smc

Let $d$ be the distance already traveled and $s_n$ be the speed for the $n^{\text{th}}$ mile. Because the speed for the second and subsequent miles varies inversely as the integral number of miles already traveled, $$d{s}_{d+1}=k$$ If the second mile is traversed in $2$ hours, then the speed in the second mile ($s_2$) equals $\frac{1}{2}$ miles per hour. Since one mile has been traveled already, $k=\frac{1}{2}$. Now solve for the speed the $n^{\text{th}}$ mile, noting that $n-1$ miles have already been traveled. $$(n-1)s_{d+1}=\frac{1}{2}$$ $$s_{d+1}=\frac{1}{2(n-1)}$$ Thus, the time it takes to travel the $n^{\text{th}}$ mile is $1÷\frac{1}{2(n-1)}=\boxed{2(n-1)}$ hours.