# Proving $\lim _{n\to \infty }a_{n+1}=\lim _{n\to \infty }b_{n+1}$ where $a_{n+1}=\frac{a_n+b_n}{2}\:$, $b_{n+1}=\sqrt{a_n\cdot \:b_n}$

$a_1,\:b_1>0$

$a_{n+1}=\frac{a_n+b_n}{2},\:b_{n+1}=\sqrt{a_n\cdot b_n}$

The question asks to prove that: $\lim _{n\to \infty }\left(a_n\right)=\lim \:_{n\to \:\infty \:}\left(b_n\right)$.

Proving it once I show that both sequences converge is obviously the easy part. However, I can't get to that point.

• Check whether the sequences are monotonic (except maybe at the first steps). Nov 20 '14 at 21:57
• An interesting side note: the closed form for the limit is given in terms of complete elliptic integral of the first kind. Jul 30 '17 at 12:30

By the AM-GM inequality we have: $$b_n \leq b_{n+1} \leq \ldots \leq a_{n+1}\leq a_n$$ so both the sequences $\{a_n\}_{n\in\mathbb{N}}$ and $\{b_n\}_{n\in\mathbb{N}}$ are converging since they are monotonic and bounded. Since: $$a_{n+1}-b_{n+1} = \frac{1}{2}(\sqrt{a_n}-\sqrt{b_n})^2 \leq \frac{1}{2}(a_n-b_n)$$ the limit is the same.
Try using the AMGM inequality: if $c,d>0$ then: $$\frac{c+d}{2}\ge \sqrt{cd}$$
More specifically - Notice that we have the following inequalities for all $n\ge 1$:
1. $a_n \ge b_n$ (AMGM inequality)
2. $a_{n+1} \le a_n$ (by the previous inequality and the definition of $a_n$)
3. $b_{n+1} \ge b_{n}$ (by the first inequality and the definition of $b_n$)
This implies that $(a_n)$ is nonincreasing and bounded from below by $b_1$ (use 1 and 3) and that $(b_n)$ is nondecreasing and bounded from above by $a_1$ (use 1 and 2), thus both converge.