Sequences converting to a common limit

Question

Let a₁ > b₁ > 0 and let a_n+1 = (a_n + b_n)/2, b_n+1 = 2a_nb_n/(a_n + b_n) for n ≥ 1. Show that a_n > a_n+1 > b_n+1 > b_n and deduce that the two sequences converge to a common limit. What limit?

Answer 1

a₁>b₁>0; a_n+1=(a_n+b_n)/2; b_n+1=2a_nb_n/(a_n+b_n).

a_n=2a_n+1-b_n, so a_n-a_n+1=a_n+1-b_n; b_n+1=a_nb_n/a_n+1, by substituting 1/a_n+1 for 2/(a_n+b_n), therefore b_n+1/b_n=a_n/a_n+1 or a_n+1b_n+1=a_nb_n.

When n=1, a₂=(a₁+b₁)/2; b₂=b₁(a₁/a₂).

Since a₁>b₁, let a₁=b₁+s₁ where s₁>0, then b₁=a₁-s₁ and a₂=(2a₁-s₁)/2=a₁-s₁/2, therefore a₂<a₁ and a₁/a₂>1, so b₂>b₁. 

a_n+1/b_n+1=[(a_n+b_n)/2]/[2a_nb_n/(a_n+b_n)]=[(a_n+b_n)/2][(a_n+b_n)/(2a_nb_n)]=(a_n+b_n)²/(4a_nb_n).

When n=1, a₂/b₂=(2a₁-s₁)²/(4a₁(a₁-s₁))=(4a₁²-4a₁s₁+s₁²)/(4a₁²-4a₁s₁)=1+s₁²/(4a₁²-4a₁s₁).

s₁²/(4a₁²-4a₁s₁)>0 so a₂/b₂>1, a₂>b₂.

So we have (for n=1), a₁>a₂>b₂>b₁, which is the special case of a_n>a_n+1>b_n+1>b_n.

But we need to show that this is the case for general n.

More to follow in due course...