Prove that the following sequence defined by the given recurrence relation is convergent:

\begin{align*} a_1 &= \sqrt{2} \\ a_{n+1} &= \sqrt{2 + \sqrt{a_n}} \\ \end{align*}

First, I can prove that the sequence is bounded such that for all $n$, $\sqrt{2} \le a_n \le 2$.

If I can prove that the sequence is monotonically increasing, then textbook theorems guarantee that bounded monotone sequences must converge.

Is there an easy way to show that this sequence is increasing?

I can solve for the fixed point, which I presume is the convergence point, at approximately $1.8311772$. I can casually see that $a_{n+1} > a_n$ for any $a_n$ less than this fixed point value, but that doesn't quite prove monotonicity, and and doesn't disprove the possibility of the sequence oscillating or cycling around the fixed point.

UPDATE: The linked question features the exact same sequence and recurrence relation, but is asking a very different question. I asked how to prove that the sequence is increasing, while the linked answer asked to find the value of convergence. I found that question quite simple, and it looks like the person who posted the linked question found my question easy. greedoid's answer completely solved my problem. So, from my perspective, I'm finished, this can be archived for future searches or deleted/closed. But it's definitely not the same question as the linked question.


Prove by induction, that $a_{n} - a_{n-1}>0$ for all $n$:

Induction step:

$$ a_{n+1}^2 - a_{n}^2 = \sqrt{a_n}-\sqrt{a_{n-1}}>0$$

since $a_n>0$ for all and we have $$\underbrace{(a_{n+1} + a_{n})}_{>0}(a_{n+1} - a_{n})>0$$

we are done.

  • 1
    $\begingroup$ amazing. thank you :) $\endgroup$ – clay Jan 25 at 18:59

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