= Solution
A constant orbit is a <fixed point> of $F(u)=u(1+u)/2$. Solving $F(u)=u$ gives $u(u-1)=0$, hence
$$
\boxed{u_n\equiv0\quad\text{and}\quad u_n\equiv1}.
$$
For <fixed point stability for an iteration>, perturb a fixed point $u_*$ by $v_n$. Its local evolution is $v_{n+1}=F'(u_*)v_n+O(v_n^2)$, where $F'(u)=u+1/2$. At zero, $F'(0)=1/2$, so perturbations contract: \b[zero is locally asymptotically stable]. For example, if $|u|\leq r<1$, then $|F(u)|\leq(1+r)|u|/2$; this gives an invariant neighborhood and geometric convergence to zero.
At one, $F'(1)=3/2$, so perturbations expand: \b[one is unstable]. More explicitly, putting $u_n=1+v_n$ gives $v_{n+1}=3v_n/2+v_n^2/2$, whose magnitude increases while a nonzero perturbation remains sufficiently small. The derivative test concerns local stability; it does not assert attraction from every real initial value.
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