How to find a $k$ and $B$ such that $\forall n\in\mathbb N, n\ge 1,|u_n-\sqrt{a}|\le Bk^{2^n}$? Let $a\in\mathbb R_+^*$ and $(u_n)_{n\in\mathbb N}$ a sequence such that : $u_0=1$ and
$\forall n \in \mathbb N, u_{n+1}=\dfrac{1}{2}\left( u_n+\dfrac{a}{u_n}\right)$
Here the goal is to find $k$ and $B$ such that :
$$\forall n\in\mathbb N, n\ge 1,|u_n-\sqrt{a}|\le Bk^{2^n} \text{ with }k\in[0,1[\text{ and }B\in\mathbb R_+^*$$
After some work, i found out that : $$\forall n \ge 1, u_{n}\ge\sqrt{a} \text{ and } (u_n)_{n\in\mathbb N^*}\text{ is decreasing}$$
So : $$B\ge \dfrac{u_1-\sqrt{a}}{k^2} = \dfrac{\dfrac{1+a}{2}-\sqrt{a}}{k^2}=\dfrac{(\sqrt{a}-1)^2}{2k^2} = \left(\dfrac{\sqrt{a}-1}{\sqrt{2}k}\right)^2$$
After that i don't know what i can do to just find something.
Yet even if i find something, i don't know how i would prove the statement.
Thanks for your future help !
EDIT : I notice that : $$\forall n\in\mathbb N, \dfrac{u_{n+1}-\sqrt{a}}{u_{n+1}+\sqrt{a}}=\left(\dfrac{u_{n}-\sqrt{a}}{u_{n}+\sqrt{a}}\right)^2$$
$$\lim_{n\to \infty}u_n=\sqrt{a} \text{ so } \forall \varepsilon\in\mathbb R^+, \exists n_0\in \mathbb N,\forall n\ge n_0,|u_n-\sqrt{a}|\le\varepsilon$$
$$\text{So }, \forall k\in[0,1[, \forall l\in\mathbb N^*,\exists n_0\in \mathbb N,\forall n\ge n_0,|u_n-\sqrt{a}|\le k^{2^l}$$
I don't know but maybe it could be useful ?
 A: Thanks to Neat Math, i found a $k$ and a $B$. Because $\forall n\in\mathbb N, \dfrac{u_{n+1}-\sqrt{a}}{u_{n+1}+\sqrt{a}}=\left(\dfrac{u_{n}-\sqrt{a}}{u_{n}+\sqrt{a}}\right)^2$, we can say :$\forall n\in\mathbb N, \dfrac{u_{n}-\sqrt{a}}{u_{n}+\sqrt{a}}=\left(\dfrac{u_{0}-\sqrt{a}}{u_{0}+\sqrt{a}}\right)^{2^n} \text{ or } \forall n\in\mathbb N^*, \dfrac{u_{n+1}-\sqrt{a}}{u_{n+1}+\sqrt{a}}=\left(\dfrac{u_{1}-\sqrt{a}}{u_{1}+\sqrt{a}}\right)^{2^{n}}$
But $\dfrac{u_{1}-\sqrt{a}}{u_{1}+\sqrt{a}}=\left(\dfrac{\sqrt{a}-1}{\sqrt{a}+1}\right)^2$ so $1\gt\dfrac{u_{1}-\sqrt{a}}{u_{1}+\sqrt{a}}\ge0$
Therefore, we can let $k = \left(\dfrac{\sqrt{a}-1}{\sqrt{a}+1}\right)^2$ and we would have :
$\forall n\in\mathbb N^*, \dfrac{u_{n+1}-\sqrt{a}}{u_{n+1}+\sqrt{a}}=k^{2^{n}}$.
Then, ${u_{n+1}-\sqrt{a}}=k^{2^{n}}(u_{n+1}+\sqrt{a})$ and because $(u_n)_{n\in \mathbb N^*}\text{ is decreasing}$, we can deduce that :
$${u_{n+1}-\sqrt{a}}=k^{2^{n}}(u_{n+1}+\sqrt{a})\le k^{2^{n}}(u_{1}+\sqrt{a}) \Longleftrightarrow k^{2^{n}}(u_{1}+\sqrt{a})\ge \dfrac{1}{2}\left( u_n+\dfrac{a}{u_n}\right)-\sqrt{a}$$
$$\Longleftrightarrow 2k^{2^{n}}(u_{1}+\sqrt{a})\ge u_n+\dfrac{a}{u_n}-\sqrt{a}\ge u_n-\sqrt{a}=|u_n-\sqrt{a}| \text{ because } \forall n \in \mathbb N^*, u_n\ge\sqrt{a}$$
$2(u_1+\sqrt{a})=(\sqrt{a}+1)^2\gt0$ so let $B = (\sqrt{a}+1)^2$
In final, we have : $\forall n \in \mathbb N^*, Bk^{2^n}\ge |u_n-\sqrt{a}|$
And that's what we wanted !
