prove $\sqrt{a_n b_n}$ and $\frac{1}{2}(a_n+b_n)$ have same limit I am given this problem:
let $a\ge0$,$b\ge0$, and the sequences $a_n$ and $b_n$ are defined in this way: $a_0:=a$, $b_0:=b$ and $a_{n+1}:= \sqrt{a_nb_n}$ and $b_{n+1}:=\frac{1}{2}(a_n+b_n)$ for all $n\in\Bbb{N}$  
To prove is that both sequences converge and that they have the same limit. I don't know how to show this. I have spent 2 hours on this, no sign of success
 A: If $a_n\to l$ and $b_n \to s$ then from $b_{n+1}=\frac{1}{2}(a_n+b_n)$ we get $s=\frac{1}{2}(l+s)\cdots$
To prove that they converge use arithmetic and geometric mean inequality to show (that $a_n\leq b_n$ so ...) they are monotonic ($a_n$ is increasing and $b_n$ decreasing). From $a_n\leq b_n$ conclude they are bounded.
A: The easy way to proceed, is to show that $ b_{n+1} - a_{n+1} = \frac {1}{2} (\sqrt{b_{n}} -\sqrt{ a_{n}})^2$ so $ b_{n} \geq a_{n} \forall n \geq 2$.
 Then, $a_{n+1} = \sqrt{a_n b_n} \geq a_n$ is an monotonically increasing sequence (after $n=2$).
$b_{n+1} = \frac {1}{2} (a_n + b_n) \leq b_n$ is a monotomically decreasing sequnece (after $n=2$).
Finally, $$ b_{n+1} - a_{n+1} = \frac {1}{2} (\sqrt{b_{n}} -\sqrt{ a_{n}})^2 \leq \frac {1}{2} (\sqrt{b_n} - \sqrt{a_n} ) ( \sqrt{b_n} + \sqrt{a_n} ) = \frac {1}{2} ( b_n - a_n) \leq \frac {1}{2^n} (b_1-a_1),$$ so the difference between the sequences go to 0. Hence, these sequences converge to the same limit.

Note: Of course we could show that since $a_i \leq b_2$, the limit of $a_i$ exists (since monotonic+bounded). But I think it's more fun to jump directly to the conclusion with the final step.
A: This is not an answer. Just a long comment which gives a classical formula of Gauss for this common limit.
Let $a$, $b$ be real numbers with $a<b$. We define their arithmetic-geometric mean  $M(a,b)$ using the sequences  $a_n$, $b_n$ defined as follows:
  $a_0=a$, $b_0=b$ and inductively
$$ a_{n+1}={ a_n +b_n \over 2},     \qquad b_{n+1}=\sqrt {a_n b_n
}\ . $$
The two sequences converge to a unigue common limit  $M(a,b)$.
 Gauss discovered a beautiful formula for  $M(a,b)$  which can be
expressed as an elliptic integral
$$ { 1 \over  M(a,b)} = { 2 \over \pi} \int_0^{ { \pi \over 2}} {
d \theta \over \sqrt{ a^2 \cos^2 \theta + b^2 \sin^2 \theta}} \ .
$$
A: Relating to PAD's answer.
If
$I(a, b)
=\int_0^{\pi/2} \dfrac{dt}{\sqrt{a^2\cos^2t+b^2\sin^2t}}
$,
using Landen's transformation
(https://en.wikipedia.org/wiki/Landen%27s_transformation)
you can show that
$I(a, b)
=I(\dfrac{a+b}{2}, \sqrt{ab})
$.
Using the well-known inequalities
in Calvin Lin's answer,
this shows that
$I(a, b)
=I(AGM(a, b), AGM(a, b))
=\dfrac{\pi}{2AGM(a, b)}
$,
which gives a quickly converging
way of computing
the complete elliptic integral
of the first kind.
