Let $a_1 > b_1 >0$ and set $$a_{n+1} = \frac{a_n + b_n}{2}$$ and $$b_{n+1} = \frac{2a_nb_n}{a_n+b_n}.$$ I've shown that $a_n > a_{n+1} >b_{n+1}>b_{n}$ by induction. Hence $a_n$ is a decreasing sequence bounded below and $b_n$ is an increasing sequence bounded above. Hence, both sequences converge by the monotone convergence theorem. Moreover, since $$\lim a_{n+1} = \lim a_{n} = \frac{\lim a_n + \lim b_n}{2}$$ 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 $b_1$.



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 $$ L= \sqrt { a_1b_1}$$

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.