Inverting Cauchy

The matrix $X(n)$ is a special case of what is called a Cauchy matrix, namely one of the form $H_{ij} \; = \; \frac{1}{a_i - b_j} \hspace{2cm} 1 \le i,j \le n$ where $a_1,a_2,\ldots,a_n,b_1,b_2,\ldots,b_n$ are all distinct. It is possible to find an explicit formula for the inverse of $H$ . To do this, define the interpolation polynomials $\begin{array}{rclcrclcl} A(X) &\equiv& \displaystyle \prod_{j=1}^n (X - a_j) &\hspace{1cm}& B(X) &\equiv& \displaystyle \prod_{j=1}^n (X - b_j) \\[2ex] A_i(X) &\equiv& \displaystyle \frac{1}{A'(a_i)}\prod_{j \neq i}(X - a_j) && B_i(X) &\equiv& \displaystyle \frac{1}{B'(b_i)}\prod_{j \neq i}(X - b_j) &\hspace{2cm}& 1 \le i \le n \end{array}$ so that $A_i(a_j) = B_i(b_j) = \delta_{ij}$ for $1 \le i,j \le n$ and $Q(X) \; \equiv \; \sum_{i=1}^n Q(a_i)A_i(X) \; \equiv \; \sum_{i=1}^n Q(b_i)B_i(X)$ for any polynomial $Q(X)$ of degree $n-1$ or less.

Define the $n \times n$ matrix $G$ by $G_{ij} \; = \; -\frac{A(b_i) B_i(a_j)}{A'(a_j)} \hspace{2cm} 1 \le i,j \le n$ Then, for $1 \le i \le n$ , $\begin{aligned} -A(b_i)B_i(X) & \equiv \; -\sum_{j=1}^n A(b_i)B_i(a_j)A_j(X) \; = \; \sum_{j=1}^n G_{ij}\prod_{k \neq j}(X - a_k) \\ \frac{A(b_i)B_i(X)}{A(X)} & \equiv \; -\sum_{j=1}^n \frac{G_{ij}}{X - a_j} \; \equiv \; \sum_{j=1}^n \frac{G_{ij}}{a_j - X} \end{aligned}$ so that $\sum_{j=1}^n G_{ij}H_{jk} \; = \; \sum_{j=1}^n \frac{G_{ij}}{a_j - b_k} \; = \; \frac{A(b_i)B_i(b_k)}{A(b_k)} \; = \; \delta_{ik} \hspace{2cm} 1 \le i,k \le n$ Thus $G = H^{-1}$ ; moreover $H^{-1}_{ij} \; = \; G_{ij} \; = \; -\frac{B_i(a_j)}{A'(a_j)}(b_i-a_j)\prod_{k \neq j}(b_i - a_k) \; = \; (a_j - b_i)A_j(b_i)B_i(a_j) \hspace{2cm} 1 \le i,j \le n$ We also note that $\sum_{i=1}^n A_j(b_i)B_i(a_j) \; = \; A_j(a_j) \; = \; 1 \hspace{2cm} \sum_{j=1}^n A_j(b_i)B_i(a_j) \; = \; B_i(b_i) \; = \; 1$ and hence $\sum_{i,j=1}^n H^{-1}_{ij} \; = \; \sum_{j=1}^n a_j\left(\sum_{i=1}^n A_j(b_i)B_i(a_j)\right) - \sum_{i=1}^n b_i\left(\sum_{j=1}^n A_j(b_i)B_i(a_j)\right) \; = \; \sum_{i=1}^n(a_i - b_i)$

In the particular case of $X(n)$ , we have $a_i = i^2$ and $b_j = -j^2$ for $1 \le i,j \le n$ , and hence $\sigma_n \; = \; \sum_{i=1}^n 2i^2 \; = \; \tfrac13n(n+1)(2n+1) \hspace{2cm} n \ge 1$ Thus $\begin{aligned} \sum_{n=1}^N \frac{1}{\sigma_n} & = \; 3\sum_{n=1}^N \frac{1}{n(n+1)(2n+1)} \; = \; 3\sum_{n=1}^N \left(\frac{1}{n} + \frac{1}{n+1} - \frac{4}{2n+1}\right) \\ & = \; 3\left[3 + 4\sum_{n=1}^N \frac{1}{n} - 4\sum_{n=1}^{2N}\frac{1}{n} + \frac{1}{N+1} - \frac{4}{2N+1}\right] \\ & = \; 3\left[3 + 4S_N - 4S_{2N} - 4\ln2 + \frac{1}{N+1} - \frac{4}{2N+1}\right] \end{aligned}$ where $S_N \; = \; \sum_{n=1}^N \frac{1}{n} - \ln N$ is well-known to converge to the Euler-Mascheroni constant $\gamma$ as $N \to \infty$ . Thus we deduce that $\sum_{n=1}^\infty \frac{1}{\sigma_n} \; = \; 3(3 -4\ln2) \; = \; 9 - 12\ln2$ making the answer $9 +12 + 2 = \boxed{23}$ .