## anonymous 3 years ago Can anybody help me with this sequence question?

1. anonymous

I tried to use proof by induction by I don't know how to show its bounded above x.x I used the rule $a _{1}= \sqrt{2} , a _{n+1}= \sqrt{2a _{n}}$

2. anonymous

did you try the ratio test?

3. anonymous

$L=\lim_{n\to\infty}\left|a_{n+1}\over a_n\right|=\lim_{n\to\infty}\sqrt{2}>1$

4. anonymous

Oops my post erased. My book / cal class does not and will not go over the ratio test I believe. I found the equation myself ( i mean the formula), when I use $\lim_{n \rightarrow \infty} f(a _{n} ) = f(l)$, the answer it gives me is right but induction doesn't seem to work because I don't know if its bounded above ( well I'm assuming its above because it's increasing)

5. Zarkon

use induction to show that $a_n\le 2$ for all $$n$$

6. anonymous

So I just calculate its lowest upper bound limit by plugging in n?

7. anonymous

oh nvm it cant be greater than 2 because 2= root of 4, right?

8. Zarkon

$\sqrt{2}\le 2$ assume $$a_n\le 2$$ $$a_{n+1}=\sqrt{2\cdot a_n}\le \sqrt{2\cdot 2}=2$$

9. Zarkon

then use monotone convergence theorem

10. anonymous

makes very much sense. Thanks a lot, I forgot to think of that. ^^ and yeah, a function is bounded above if its increasing.

11. Zarkon

"...a function is bounded above if its increasing" that is not true

12. anonymous

I mean sequence* I thought if it was a sequence that was only increasing it was bounded above

13. Zarkon

$a_n=n$ is an increasing sequence

14. Zarkon

it is not bounded above

15. anonymous

So, is it IF a sequence is bounded, it is converging? Im sorry Im pretty new to this stuff

16. Zarkon

if the sequence is bounded and it is either increasing or decreasing then the sequence converges.

17. anonymous