# Find the limit of the sequence

• KungPeng Zhou
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?

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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##.

nuuskur said:
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.

nuuskur said:
For starters, how do you know this sequence is bounded?
By Herschfeld's Convergence Theorem.

weirdoguy and nuuskur
KungPeng Zhou said:
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.

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 $x \geq 0$, $$\left|\sqrt{2x} -2\right| < |x - 2|$$ from which the result follows.

nuuskur
nuuskur said:
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.

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## What is the limit of the sequence as n approaches infinity?

The limit of a sequence as n approaches infinity is the value that the terms of the sequence get closer to as n becomes very large. If the terms of the sequence do not approach any particular value, then the sequence does not have a limit.

## How do you determine if a sequence is convergent or divergent?

A sequence is convergent if it approaches a specific limit as n approaches infinity. It is divergent if it does not approach any limit. To determine this, you can use various mathematical techniques such as analyzing the general term of the sequence, applying limit theorems, or using comparison tests.

## What is the limit of the sequence a_n = 1/n?

The limit of the sequence a_n = 1/n as n approaches infinity is 0. This is because as n gets larger and larger, the value of 1/n gets smaller and smaller, approaching zero.

## Can a sequence have more than one limit?

No, a sequence can have at most one limit. If a sequence has a limit, it is unique. If a sequence appears to have more than one limit, it is actually divergent.

## What methods can be used to find the limit of a sequence?

Methods to find the limit of a sequence include direct substitution, applying limit laws, using L'Hôpital's rule (for sequences that can be expressed as functions), and employing comparison tests or squeeze theorem. Each method is suitable for different types of sequences.

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