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
The easy way to proceed, is to show that $a_n, b_n \geq 0$, and then $ 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.
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
$$ { 1 \over M(a,b)} = { 2 \over \pi} \int_0^{ { \pi \over 2}} { d \theta \over \sqrt{ a^2 \cos^2 \theta + b^2 \sin^2 \theta}} \ . $$
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.
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.
