# limit of a recurrent serie

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

• Hint 1) $u_{n+1}v_{n+1} = u_nv_n$. 2) $\lim\limits_{n\to\infty} (u_n - v_n ) = 0$. From this, you can deduce the limit is the GM. Oct 7, 2019 at 5:14

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

• Got it , all what I need was digging deeper to get the limit. $Thanks$ Oct 7, 2019 at 8:10