Studying the convergence of the series $$\sqrt{2}+\sqrt{2+\sqrt{2}}+\sqrt{2+\sqrt{2+\sqrt{2}}}+ \cdots$$

I saw this problem and I tried to do it my own way but I don't know what I'm doing wrong because I'm getting a divergent series while in the solution it says that it's convergent.

This is what I tried to do

I can write $\sqrt{2}=2\cos\frac{\pi}{4}=2\cos\frac{\pi}{2^2}$


Continuing like this I get $$\underbrace{\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}_{n\text{ times}}=2\cos\frac{\pi}{2^{n+1}}$$

So I can write the series $$\sqrt{2}+\sqrt{2+\sqrt{2}}+\sqrt{2+\sqrt{2+\sqrt{2}}}+ \cdots=2\sum_{n=1}^\infty\cos\frac{\pi}{2^{n+1}}$$

Since $$\lim_{n\to\infty}a_{n}=\lim_{n\to\infty}\cos\frac{\pi}{2^{n+1}}=1\ne0$$ the series diverges

  • 10
    $\begingroup$ $\sqrt{2}+\sqrt{2+\sqrt{2}}+\sqrt{2+\sqrt{2+\sqrt{2}}}+ \cdots$ cannot be convergent, since each term is $\ge \sqrt 2$. So your conclusion is correct. $\endgroup$ – Martin R Feb 11 at 18:56
  • 1
    $\begingroup$ I suspect they are talking about the convergence of the sequence $\sqrt{2},\sqrt{2+\sqrt{2}},\sqrt{2+\sqrt{2+\sqrt{2}}}, \cdots$ It converges nicely, but the sum does not. $\endgroup$ – Ross Millikan Feb 11 at 19:00
  • $\begingroup$ Okay, thank you. I wasn't very sure because in the solution they tried to solve it with induction and it got complicated so I tried my own way $\endgroup$ – J.Dane Feb 11 at 19:00
  • $\begingroup$ @RossMillikan No I think that they just made a mistake because in that section where I'm studiyng it only has series $\endgroup$ – J.Dane Feb 11 at 19:02

Your conclusion that $\sum_{n=1}^\infty a_n$ diverges is correct. You don't need the explicit formula, it would suffice to observe that $$ a_n = \underbrace{\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}_{n\text{ times}} \ge \sqrt 2 $$ or that $$ 0 < a_n < a_{n+1} $$ for all $n \in \Bbb N$.


I think what is meant is that the sequence $a_0 = \sqrt{2}, a_{n+1} = \sqrt{2 + a_n}$ converges. This can be shown by proving that for all $n \in \mathbb{N}$ we have $a_n \leq 2$ by induction. It is clear that the sequence is monotonically increasing. Since the sequence is monotonically increasing and has upper bound $2$ it follows that the sequence converges.

Induction step is: $a_{n+1} = \sqrt{2 + a_n} \leq \sqrt{2+2} = 2$

  • $\begingroup$ The sequence converges, but the series does not. $\endgroup$ – N. F. Taussig Feb 11 at 20:36
  • $\begingroup$ True, thanks. I mixed up the terms (had calculus taught in german ;-) ). Changed it now. $\endgroup$ – araomis Feb 11 at 20:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.