How to prove two sequence have a common limit. We have
\begin{align}
U_{0} &= 1 &&\text{and} & V_{0} &= 2 \\
U_{n+1} &= \frac{U_{n}+V_{n}}{2} &&\text{and} & V_{n+1} &= \sqrt{U_{n+1}V_n}
\end{align}
How to prove two sequence have a common limit?
I found $(U_n)$ is increasing and $(V_n)$  is decreasing but I don't know how to do with  $\lim (V_n) - (U_n)$
Can someone help me, please?
 A: Since $U_n$ is increasing and $V_n$ is decreasing, all you have to show is that the difference $V_n-U_n$ has limit $0$. But, the limit exists since both sequences are monotone and bounded, and rewriting the equation $U_{n+1}=(U_n+V_n)/2$ to $\,V_n = 2 U_{n+1}-U_n \,$ shows the two limits must be equal.
Given a diameter one circle, its circumference is $\,\pi.\,$ Archimedes calculated the perimeters of inscribed and circumscribed regular polygons to find upper and lower bounds for $\,\pi.\,$ Let $\,a(n)\,$ be the perimeter
of the circumscribed regular polygon of $\,n\,$ sides,
and $\,b(n)\,$ the perimeter of the inscribed regular polygon of $\,n\,$ sides. We have that $\,a(2n)\,$ is the harmonic mean of $\,a(n)\,$ and $\,b(n)\,$ and $\,b(2n)\,$ is the geometric mean of $\,a(2n)\,$ and $\,b(n).\,$
We can start with triangles where $\,n=3\,$ and find
$\,a(3) = 3\sqrt{3}\,$ and $\,b(3) = a(3)/2.\,$ We then keep doubling the number of sides indefinitely. The connection between the two recursions is that 
$\, U_n = 3\sqrt{3}/a(3\,2^n)\,$ and
$\, V_n = 3\sqrt{3}/b(3\,2^n)\,$ since the recursions and initial values for $\, U_n, V_n \,$ come from those for $\,a(n), b(n).\,$
The common limit of $\,a(n), b(n)\,$ is $\,\pi,\,$ thus the common limit of $\, U_n, V_n\,$ is $\,3\sqrt{3}/\pi.\,$
If we start with squares where $\,n=4\,$ we find that
$\, a(4) = 4\,$ and $\,b(4) = 2\sqrt{2}.\,$ Now let
$\, U_n := 4/a(4\, 2^n)\,$ and 
$\, V_n := 4/b(4\, 2^n).\,$ Then 
$\, U_0 = 1, \,\, V_0 = \sqrt{2} \,$ and the common limit is $\, 4/\pi.\,$ The general values are
$\, b(n) = n\,\sin(\pi/n),\,$ and
$\, a(n) = n\,\tan(\pi/n).$
A: Given that
$$
u_{n+1}=\frac{u_n+v_n}2\quad\text{and}\quad v_{n+1}=\sqrt{\vphantom{1}u_{n+1}v_n}\tag1
$$
it is easy to derive that
$$
u_{n+1}^2-v_{n+1}^2=\frac{u_n^2-v_n^2}4\tag2
$$
Thus, induction says that
$$
u_n^2-v_n^2=\frac{u_0^2-v_0^2}{4^n}\tag3
$$
Furthermore,
$$
u_{n+1}-u_n=\frac{v_n-u_n}2\quad\text{and}\quad v_{n+1}^2-v_n^2=\frac{u_n-v_n}2v_n\tag4
$$
Case $\boldsymbol{u_0\gt v_0}$
Equation $(3)$ says $u_n\gt v_n$, and $(4)$ says that $u_n$ is decreasing and $v_n$ is increasing. Since, $u_n\gt v_n\gt v_0$, $u_n$ is decreasing and bounded below. Since $v_n\lt u_n\lt u_0$, $v_n$ is increasing and bounded above.
Case $\boldsymbol{u_0\lt v_0}$
Equation $(3)$ says $u_n\lt v_n$, and $(4)$ says that $u_n$ is increasing and $v_n$ is decreasing. Since, $u_n\lt v_n\lt v_0$, $u_n$ is increasing and bounded above. Since $v_n\gt u_n\gt u_0$, $v_n$ is decreasing and bounded below.
In either case, both $u_n$ and $v_n$ converge. $(3)$ says that their limits are the same.
A: If $U_n,V_n\in[1,2]$,
$$U_n+V_n\in[2,4]\implies U_{n+1}\in[1,2]$$
and
$$U_{n+1}V_n\in[1,4]\implies V_n\in[1,2].$$
Hence both sequences are bounded and both converge.
And if the limits exist, they are equal. Indeed,
$$U=\frac{U+V}2,V=\sqrt{UV}$$ imply $U=V$.
