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did you try the ratio test?

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

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

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

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

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

then use monotone convergence theorem

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

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

\[a_n=n\]
is an increasing sequence

it is not bounded above

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

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

thank you for your patience!

Monotone convergence theorem