Question

Can anybody help me with this sequence question?

Answer 1

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

Answer 2

did you try the ratio test?

Answer 3

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

Answer 4

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)

Answer 5

use induction to show that \[a_n\le 2\] for all \(n\)

Answer 6

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

Answer 7

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

Answer 8

\[\sqrt{2}\le 2\]

assume \(a_n\le 2\)

\(a_{n+1}=\sqrt{2\cdot a_n}\le \sqrt{2\cdot 2}=2\)

Answer 9

then use monotone convergence theorem

Answer 10

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

Answer 11

"...a function is bounded above if its increasing"

that is not true

Answer 12

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

Answer 13

\[a_n=n\]

is an increasing sequence

Answer 14

it is not bounded above

Answer 15

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

Answer 16

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

Answer 17

thank you for your patience!

Answer 18

Monotone convergence theorem