Finding Nonsingular Matrices

Consider the vectors $\mathbf{s}(m) \in \mathbb{R}^n$ defined by $\mathbf{s}(m) \; = \; \big(j^m\big)_{1 \le j \le n} \hspace{2cm} 0 \le m \le k$ If $\mathbf{r}_u$ , for $1 \le u \le n$ , are the $n$ rows of $A_k(n)$ , then $\mathbf{r}_u = \big((u+j)^k\big)_{1\le j \le n}$ , and so $\mathbf{r}_u \; = \; \sum_{m=0}^k \binom{k}{m} u^{k-m}\mathbf{s}(m) \hspace{2cm} 1 \le u \le n$ and hence $\mathbf{r}_u$ belongs to the linear span $\mathcal{S} = \langle \mathbf{s}(0),\mathbf{s}(1),...,\mathbf{s}(k)\rangle \le \mathbb{R}^n$ . Of course, $\mathcal{S}$ is at most $(k+1)$ -dimensional. Thus, if $n \ge k+2$ , then the vectors $\mathbf{r}_1,\mathbf{r}_2,...,\mathbf{r}_n$ must be linearly dependent, which implies that $A_k(n)$ is singular. In other words, $S(k) \subseteq \{1,2,...,k+1\}$ .

Suppose that $n = k+1$ , and consider the $k+1$ vectors $\mathbf{r}_1,\mathrm{r}_2,...,\mathbf{r}_{k+1} \in \mathcal{S}$ . If there exist constants $\alpha_1,\alpha_2,...,\alpha_{k+1}$ such that $\sum_{u=1}^{k+1}\alpha_u \mathbf{r}_u \; = \; \mathbf{0}$ , then $\sum_{u=1}^{k+1}\alpha_u(u+j)^k \; = \; 0 \hspace{2cm} 1 \le j \le n$ and hence $F(X) \; = \; \sum_{u=1}^{k+1}\alpha_u(u+X)^k$ is a polynomial of degree at most $k$ such that $F(j) = 0$ for $1 \le j \le k+1$ . This implies that $F(X) \equiv 0$ . Now $0 \; \equiv \; F(X) \; \equiv \; \sum_{u=1}^{k+1}\alpha_u \sum_{m=0}^k \binom{k}{m} u^{k-m}X^m \; \equiv \; \sum_{m=0}^k \binom{k}{m} \left(\sum_{u=1}^{k+1} u^{k-m}\alpha_u\right)X^m$ and hence we deduce that $\sum_{u=1}^{k+1} u^m \alpha_u \; = \; 0 \hspace{2cm} 0 \le m \le k$ Since the Vandermonde matrix $\Vert u^m\Vert_{1 \le u \le k+1,0 \le m \le k}$ is nonsingular, we deduce that $\alpha_1 = \alpha_2 = \cdots = \alpha_{k+1} = 0$ . Thus we deduce that $\mathbf{r}_1,\mathbf{r}_2,...,\mathbf{r}_{k+1}$ form a linearly independent set, which implies that $A_k(k+1)$ is nonsingular. In fact it is possible to show that $\mathrm{det}\,\big[A_k(k+1)\big] \; = \; (-1)^{\lfloor \frac12(k+1)\rfloor} (k!)^{k+1} \; \neq 0$ Thus $N(k) = k+1$ , and so $\sum_{k=1}^\infty \frac{1}{kN(k)^2} \; = \; \sum_{k=1}^\infty \frac{1}{k(k+1)^2} \; = \; \sum_{k=1}^\infty \left(\frac{1}{k} - \frac{1}{k+1} - \frac{1}{(k+1)^2}\right) \; = \; 1 - \sum_{k=2}^\infty \frac{1}{k^2} \; = \; 2 - \tfrac16\pi^2$ making the answer $2 + 2 + 6 = 10$ .