A little help proving this statement? If $a_n$ is a sequence that implies $a_{n+1}\lt a_n$ and $b_n$>0 is a sequence which implies 
$b_{n+1}\gt b_n$ and also $b_{n+1}=\sqrt{a_n\cdot b_n}$ then I need to prove that $a_n$ and $b_n$ has a limit and they have the same limit.  I think I need to use the lemma of Cantor.I proved one of the two conditions I just need to prove that the limit of the subtraction of both sequences is zero.
 A: Since we are given that $b_n>0$ and for $\sqrt{a_nb_n}$ to exist, we need $a_n > 0$. Now since $a_n$ is monotone decreasing sequence bounded below, we have that $\lim_{n \to \infty} a_n$ exists. 
We shall now prove that $b_n$ is bounded above by $a_1$.
If $b_1 > a_1$, we then obtain that $b_2 \in (a_1,b_1)$, which gives us $b_2 < b_1$ contradicting the fact that $b_n$ is a monotone increasing sequence. Hence, $b_1 < a_1$. Now by induction if $b_k < a_1$, then $b_{k+1} = \sqrt{a_kb_k} < \sqrt{a_1 b_k} < a_1$. Hence, we now have that $b_n$ to be a monotone increasing sequence bounded above by $a_1$. Hence, $\lim_{n \to \infty} b_n$ exists.
Now lets prove that the limits are equal.
Let $\lim_{n \to \infty} a_n = A$ and $\lim_{n \to \infty} b_n = B$. Since we have
$$b_{n+1} = \sqrt{a_n b_n} \implies \lim_{n \to \infty} b_{n+1} = \lim_{n \to \infty} \sqrt{a_n b_n} \implies \lim_{n \to \infty} b_{n+1} = \sqrt{\lim_{n \to \infty} a_n \lim_{n \to \infty} b_n}$$
This gives us that
$$B = \sqrt{AB} \implies B =0 or B=A$$
However, $B=0$ is not possible, since $b_n$ is a positive montone increasing sequence. Hence, we obtain that $B=A$, i.e.,
$$\lim_{n \to \infty} b_{n} = \lim_{n \to \infty} \sqrt{a_n}$$
