AHSME 1969 Problem 27

A particle moves so that its speed for the second and subsequent miles varies inversely as the integral number of miles already traveled. For each subsequent mile the speed is constant. If the second mile is traversed in $2$ hours, then the time, in hours, needed to traverse the $n$th mile is: $\text{(A) } \frac{2}{n-1} \text{(B) } \frac{n-1}{2} \text{(C) } \frac{2}{n} \text{(D) } 2n \text{(E) } 2(n-1)$