 Solution

Every convergent member is also A-stable. For $0<\alpha <2$, put $\zeta =e^{i\theta }$ and calculate its boundary-locus test for multistep A-stability:The denominator never vanishes on the unit circle in this parameter interval. It follows thatTherefore no amplification root can cross the unit circle as $z=h\lambda$ moves in the open left half-plane. For small negative real $z$, the root near one moves to $1+z+O(z^2)$, inside the circle; the other begins at $\alpha -1$, also strictly inside. The leading coefficient $1-(1-\alpha /4)z$ never vanishes in the left half-plane. Continuity of polynomial roots and connectedness of that half-plane now keep both roots inside everywhere. On its imaginary boundary the same locus has zero real part only at $\zeta =1$, corresponding to $z=0$; that root is simple because $2-\alpha >0$.

At $\alpha =0$, the two interlaced subsequences follow the trapezoidal rule with step $2h$:The right side has modulus at most one for $\mathrm{Re}z\le 0$. Its unit-circle roots are simple; the double zero at $z=-1$ lies strictly inside and is harmless. Thus the endpoint is also A-stable. Combining this with part 1, the requested range is $0\le \alpha <2$, including the BDF2 method at $\alpha =4/3$.