如图，其中 ∠AA1A2=∠AA2A3=∠AA3A4=∠AA4A5=⋯=∠AAnAn+1=Rt∠\angle AA_1A_2=\angle AA_2A_3= \angle AA_3A_4 =\angle AA_4A_5 = \cdots = \angle AA_nA_{n+1} = Rt\angle∠AA1A2=∠AA2A3=∠AA3A4=∠AA4A5=⋯=∠AAnAn+1=Rt∠ AA1=A1A2=A2A3=A3A4=A4A5=⋯=AnAn+1=1AA_1=A_1A_2=A_2A_3=A_3A_4=A_4A_5=\cdots =A_nA_{n+1}=1AA1=A1A2=A2A3=A3A4=A4A5=⋯=AnAn+1=1 当一直以此规律画三角形，求 limn→∝∠A1AAn\displaystyle \lim_{n \to \propto }\angle A_1AA_nn→∝lim∠A1AAn 也即
请问这个式子收敛吗？若是，收敛到几？若否能否证明
