limit of a recurrent serie

Question

I would like to find to determine the limit for $u_n$ which I determined that it's a decreasing series and that $v_n$ is an increasing one.

let $a >0$ And $b>0$ .

$u_0=a And v_0=b$

$u_{n+1}=\frac{u_n+v_n}{2}$ and $v_{n+1}=\frac{2}{\frac{1}{u_n}+\frac{1}{v_n}}$

Answer 1

Expanding achille hui's comment.

$\begin{array}\\ u_{n+1} &=\dfrac{u_n+v_n}{2},\\ v_{n+1} &=\dfrac{2}{\frac{1}{u_n}+\frac{1}{v_n}}\\ &=\dfrac{2u_nv_n}{u_n+v_n}\\ &=\dfrac{2u_nv_n}{2u_{n+1}}\\ u_{n+1}v_{n+1} &=u_nv_n\\ v_{n+1} &=\dfrac{u_nv_n}{u_{n+1}}\\ u_1, v_1 &=a, b\\ u_1v_1 &=ab\\ &=u_nv_n\\ v_n &=\dfrac{ab}{u_n}\\ u_{n+1}-v_{n+1} &=\dfrac{u_n+v_n}{2}-\dfrac{ab}{u_{n+1}}\\ &=\dfrac{u_n+v_n}{2}-\dfrac{2ab}{u_{n}+v_n}\\ &=\dfrac{(u_n+v_n)^2-4ab}{2(u_{n}+v_n)}\\ &=\dfrac{(u_n-v_n)^2}{2(u_{n}+v_n)}\\ &=\dfrac{u_n-v_n}{2}\dfrac{u_n-v_n}{u_{n}+v_n}\\ |u_{n+1}-v_{n+1}| &=|\dfrac{u_n-v_n}{2}||\dfrac{u_n-v_n}{u_{n}+v_n}|\\ &<|\dfrac{u_n-v_n}{2}|\\ &\to 0\\ \end{array} $

Therefore $u_n, v_n \to \sqrt{ab} $.