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Let $a_1 = \sqrt{2}$ and $a_{n+1} = \sqrt{2 + \sqrt{a_n}}$. Now I want to show by induction that $\sqrt{2} \leq a_n \leq 2$ for all $n$.

The base case is $n=1$ and it is clear $\sqrt{2} \leq a_1 \leq 2$. Then I assume that $\sqrt{2} \leq a_n \leq 2$ holds and I want to show $\sqrt{2} \leq a_{n+1} \leq 2$.

Then I note that $\sqrt{2} \leq a_{n+1} \leq 2 \implies \sqrt{2} \leq \sqrt{2 + \sqrt{a_n}} \leq 2$. By squaring both sides I get $2 \leq 2 + \sqrt{a_n} \leq 4$. Then by subtracting $2$, I get $0 \leq \sqrt{a_n} \leq 2$. This means that $0 \leq a_n \leq 4$. This is ok because I assumed $0 \leq \sqrt{2} \leq a_n \leq 2 \leq 4$.

Edit How about this: Since I know $\sqrt{2} \leq a_n \leq 2$. Then it is clear that $0 \leq a_n \leq 4$. By taking square roots I get $ 0 \leq \sqrt{a_n} \leq 2$. Now if I add 2, $2 \leq \sqrt{a_n} + 2 \leq 4$. Taking another square root I get $\sqrt{2} \leq \sqrt{\sqrt{a_n} + 2} \leq 2$. So $\sqrt{2} \leq a_{n+1} \leq 2$.

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Note that you want to show the case $n$ implies $n+1$, but you're going the other way.

The good thing about this sequence is the inductive step is indeed easy.

The base case is $\sqrt 2 \leq a_1 \leq 2$

We now assume $\sqrt 2 \leq a_n \leq 2$

Take a square root, then add two, then take a square root.

$$\sqrt{\root 4 \of 2+2} \leq \sqrt{2+\sqrt a_n}\leq \sqrt{\sqrt2+2}$$

$$\sqrt{\root 4 \of 2+2} \leq a_{n+1}\leq \sqrt{\sqrt2+2}$$

Note that since $\sqrt 2 + 2 < 4 \Rightarrow \sqrt {\sqrt 2 + 2} < 2$ and similarily $2 < \root 4 \of 2 + 2 \Rightarrow \sqrt 2 < \sqrt {\root 4 \of 2 + 2} $

Add: Your edit makes perfect sense.

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    I thought you start with the ($n+1$)th case and break it down to the $n$th case when doing induction.2012-04-29
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    Think about the definition of an inductive set $S$. a. $1\in S$ b. $n \in S \Rightarrow n+1\in S$ What would you get if you prove $n+1$ implies $n$? Induction works because we move from $n$ to $n+1$, which, having proven the base case $1$, transmits the property to $2$, $2$ to $3$, and so on...2012-04-29
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    @Peter : What he meant is that he thought that he had to write what the $(n+1)$th case meant, and then make its statement somehow equivalent to the $n$th statement in order to conclude that the $(n+1)$th is true because the $n$th is.2012-04-29
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    @PatrickDaSilva: Could you take a look at my edit?2012-04-29
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    @Student : as I mentioned in my comment to Peter, it doesn't matter how you decide to prove that the $n$th case implies the $(n+1)$th case ; the only thing that is important is that you do prove it. Peter decided to construct the $(n+1)$th case out of the $n$th, i.e. he wrote the $n$th, assumed it true, and then did some computations to deduce that the $(n+1)$th is true by making it appear little by little. Let's say his technique is "a little harder to think to do in general, but is more clean mathematically speaking".2012-04-29
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    @Student ; this is essentially what "getting the $(n+1)$th case out of the $n$th" is about.2012-04-29
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    @PatrickDaSilva That was what I was trying to explain.2012-04-29
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    @Peter : I meant that he understood the logic behind induction, but his mechanics of "how to prove $n \rightarrow n+1$" were different and that is what his comment was about.2012-04-29
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    @PatrickDaSilva I totally agree with you. Thanks for explaining it.2012-04-29