)) = F_{n-1}(0)$ . We thus obtain the recurrence relations $F_n(0) \; = \; \tfrac{1}{n(n-1)}F_{n-1}(0) \hspace{2cm} F_n'(0) \; = \; \tfrac{1}{n(n-1)}F_{n-1}'(0) + 2F_{n-1}(0) \hspace{2cm} n \ge 3$ We can solve these recurrence relations to deduce that $n!(n-1)! F_n(0) \; = \; -1 \hspace{1cm} , \hspace{1cm} n!(n-1)!F_n'(0) \; =\; 1 - \tfrac23(n-1)n(n+1) \hspace{2.5cm} n \ge 1$

Since $A(n)$ is a symmetric matrix, it is diagonalisable, and hence, if $F_n(t) \; = \; t^n + a_{n-1}t^{n-1} + a_{n-2}t^{n-2} + \cdots + a_1t + a_0$ then $\mathrm{Tr}\,A(n)$ is just the sum of the eigenvalues of $A(n)$ , and hence equal to $-a_{n-1}$ . The matrix $A(n)^{-1}$ is also symmetric, and its characteristic polynomial is the reverse of $F_n(t)$ , namely $\frac{1}{a_0}\left(a_0t^n + a_1t^{n-1} + \cdots + a_{n-1}t + a_n\right)$ and hence we deduce that $t_n \; = \; \mathrm{Tr}\,A(n)^{-1} \; = \; -\frac{a_1}{a_0} \; = \; -\frac{F_n'(0)}{F_n(0)} \; = \; 1 - \tfrac23(n-1)n(n+1) \hspace{2cm} n \ge 1$ and hence, finally, $\sum_{n=1}^\infty \frac{1}{|t_n|+1} \; = \; \frac12 + \sum_{n=2}^\infty \frac{3}{2(n-1)n(n+1)} \; = \; \frac12 + \frac34\sum_{n=2}^\infty\left(\frac{1}{(n-1)n} - \frac{1}{n(n+1)}\right) \; = \; \frac12 + \frac38 \; = \; \boxed{\frac78}$

An alternative approach to calculating $\mathrm{Tr}\big[A(n)^{-1}\big]$ relies on obtaining an inductive formula for the inverse matrix $A(n)^{-1}$ . Note that $\begin{aligned} \left(\begin{array}{ccccc|c} & & & & & 0 \\ & & & & & \vdots \\ & & A(n)^{-1} & & & \vdots \\ & & & & & 0 \\ & & & & & 0 \\\hline 0 & 0 & \cdots & & n(n+1)& - n(n+1) \end{array} \right)A(n+1) & = \; \left(\begin{array}{ccccc|c} & & & & & 0 \\ & & & & & \vdots \\ & & A(n)^{-1} & & & \vdots \\ & & & & & 0 \\ & & & & & 0 \\\hline 0 & 0 & \cdots & & n(n+1)& - n(n+1) \end{array} \right)\left(\begin{array}{ccccc|c} & & & & & 1 \\ & & & & & \frac12 \\ & \; A(n) & & & \vdots \\ & & & & & \vdots \\ & & & & & \frac{1}{n} \\\hline 1 & \frac12 & \cdots & \cdots & \frac{1}{n} & \frac{1}{n+1} \end{array}\right) \\ & = \; \left(\begin{array}{ccccc|c} & & & & & 0 \\ & & & & & 0 \\ & & I_n & & & \vdots \\ & & & & & 0 \\ & & & & & 1 \\\hline 0 & 0 & \cdots & 0 & 0 & 1 \end{array}\right) \end{aligned}$ so that $\begin{aligned} A(n+1)^{-1} & = \; \left(\begin{array}{ccccc|c} & & & & & 0 \\ & & & & & 0 \\ & & I_n & & & \vdots \\ & & & & & 0 \\ & & & & & -1 \\\hline 0 & 0 & \cdots & \cdots & 0 & 1 \end{array}\right)\left(\begin{array}{ccccc|c} & & & & & 0 \\ & & & & & \vdots \\ & & A(n)^{-1} & & & \vdots \\ & & & & & 0 \\ & & & & & 0 \\\hline 0 & 0 & \cdots & 0 & n(n+1)& - n(n+1) \end{array} \right) \\ & = \; \left(\begin{array}{ccccc|c} & & & & & 0 \\ & & & & & 0 \\ & & A(n)^{-1} - n(n+1)K(n) & & & \vdots \\ & & & & & 0 \\ & & & & & n(n+1) \\\hline 0 & 0 & \cdots & 0 & n(n+1) & -n(n+1) \end{array}\right) \end{aligned}$ where $K(n)$ is the $n \times n$ matrix with zeros in all components except the bottom right one, which has the value $1$ . From this it seems that $A(n)^{-1}$ is always a tridiagonal matrix, but we do not need to identify it orecisely. We see that $\mathrm{Tr}\big[A(n+1)^{-1}\big] \; =\; \mathrm{Tr}\big[A(n)^{-1} - n(n+1)K(n)\big] - n(n+1) \; = \; \mathrm{Tr}\big[A(n)^{-1}\big] - 2n(n+1) \hspace{1cm} n \ge 1$ and hence we deduce that $\mathrm{Tr}\big[A(n)^{-1}\big] \; =\; 1 - 2\sum_{r=1}^{n-1}r(r+1) \; = \; 1 - \tfrac23(n-1)n(n+1) \hspace{2cm} n \ge 1$