Prove $a_n = \frac{a_{n-1} + b_{n-1}}{2}$ and $b_n = \frac{2 \times a_{n-1} \times b_{n-1}}{a_{n-1} + b_{n-1}}$ are bounded, monotonic and have limit Title is too small for providing all needed information. So here it is:
$a_0 > b_0 > 0$. $a_0$ and $b_0$ are fixed numbers. Prove that $a_n = \frac{a_{n-1} + b_{n-1}}{2}$ and $b_n = \frac{2 \times a_{n-1} \times b_{n-1}}{a_{n-1} + b_{n-1}}$ are bounded, monotonic and have the same limit
Before we can state that $\lim a_n$ or $\lim b_n$ even exists, we must prove, that this sequences are monotonic and bounded.
Obviously, $a_n$ is decreasing because it's arithmetic mean and $a_n$ is the largest number we have every time.
I can't say for sure about $b_n$, but it is increasing (did small tests).
In order to prove that, for example, $a_n$ is monotonic, I need 1) check difference $a_n - a_{n-1}$ or 2) look at $\frac{a_n}{a_{n-1}}$.
The problem is $a_n$ is defined by "mix" of $a_{n-1}$ and $b_{n-1}$. Same about $b_n$.
Unfortunately, I can't find the way to express $a_n$ only by $a_{n-1}$ or $b_{n-1}$ (again, same about $b_n$).
Let's see an example.
$a_n - a_{n-1} = \frac{a_{n-1} + b_{n-1} - a_{n-2} - b_{n-2}}{2}$
Well, we didn't prove that $a_{n-2} > a_{n-1}$ or vice versa, hence, we can't say nothing about $a_n - a_{n-1}$
Perhaps, it is easily done by induction. However, I have no clue how to properly do this.
Maybe, like this(?):
(1) $a_1 - a_0$ < 0, because $a_1 - a_0 = \frac{a_0 + b_0}{2} - a_0 = \frac{a_0 + b_0 - 2 a_0}{2} = \frac{- a_0 + b_0}{2} < 0$, since $a_0 > b_0$
and $b_1 - b_0$ > 0, because $b_1 - b_0 = \frac{2 \times a_0 \times b_0 - a_0 b_0 - b_0^2}{a_0 + b_0} = \frac{a_0 b_0 - b_0^2}{a_0+b_0} > 0$, since $a_0 > b_0$.
(2) Suppose $a_n - a_{n-1}$ < 0 is true and $b_n - b_{n-1} > 0$ is true.
Then $a_{n+1} - a_n = \frac{a_n + b_n - a{n-1} - b{n-1}}{2}$. Can we compare absolute values $|a_n - a_{n-1}|$ and $|b_n - b_{n-1}|$? If first is bigger, then $a_{n+1} - a_n < 0$.
So many questions.
 A: Let $a>b>0$. Then $A = \frac{a+b}{2}$ is the arithmetic mean of $a,b$, and $B = \frac{2ab}{a+b}$ is the harmonic mean of $a,b$. We always have
$$
b < B < A < a.
$$
This is the most important inequality, and I will leave it to you to prove it.
For any $n\in \mathbb{N}$, set $(a,b)\leftrightarrow (a_n,b_n)$ and $(A,B) \leftrightarrow (a_{n+1},b_{n+1})$.
By the above inequality it follows that the sequences $\{a_n\}_{n\in\mathbb{N}}$,$\{b_n\}_{n\in\mathbb{N}}$ are bounded (by $a$ and $b$), and monotonic.
Since $|A-B| < \frac{|a-b|}{2}$, we have that $\lim_{n \to \infty}{(a_n-b_n)}=0$.
Let $\varepsilon>0$, and let $n_0$ such that $|a_n-b_n| < \varepsilon$. We have $a_k \in (b_n, a_n)$ for all $k> n_0$. Therefore the series ${a_n}$ is Cauchy, and the same can be said for ${b_n}$, which leads us to conclude that both series are convergent.
Their limit is equal because
$$
\lim_{n\to\infty}(a_n-b_n) = 
\lim_{n\to\infty}a_n - \lim_{n\to\infty}b_n = 0.
$$
EDIT: I wrote this solution before noticing that you asked a proof of monotonicity. Originally I didn't include that, but I edited the answer. However, once you have $b_n < b_{n+1} < a_n < a_{n+1}$, you instantly have that both sequences are bounded and monotonic, therefore convergent, which wraps up the exercise without touching Cauchy convergence.
