proving that two recursive series converge to the same limit given two groups:$$a_{n+1}=\frac{a_n+b_n}2, b_{n+1}=\frac{2a_n*b_n}{a_n+b_n}, a_1=x,b_1=y$$
We need to prove that both series converge to the same limit, and show what the limit is, without knowledge of which value is greater at the start.
I managed to prove that for every $n$, $a_n*b_n=x*y$, which would mean that $b_n=\frac{xy}{a_n}$. I also know that the two series are bounded by x and y, but I still cannot figure out how to carry on from there. Any advice?
 A: This is one of the situations where we can profit from Polya's adivise 'first guess, then prove'. In this case: guess what the limit of the $a_n$ is and then prove that you are right.
In this case I would guess...
Spoiler alert...
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$$\lim_{n\to \infty}  a_n = \sqrt{xy}$$
Reason: you told me (I wouldn't have noticed otherwise) that $a_n b_n = xy$. It follows that IF $\lim a_n$ and $\lim b_n$ both exist we will have $(\lim a_n)(\lim b_n) = xy$. The only way in which they might be equal is if both equal $\sqrt{xy}$.
Now this reason for guessing does not gives us any hints at how to prove that $\lim_{n\to \infty}  a_n = \sqrt{xy}$ since obviously we cannot use that both $a_n$ and $b_n$ converge to the same limit. But now that we know what we WANT to prove we can find an other way to do it.
$a_{n+1}$ is the arithmetic mean of $a_n$ and $\frac{xy}{a_n}$. If $a_n > \sqrt{xy}$ then $\frac{xy}{a_n} < \sqrt{xy}$ and if $a_n < \sqrt{xy}$ then $\frac{xy}{a_n} > \sqrt{xy}$ and in both case $a_{n+1}$ will be closer to $\sqrt{xy}$ then either of them. Formalize this a little bit and you have a proof that $\lim_{n\to \infty}  a_n = \sqrt{xy}$.
From this we can go back to your insight that $b_n = xy/a_n$ and show that $\lim_{n\to \infty}  b_n = \sqrt{xy}$ as well.
With both limits equal to the same very concrete value we can conclude that they must be equal to eachother as well.
