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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 $b_1$.

Thanks.

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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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