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Given a sequence $(a_{n})$ where $a_{0}=1, a_{n+1} = \sqrt{a_{n}+2}$ I'm trying to show using induction that $1 \leq a_{n} \leq 2$, for all $n$. (I've shown the limit is actually $2$).

I've never actually done an inequality induction before, but an attempt is below:

Base case: For $a_{0}$ we clearly have $1\leq a_{0} \leq 2$

Inductive step: Assume the result holds for $n=k$ i.e. $1\leq a_{k} \leq 2$

Then I need to show that $1 \leq a_{k+1} \leq 2$. Because this is a sequence I'm not sure if I can say $a_{k+1} = a_{k}+a_{1}$ (I'm assuming not).

I think the recursion is confusing me for some reason.

Any hints on how to proceed would be helpful.

Thanks.

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We know that $a_k \leq 2$, so

$$a_{k+1}=\sqrt{a_k+2} \leq \sqrt{2+2} = \sqrt{4} = 2$$

The other inequality can be achieved in the following way:

$$a_{k+1}=\sqrt{a_k+2}\geq\sqrt{2}>1$$

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With the substitution $a_n=2\cos\theta_{n}$ we get $a_{n+1}=2\cos\left(\frac{\theta_n}{2}\right)$. Since $\theta_0=\frac{\pi}{3}$, $$ a_n = 2\cos\left(\frac{\pi}{3\cdot 2^n}\right) $$ is blatantly convergent to $2$.

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  • $\begingroup$ Reminds me of this one for Heron's method of approximating $ \sqrt A:$ For $A>0$ and $x_0>0$, let $x_{n+1}=(x_n+Ax_n^{-1})/2.$ Now if $x_0\ne \sqrt A$ then $x_1=\sqrt A\coth T$ with $T\in \Bbb R^+$ and we have $x_{n+1}=\sqrt A \coth (2^nT)$ for all $n\ge 0$. $\endgroup$ – DanielWainfleet Mar 21 '20 at 9:17
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Is easy to see that $a_n \geq 0,\forall n \in \Bbb{N}$

So $a_{n+1} \geq \sqrt{2} >1$

Also since you assume that $a_n \leq 2$ then $a_{n+1} \leq \sqrt{4}=2$

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