# Prove sequences are both convergent, given $a_{n+1} = \sqrt{a_{n}b_{n}}$ and $b_{n+1} = \frac{a_{n} + b_{n}}{2}$ [duplicate]

Define two sequences $a_{n}$ and $b_{n}$ as $a_{1} = 1$, $b_{1} = \sqrt{2}$, and

$a_{n+1} = \sqrt{a_{n}b_{n}}$ and $b_{n+1} = \frac{a_{n} + b_{n}}{2}$

How can we show that the sequences are convergent. I tried to use the monotonic sequence theorem by showing that $a_{n} > a_{n+1}$ for all $n$ and $b_{n} > b_{n+1}$ for all $n$ . But the sequence $a_{n}$ and $b_{n}$ are defined as a product of $a_{n}, b_{n}$ and the average of $a_{n}, b_{n}$ [They consist of two sequence $a_{n}$ and $b_{n}$.] And then I don't know where to start.

My question is how can we solve this kind of problem systematically? Please give me some suggestions on where can I get started.

• Asked n times on the site.
– Did
Oct 3, 2015 at 7:49

By the AM-GM inequality we have $a_n=\sqrt{a_{n-1}b_{n-1}}\leqslant\dfrac{a_{n-1}+b_{n-1}}{2}=b_n$ for each $n>1$ and since $a_1<b_2,$ we conclude that $a_n\leqslant b_n$ for each $n\geqslant1$ and hence $b_{n+1}=\dfrac{a_n+b_n}{2}\leqslant b_n$ for all $n.$ Since every $b_i$ is positive, we conclude that $\{b_n\}$ converges (because it is monotonically decreasing and bounded from below) and so $\{a_n\}$ converges as well (because the $a_i$'s are bounded from above by the $b_i$'s).
• Then I was just wondering that is there any way to find the limit of this kind of sequence like $a_{n}$ and $b_{n}$ in this problem? (I mean.. the sequence that is defined as a sum or product or etc. of another sequence) Oct 3, 2015 at 7:57
You can start by showing both sequences are bounded, $a_n$ from above and $b_n$ from below (using AM-GM inequality)
$a_{n+1} = \sqrt {a_nb_n} \leq \dfrac {a_n + b_n} {2} = b_{n+1}$
Now, if you'll show they are monotonic ($a_n$ non-decreasing and $b_n$ non-increasing) - you're done. The approach to this problem is no different than the 1 we take when working out any real sequences problem.