Consider the sequence defined by
a1=√2, a2=√2+√2, so that in general, an=√2+an−1 for n>1.
I know 2 is an upper bound of this sequence (I proved this by induction). Is there a way to show that this sequence converges to 2? What I think is that the key step is to prove 2 is the least upper bound of this sequence. But how?
Answer
Let x=√2+√2+√2+⋯. Then, note that x2=2+√2+√2+⋯=2+x⟹x2−x−2=0.Note that the two solutions to this equation are x=2 and x=−1, but since this square root cannot be negative, it must be 2.
No comments:
Post a Comment