Prove that both sequences converge to the same limit.

Let ${a_n}$ be a monotonic decreasing sequence. and Let ${b_n}$ be a monotonic increasing sequence.

Its given that for each $n:$
$$b_{n+1}=\sqrt{a_nb_n}$$

Prove that both sequences converge to the same limit.

by doing some algebra, using the fact that: $a_{n+1}\le a_n$ and $b_n \le b_{n+1}$ I got that the following inequality:

$b_n \le b_{n+1} \le a_{n+1} \le a_n$.

now in order to prove that $a_n$ and $b_n$ converge to the same limit I thought about using the Lemma of Cantor. but there is one thing else that I need to prove first, which is that: $\lim \limits_{n \to \infty}$ $(a_n - b_n)=0$.

do you guys have any hints for how to do that?

• Presumably you mean strict when talking about increasing/decreasing.If not, then $a_n = 1$, $b_n = 0$ would be a solution. – copper.hat Nov 10 '15 at 17:28
• Or at least positive. @copper.hat – Thomas Andrews Nov 10 '15 at 17:28
• @ThomasAndrews: Yes I am certain about that :-). – copper.hat Nov 10 '15 at 17:29
• Thomas' comment contains a key element, that is, at least one $b_n >0$ (and hence all from there on). (I find it remarkably difficult to make a brief, mathematically correct statement.) – copper.hat Nov 10 '15 at 17:30
• @copper.hat yes, but Im not trying to find $a_n$ and $b_n$, i'm trying to prove in general, and that its true for every two sequences with the same qualities. – F1sargyan Nov 10 '15 at 17:34

Hint: $$a_n=\frac{b_{n+1}^2}{b_n}$$
Since you know $b_n$ converges, what is the limit of the right side?
• But Op hasnot said, $b_n$ converges – R.N Nov 10 '15 at 17:28
• Ok, now I noticed that $b_n$ is bounded from above and monotonic increasing, then it converges to its supremum. which is $a_{n+1}$, but how do I calculate the of the right side where I got no numbers at all, I got the limit of $b_n$ as a description of $a_n$. – F1sargyan Nov 10 '15 at 17:31
• No, $a_{n+1}$ is not the supremum of $b_n$. You only know that all $a_k$ are an upper bound for $b_1,\dots,b_n,\dots$, not that $a_k$ is a least upper bound (supremum.) @F1sargyan – Thomas Andrews Nov 10 '15 at 17:32