# Limit and rate of convergence of the sequence $a_{n+1}=\frac{a_n^2+b_n^2}{a_n+b_n},~~b_{n+1}=\frac{a_n+b_n}{2}$

Define the sequence the following way for some $x,y \geq 0$:

$$a_0=x,~~~~~~~b_0=y$$

$$a_{n+1}=\frac{a_n^2+b_n^2}{a_n+b_n},~~~~~~b_{n+1}=\frac{a_n+b_n}{2}$$

Obviously:

$$a_n \geq b_n,~~~~n \geq 1$$

For convergence rate we have:

$$a_{n+1}-b_{n+1}=\frac{(a_n-b_n)^2}{2(a_n+b_n)} \tag{1}$$

We have a weak inequality:

$$\frac{(a_n-b_n)^2}{2(a_n+b_n)} \leq \frac{(a_n-b_n)^2}{2(a_n-b_n)}=\frac{a_n-b_n}{2}$$

So our convergence rate is at least linear:

$$\frac{a_{n+1}-b_{n+1}}{a_n-b_n} \leq \frac{1}{2} \tag{2}$$

But shouldn't $(1)$ imply faster (quadratic) convergence?

(I know we need to subtract the limit to find the convergence, but I don't know the closed form, see below).

Now for the limit.

We have the following relations:

$$a_{n+1}b_{n+1}=\frac{a_n^2+b_n^2}{2} \geq a_nb_n$$

$$a_{n+1}b_{n+1}-a_nb_n=\frac{(a_n-b_n)^2}{2}$$

Can we find the limit of this sequence in closed form?

What is the true rate of convergence for this sequence?

• Using an analogy with the AGM mean (en.wikipedia.org/wiki/Arithmetic%E2%80%93geometric_mean), I would try to find a parametric integral $I(a,b)$ that fulfils $$I(a,b)=I\left(\frac{a+b}{2},\frac{a^2+b^2}{a+b}\right)$$ as a consequence of Lagrange's identity and Glasser's master theorem. – Jack D'Aurizio Jun 14 '16 at 17:12
• Using the notations of rgmia.org/papers/monographs/Grec.pdf , we are looking for $\mathcal{A}\otimes\mathcal{C}$. – Jack D'Aurizio Jun 14 '16 at 17:20
• It is also known as the Neumann mean $N_{AC}$: hindawi.com/journals/aaa/2014/914242 – Jack D'Aurizio Jun 14 '16 at 17:38