## MarcLeclair Group Title Can anybody help me with this sequence question? one year ago one year ago

1. MarcLeclair Group Title

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. electrokid Group Title

did you try the ratio test?

3. electrokid Group Title

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

4. MarcLeclair Group Title

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 Group Title

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

6. MarcLeclair Group Title

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

7. MarcLeclair Group Title

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

8. Zarkon Group Title

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

9. Zarkon Group Title

then use monotone convergence theorem

10. MarcLeclair Group Title

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 Group Title

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

12. MarcLeclair Group Title

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

13. Zarkon Group Title

$a_n=n$ is an increasing sequence

14. Zarkon Group Title

it is not bounded above

15. MarcLeclair Group Title

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

16. Zarkon Group Title

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

17. MarcLeclair Group Title