Trickier Trace

Let $F_n(t)$ be the characteristic polynomial of $A(n)$ . In the first round of this question, I showed that $F_n(t) \; = \; \left(t + \tfrac{1}{n(n-1)}\right)F_{n-1}(t) + t\,\mathrm{det} B_{n-1}(t) \hspace{2cm} n \ge 3$ where $B_{n-1}(t)$ is the $(n-1)\times(n-1)$ matrix that only differs from $t I_{n-1} - A_{n-1}$ by having $-\tfrac{1}{n-1}$ , instead of $t-\tfrac{1}{n-1}$ , in the bottom right corner. This, together with the initial values $F_1(t) = t-1$ , $F_2(t) = t^2 - \tfrac32t -\tfrac12$ implied that $n!(n-1)! F_n(0) \; = \; -1 \hspace{1cm} \mathrm{Tr}\big[A(n)^{-1}\big] \; = \; n!(n-1)!F_n'(0) \; = \; 1 - \tfrac23(n-1)n(n+1) \hspace{2cm} n \ge 1$ Following these ideas further, the fact that the determinant or a matrix is a multilinear function of its rows means that $\begin{aligned} \mathrm{det}\,B_{n-1}(t) & = \; \left| \begin{array}{ccccc|c} & & & & & -1 \\ & & & & & -\frac12 \\ & & tI_{n-2} - A(n-2) & & & \vdots \\ & & & & & \vdots \\ & & & & & -\frac{1}{n-2} \\ \hline -1 & -\frac12 & \cdots & & -\frac{1}{n-2} & -\frac{1}{n-1} \end{array} \right| \\[4ex] & = \;\left| \begin{array}{ccccc|c} & & & & & -1 \\ & & & & & -\frac12 \\ & & tI_{n-2} - A(n-2) & & & \vdots \\ & & & & & \vdots \\ & & & & & -\frac{1}{n-2} \\ \hline -1 & -\frac12 & \cdots & & -\frac{1}{n-2} & t-\frac{1}{n-1} \end{array} \right| - \left| \begin{array}{ccccc|c} & & & & & -1 \\ & & & & & -\frac12 \\ & & tI_{n-2} - A(n-2) & & & \vdots \\ & & & & & \vdots \\ & & & & & -\frac{1}{n-2} \\ \hline 0 & 0 & \cdots & & 0 & t \end{array} \right| \\[4ex] & = \; F_{n-1}(t) - tF_{n-2}(t) \end{aligned}$ for $n \ge 3$ , so we deduce that $F_n(t) \; = \; \left(2t + \tfrac{1}{n(n-1)}\right)F_{n-1}(t) - t^2F_{n-2}(t) \hspace{2cm} n \ge 3$ and hence that $F_n''(0) \; = \; \tfrac{1}{n(n-1)}F_{n-1}''(0) + 4F_{n-1}'(0) -2F_{n-2}(0) \hspace{2cm} n \ge 3$ This implies that $\begin{aligned} n! (n-1)! F_n''(0) & = \; (n-1)!(n-2)!F_{n-1}''(0) + 4n!(n-1)!F_{n-1}'(0) - 2n!(n-1)!F_{n-2}(0) \\ & = \; (n-1)!(n-2)!F_{n-1}''(0) + 4n(n-1)\left[1 - \tfrac23(n-2)(n-1)n\right] + 2n(n-1)^2(n-2) \\ & = \; (n-1)!(n-2)!F_{n-1}''(0) - \tfrac23n(n-1)\big[4n^3 - 15n^2 + 17n - 12\big] \end{aligned}$ for $n \ge 3$ , and hence $n!(n-1)! F_n''(0) \; = \; 4 -\tfrac23\sum_{r=3}^n r(r-1)\big[4r^3 - 15r^2 + 17r - 12\big] \; = \; -\tfrac49n^6 + \tfrac65n^5 - \tfrac19n^4 + \tfrac59n^2 - \tfrac65n - 4 \hspace{1.5cm} n \ge 2$ Using Vieta's formulae, if $u_{n1},u_{n2}, \ldots,u_{nn}$ are the eigenvalues of $A(n)$ , then $\begin{aligned} \mathrm{Tr}\big[A(n)^{-1}\big] \; = \; \sum_{j} u_{nj}^{-1} & = \; -\frac{F_n'(0)}{F_n(0)} \; = \; 1 - \tfrac23(n-1)n(n+1) \\ \sum_{j<k} u_{nj}^{-1}u_{nk}^{-1} & = \; \frac{F_n''(0)}{2 F_n(0)} \; = \; \tfrac29n^6 - \tfrac35n^5 + \tfrac{1}{18}n^4 - \tfrac{5}{18}n^2 + \tfrac35n + 2 \end{aligned}$ for $n \ge 2$ , from which we deduce that $\mathrm{Tr}\big[A(n)^{-2}\big] \; = \; \sum_j u_{nj}^{-2} \; = \;\left(\sum_ju_{nj}^{-1}\right)^2 - 2\sum_{j<k}u_{nj}^{-1}u_{nk}^{-1} \; = \; \tfrac{1}{15}(n-1)n(n+1)(18n^2 - 15n - 2) - 3$ for all $n \ge 2$ . But then we deduce that $X(n) \; = \; \left\{ \begin{array}{lll} \tfrac65(n-2)(n-1)n(n+1)(n+2) & \hspace{1cm}& n \ge 2 \\[2ex] 4 & & n = 1 \end{array}\right.$ and hence we deduce that $\sum_{n=3}^\infty X(n)^{-1} \; = \; \frac{5}{24}\sum_{n=3}^\infty \left(\frac{1}{(n-2)(n-1)n(n+1)} - \frac{1}{(n-1)n(n+1)(n+2)}\right) \; = \; \frac{5}{576}$ making the answer $5 + 576 = \boxed{581}$ .

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Just as there was an alternate approach to the first part of this question, there is an alternate approach to the second part. Recall that

$A(n+1)^{-1} \; = \; \left(\begin{array}{ccccc|c} & & & & & 0 \\ & & & & & 0 \\ & & A(n)^{-1} - n(n+1)K(n) & & & \vdots \\ & & & & & 0 \\ & & & & & n(n+1) \\\hline 0 & 0 & \cdots & 0 & n(n+1) & -n(n+1) \end{array}\right)$ where $K(n)$ is the $n \times n$ matrix with zeros in all components except the bottom right one, which has the value $1$ . This implies that $A(n+1)^{-2} \; = \; \left(\begin{array}{ccccc|c} & & & & & \star \\ & & & & & \star \\ & & \big(A(n)^{-1} - n(n+1)K(n)\big)^2 & & & \vdots \\ & & + n^2(n+1)^2K(n)& & & \star \\ & & & & & \star \\\hline \star & \star & \cdots & \star & \star & 2n^2(n+1)^2 \end{array}\right)$ where the stars indicate that the values in those places are unimportant. Thus $\mathrm{Tr}\big[A(n+1)^{-2}\big] \; = \; \mathrm{Tr}\big[\big(A(n)^{-1} - n(n+1)K(n)\big)^2\big] + 3n^2(n+1)^2 \; = \; \mathrm{Tr}\big[A(n)^{-2}\big] - 2n(n+1)\,\mathrm{Tr}\big[K(n)A(n)^{-1}] + 4n^2(n+1)^2$

It is easy to calculate that $\mathrm{Tr}\big[K(n+1)A(n+1)^{-1}\big] = -n(n+1)$ for all $n \ge 1$ , and hence $\mathrm{Tr}\big[A(n+1)^{-1}\big] \; = \; \mathrm{Tr}\big[A(n)^{-2}\big] + 2(n+1)n^2(n-1) + 4n^2(n+1)^2 \; = \; \mathrm{Tr}\big[A(n)^{-2}\big] + 2n^2(n+1)(3n+1)$ for all $n \ge 2$ , and hence $\mathrm{Tr}\big[A(n)^{-2}\big] \; =\; 13 + 2\sum_{r=2}^{n-1}r^2(r+1)(3r+1) \; = \; \tfrac{1}{15}(n-1)n(n+1)(18n^2 - 15n - 2) - 3$ for $n \ge 2$ , as before.

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Simplest yet, the above observations about the matrix $A(n)$ make it easy to show that the inverse of $A(n)$ is the symmetric tridiagonal matrix $A(n)^{-1} \; = \; \left(\begin{array}{ccccccc} a_1 & b_1 & 0 & 0 & \cdots & 0 & 0 \\ b_1 & a_2 & b_2 & 0 & \cdots & 0 & 0 \\ 0 & b_2 & a_3 & b_3 & \cdots & 0 & 0 \\ 0 & 0 & b_3 & a_4 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & a_{n-1} & b_{n-1} \\ 0 & 0 & 0 & 0 & \cdots & b_{n-1} & a_n \end{array} \right)$ where $a_j \; = \; \left\{ \begin{array}{lll} -1 & \hspace{1cm} & j = 1 \\ -2j^2 & & 2 \le j \le n-1 \\ -n(n-1) & & j = n \end{array}\right. \hspace{2.5cm} b_j \; = \; j(j+1) \hspace{1cm} 1 \le j \le n-1$ and hence we can just calculate $\mathrm{Tr}\big[A(n)^{-1}\big] \; = \; \sum_{j=1}^n a_j \hspace{2cm} \mathrm{Tr}\big[A(n)^{-2}\big] \; = \; \sum_{j=1}^n a_j^2 + 2\sum_{j=1}^{n-1}b_j^2$