Proof that arithmetic and geometric mean converge I need some help with understanding a part of this proof and also writing it up correctly. Given $a_n\geq a_{n+1}\geq b_{n+1} \geq b_n$ with $a_1=a$ and $b_1=b$. I am also given that  $$a_{n+1}=\frac{a_n+b_n}{2}$$ and $$b_{n+1}=\sqrt{a_nb_n}$$
I need to show that sequences ${a_n}$ and ${b_n}$ converges and that ${a_n}$ and ${b_n}$ have the same limit.  
I am told to use the monotonic convergence theorem to prove that both sequences converges and I have the following proof:
Notice that {$a_n$} is monotonically decreasing while {$b_n$}  is monotonically increasing.  Since {$a_n$}  is bounded above by supremum $a_1$ below by its infimum $b_1$, {$a_n$} according to the monotonic convergence theorem has to converge.  
Similarly, notice that {$b_n$} is bounded below by infimum $b$ and supremum $a$.  By monotonic convergence theorem {$b_n$} must also converge as well.  
Next, I am told to show that {$a_n$} and {$b_n$} have the same limit.  In other words, if [$a_n-b_n$] as n tends to infinity must be 0.  For this part, it seems to be the case that one can prove it by just showing that  $a_{n+1} - b_{n+1} \leq (1/2) (a_n - b_n) $.  And I know you can just show this by using the definition of the arithmetic mean, which is  $a_{n+1} - b_{n+1} \leq a_{n+1} - b_n = (1/2) (a_n - b_n)$.  Why is that?  It seems incompletely and not so obvious to me.  An explanation here would help.
Please help me edit my proof (what I have already) and clarify my understanding
 A: Note that from your first inequality: $$a_n > a_{n+1} > b_{n+1} > b_n$$ we know that each sequence $\{ b_n \}$ and $\{ a_n \}$ is monotonic and bounded. Therefore they both converge.
Let $L = \lim_{n\to\infty} a_n$ and $M = \lim_{n\to\infty} b_n$.
Now we also know that $$L = \lim_{n\to \infty} a_{n+1} = \lim_{n\to \infty} \frac{a_n + b_n}{2} = \frac{L+M}{2}$$ and $$M = \lim_{n\to\infty} b_{n+1} = \lim_{n\to\infty} \sqrt{a_n b_n} = \sqrt{LM}.$$
Either equation leads to your solution. $$L = \frac{L+M}{2} \implies 2L = L+M \implies L=M$$ and $$M=\sqrt{LM} \implies M^2 = LM \implies M=L$$ provided $M\neq 0$ for the second equation. However, provided that we assume the initial condition $a>b\neq 0$, we have $M > 0$ by monotonicity.
A: For every $\quad x\epsilon N$ $\quad a_{ n }\ge 0,{ b }_{ n }\ge 0\quad $ using well known inequetion $$\sqrt { ab } \le \frac { a+b }{ 2 } \quad ,a\ge 0,b\ge 0$$ we get $${ a }_{ n+1 }=\frac { { a }_{ n }+{ b }_{ n } }{ 2 } \ge \sqrt { { a }_{ n }{ b }_{ n } } ={ b }_{ n+1 }$$
since $b_{ n+1 }=\sqrt { { a }_{ n }{ b }_{ n } } \ge \sqrt { { b }_{ n }^{ 2 } } ={ b }_{ n }$ and $\\ { a }_{ n+1 }=\frac { { a }_{ n }+{ b }_{ n } }{ 2 } \le { a }_{ n }$ and ${ b }_{ n }\le { a }_{ n }\le { a }_{ 1 },{ a }_{ n }\ge { b }_{ n }\ge { b }_{ 1 }$ so $\left( a_{ n } \right) $ and $\quad \left( { b }_{ n } \right) $ are monoton and bounded so they have finit limits A and B.when $\\ n\rightarrow \infty $ $${ a }_{ n+1 }=\frac { { a }_{ n }+{ b }_{ n } }{ 2 } $$ we get $$A=B$$ 
A: Hint: Take limit in the first relation on both sides. You might want $a$ and $b$ non-negative, otherwise $b_n$ might not be well defined as a real number.
EDIT: Also, as $a_1=a\geq b_1=b\geq 0$ and $a_{n+1}-b_{n+1}\leq 1/2(a_n-b_n)$ for all $n\geq 1$, we have:
$$ 0\leq a_{n+1}-b_{n+1}\leq \frac{1}{2^n}(a-b),$$
leading to $$\lim_{n\to \infty} (a_n- b_n) = 0.$$
