Another Putnam Problem

Suppose that $n \ge 3$ . The entries in the $j$ th row of the initial matrix $M_n$ are $r_j \;= \; \left(\cos\big(n(j-1)+1\big)\,,\,\cos\big(n(j-1)+2\big)\,,\,\cos\big(n(j-1)+3\big)\,,\,\ldots\,,\,\cos\big(n(j-1)+n\big)\right)$ Thus the $k$ th element of $r_j - \cos\big(n(j-1)\big)r_1$ is $\cos\big(n(j-1) + k\big) - \cos\big(n(j-1)\big)\cos k \; = \; -\sin\big(n(j-1)\big)\sin k$ and hence we see that $r_j - \cos\big(n(j-1)\big)r_1 \; = \; -\sin\big(n(j-1)\big) s$ where $s \; = \; \big(\sin 1\,,\,\sin2\,,\,\sin3\,,\,\ldots\,,\,\sin n\big)$ Thus, applying elementary row operations, $\mathrm{det}\,M_n$ is equal to the determinant of the matrix whose rows are $r_1\,,\, r_2 - \cos n\;r_1\,,\,r_3 - \cos2n\;r_1\,,\,r_4 - \cos3n\;r_1\,,\,\ldots\,,\,r_n - \cos\big((n-1)n\big)r_1$ namely the matrix whose rows are $r_1\,,\, - \sin n\;s\,,\,- \sin2n\;s\,,\,- \sin3n\;s\,,\,\ldots\,,\ - \sin\big((n-1)n\big)s$ Since all the rows except the first are multiples of each other, this determinant is $0$ . Thus we deduce that $d_n = 0$ for $n \ge 3$ , and hence $\lim_{n \to \infty} d_n \; = \; \boxed{0}$