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

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

$$\sqrt{2+\sqrt{2}}=\sqrt{2+2\cos\frac{\pi}{4}}=2\sqrt{\frac{1+\cos2\frac{\pi}{8}}{2}}=2\cos\frac{\pi}{8}=2\cos\frac{\pi}{2^3}$$

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

• $\sqrt{2}+\sqrt{2+\sqrt{2}}+\sqrt{2+\sqrt{2+\sqrt{2}}}+ \cdots$ cannot be convergent, since each term is $\ge \sqrt 2$. So your conclusion is correct. – Martin R Feb 11 at 18:56
• 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. – Ross Millikan Feb 11 at 19:00
• 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 – J.Dane Feb 11 at 19:00
• @RossMillikan No I think that they just made a mistake because in that section where I'm studiyng it only has series – 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: $$a_{n+1} = \sqrt{2 + a_n} \leq \sqrt{2+2} = 2$$