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 $a_n^2<a_n+2$, 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?

  • $\begingroup$ I'm sorry, an edit was made to the post that was actually incorrect. It is increasing. √2 < √(2+√2) and so on $\endgroup$ – jacksonf Mar 26 at 5:52
  • $\begingroup$ 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? $\endgroup$ – jacksonf Mar 26 at 5:58
  • $\begingroup$ $(a_n-2)(a_n+1)<0$ 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! $\endgroup$ – 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.


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