Let a1>b1>0 and set an+1=an+bn2 and bn+1=2anbnan+bn. I've shown that an>an+1>bn+1>bn by induction. Hence an is a decreasing sequence bounded below and bn is an increasing sequence bounded above. Hence, both sequences converge by the monotone convergence theorem. Moreover, since lim It follows that \lim a_n = \lim b_n =L. What I am having trouble with is actually computing the limit. I am aware of the similar question here but unless I'm mistaken no one gives an exact computation for L. It is clear that b_1 < L < a_1 but can anyone give a hint on how I might find L exactly? I assume it would be some function of the "initial conditions": a_1 and b_1.
Thanks.
Answer
Multiply both sides of the a_{n+1} = \frac{a_n + b_n}{2} by
b_{n+1} = \frac{2a_nb_n}{a_n+b_n}
to get a_{n+1} b_{n+1} = a_n b_n
You have proved that both sequences converge to the same limit L.
Thus \lim _{n\to \infty} a_nb_n=L^2 =a_1b_1
That is L= \sqrt { a_1b_1}
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