jmc

We compute the first few terms: $$a_1=1,a_2=\frac{1}{2},a_3=\frac{1}{4},a_4=\frac{3}{4},a_5=\frac{1}{2},a_6=\frac{1}{3},a_7=\frac{2}{3},a_8=\frac{1}{2}.$$The sequence appears to be converging to $\frac{1}{2}.$ In fact, every third term appears to be $\frac{1}{2}.$ So we can define a new sequence $(b_n)$ where $b_n=2a_n-1.$ Then $a_n=\frac{b_n+1}{2}.$ Substituting, we get $$\frac{b_n+1}{2}=\frac{1-\frac{1+b_{n-1}}{2}}{2\cdot \frac{1+b_{n-2}}{2}}.$$This simplifies to $$b_n=-\frac{b_{n-1}+b_{n-2}}{b_{n-2}+1}.$$Note that $b_1=1,$ $b_2=0,$ and $b_3=-\frac{1}{2}.$

Suppose $b_n=0.$ Then $$\begin{array}{rl}{b}_{n+1}& =-\frac{b_n+b_{n-1}}{b_{n-1}+1}=-\frac{b_{n-1}}{b_{n-1}+1},\\ b_{n+2}& =-\frac{b_{n+1}+b_n}{b_n+1}=-b_{n+1}=\frac{b_{n-1}}{b_{n-1}+1},\\ b_{n+3}& =-\frac{b_{n+2}+b_{n+1}}{b_{n+1}+1}=0,\\ b_{n+4}& =-\frac{b_{n+2}}{b_{n+2}+1}=\frac{b_{n+1}}{1-b_{n+1}}.\end{array}$$This tells us if $b_n=0,$ then $b_{n+3}=0.$ Hence, $b_{3m-1}=0$ for all $m\ge 1.$

Furthermore, if $b_{n+1}=-\frac{1}{k},$ then $$b_{n+4}=\frac{b_{n+1}}{1-b_{n+1}}=\frac{-1/k}{1+1/k}=-\frac{1}{k+1}.$$Hence, $b_6=-\frac{1}{3},$ $b_9=-\frac{1}{4},$ $b_{12}=-\frac{1}{5},$ and so on. In general, $$b_{3m}=-\frac{1}{m+1}.$$Then $$a_{3m}=\frac{b_{3m}+1}{2}=\frac{-1/(m+1)+1}{2}=\frac{m}{2(m+1)}.$$In particular, $$a_{120}=\frac{40}{2(40+1)}=\boxed{\frac{20}{41}}.$$