# Sequence defined by $a_{n+1} = \sqrt{2+a_n}$, $a_1 = \sqrt{2}$ is monotonically increasing

I am trying to prove that the sequence $$a_1=\sqrt{2}$$, $$a_{n+1}=\sqrt{2+a_n}$$ is monotonically increasing. My thought was that since $$a_{n+1}^2= 2+a_n> 2a_n > a_n \times a_n > a_n^2, a_{n+1} > a_n^2$$.

However, upon reflecting I think this is not actually true that just because $$a_{n+1}^2 > a_n^2, a_{n+1} > a_n$$.

Is the only way to prove this by using induction? I was trying to do something more direct. If it is using induction, could you outline the proof?

Hint: All quantities involved are positive, and $$a_{n+1}=\sqrt{2+a_n}>a_n$$ if and only if $$a_n^2, i.e. $$(a_n-2)(a_n+1)<0$$. Can you put a bound on $$a_n$$ and hence use this inequality to prove the result?

• I'm sorry, an edit was made to the post that was actually incorrect. It is increasing. √2 < √(2+√2) and so on – jacksonf Mar 26 at 5:52
• I have already proved that an is always less than 2. I have to say I'm still confused, how does this bound help prove the result? – jacksonf Mar 26 at 5:58
• $(a_n-2)(a_n+1)<0$ if and only if $a<2$ and $a_n>-1$. You have the first part, the second part is trivial. So the inequality is true! – YiFan Mar 26 at 6:00

First you check as an exercise that $$a_{n}>0$$ for all $$n\geq 1$$

We need to prove that $$a_{n+1}-a_{n}>0$$ for all $$n\geq 1$$.

We would use induction on $$n$$.

For $$n=1$$, we have $$a_{2}-a_{1}=\sqrt{2+\sqrt{2}}-\sqrt{2}$$

Now observe that, $$\sqrt{2}=(\sqrt{2+\sqrt{2}})^{2}-(\sqrt{2})^2=(\sqrt{2+\sqrt{2}}-\sqrt{2})(\sqrt{2+\sqrt{2}}+\sqrt{2})$$

And since $$\sqrt{2}>0$$ and $$\sqrt{2+\sqrt{2}}+\sqrt{2}>0$$, the above product shows that $$\sqrt{2+\sqrt{2}}-\sqrt{2}>0$$ which is equivalent to $$a_{2}-a_{1}>0$$.

So the result is true for $$n=1$$.

Now assume that the result is true for $$n=k$$,i.e. $$a_{k+1}>a_{k}$$.

$$a_{k+2}>a_{k+1} \iff \sqrt{2+a_{k+1}}>a_{k+1} \iff 2+a_{k+1}>(a_{k+1})^2 \iff 2+a_{k+1}>2+a_{k} \iff a_{k+1}>a_{k}$$.

Thus, we have proved the result by induction.