# Induction Inequality Proof

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.

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:

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

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 $$2$$.

• 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 $n\ge 0$. – DanielWainfleet Mar 21 '20 at 9:17

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