Is this a harmonic sequence?

Since we want $a_2 =\tfrac12$ and $a_1 = 1$ , we must have $b=2$ . Consider the $b=2$ case. Then $a_n = \frac{a_{n-1}}{1+a_{n-1}}$ , so that $a_na_{n-1} + a_n = a_{n-1}$ , and hence $1 + a_{n-1}^{-1} \; = \; a_n^{-1}$ With $a_1 = 1$ it is a simple induction that $a_n^{-1} = n$ , and hence that $a_n = n^{-1}$ . Thus we obtain $a_n = n^{-1}$ precisely when $b=2$ , making the answer $\boxed{1}$ .