Find the limit of the sequence

[KungPeng Zhou]

Homework Statement

${\sqrt{2},\sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}... }$

Relevant Equations

The way to calculate limit of sequence

First, we can know
$a_{n}=\sqrt{2a_{n-1}}$
When$n\rightarrow \infty$
$a_{n}=\sqrt{2a_{n-1}}$
And we can get the answer is 2.
Is this solution right? And is any other way to solve the question?

 

[nuuskur]

No, it is definitely not right. For starters, how do you know this sequence is bounded? Clearly the sequence is monotone. Show, for instance by induction, that $a_n\leqslant 2$ for every $n$. Follow up by showing that $\sup a_n = 2$. Then may we conclude that $\lim a_n= 2$.

 

[pbuk]

No, it is definitely not right.

I think that's a bit harsh: clearly $a = 2$ is the only solution to $a = \sqrt{2a}$ so if a limit exists, than it is 2.

For starters, how do you know this sequence is bounded?

By Herschfeld's Convergence Theorem.

 

[pbuk]

When$n\rightarrow \infty$ $a_{n}=\sqrt{2a_{n-1}}$

This doesn't make sense, you already stated $a_{n}=\sqrt{2a_{n-1}}$ for any $n$. Did you mean to write $a_{\infty}=\sqrt{2a_{\infty}}$? This notation is a bit of a shortcut (I would prefer " let$a = \lim_{n \to \infty} a_n$ then we have $a = \sqrt{2a}$") but it gets the job done for me.

 

[nuuskur]

I consider only what was written. Had all of that from #3 been written in #1 I'd have no problem with it. As it stands right now, it feels more like "is my solution right, because I guessed the limit?". Maybe I'm being unreasonably uncharitable.

 

[pasmith]

It's not sufficient to note the existence of a fixed point of the iteration. The fixed point might be unstable, or if it is stable then the initial value might not be in its domain of stability. Those things need to be checked.

In this case, you can show that for any x \geq 0, <br /> \left|\sqrt{2x} -2\right| &lt; |x - 2| from which the result follows.

 

[KungPeng Zhou]

No, it is definitely not right. For starters, how do you know this sequence is bounded? Clearly the sequence is monotone. Show, for instance by induction, that $a_n\leqslant 2$ for every $n$. Follow up by showing that $\sup a_n = 2$. Then may we conclude that $\lim a_n= 2$.

Ok, there is good way to proof this sequence is bounded.
From$a_{n}=\sqrt{2a_{n-1}}$
$a_{n}<2$ as $a_{n-1}<2$
However we know$a_{1}=\sqrt{2}<2$
So we kown$a_{2}<2... a_{n}<2$

 

[nuuskur]

That works. You can express this more clearly. We have $a_1\leqslant 2$. Assume $a_n\leqslant 2$, then $a_{n+1} = \sqrt{2a_n} \leqslant \sqrt{2\cdot 2} = 2.$

As for the initial problem, one may note that $1<a_n<2$ for all $n$ and $a_n$ is strictly increasing. Hence, $\sup a_n=2$ is forced.