The converges of $ \sqrt 2 +\sqrt { 2-\sqrt 2} +\sqrt { 2-\sqrt { 2+\sqrt 2} } + \cdots =$

Question

I would like to know wheather this series converge or diverge?

$\sqrt 2 +\sqrt { 2-\sqrt 2} +\sqrt { 2-\sqrt { 2+\sqrt 2} } +\sqrt { 2-\sqrt { 2+\sqrt { 2+\sqrt 2} } } +\cdots$

My work: multiplyIing both demominator and numerator to $\sqrt { 2+\sqrt { 2 } } $,$\sqrt { 2+\sqrt { 2+\sqrt { 2 } } } $,$\sqrt { 2+\sqrt { 2+\sqrt { 2+\sqrt { 2 } } } } $ respectively and so on..

$$\sqrt { 2 } +\frac { \sqrt { 2 } }{ \sqrt { 2+\sqrt { 2 } } } +\frac { \sqrt { 2-\sqrt { 2 } } }{ \sqrt { 2+\sqrt { 2+\sqrt { 2 } } } } +\frac { \sqrt { 2-\sqrt { 2+\sqrt { 2 } } } }{ \sqrt { 2+\sqrt { 2+\sqrt { 2+\sqrt { 2 } } } } } +...=\\ =\sqrt { 2 } +\frac { \sqrt { 2 } }{ \sqrt { 2+\sqrt { 2 } } } +\frac { \sqrt { 2 } }{ \left( \sqrt { 2+\sqrt { 2 } } \right) \left( \sqrt { 2+\sqrt { 2+\sqrt { 2 } } } \right) } +\frac { \sqrt { 2 } }{ \left( \sqrt { 2+\sqrt { 2 } } \right) \left( \sqrt { 2+\sqrt { 2+\sqrt { 2 } } } \right) \left( \sqrt { 2+\sqrt { 2+\sqrt { 2+\sqrt { 2 } } } } \right) } +...=\\ =\sqrt { 2 } \left[ 1+\frac { 1 }{ \sqrt { 2+\sqrt { 2 } } } +\frac { 1 }{ \left( \sqrt { 2+\sqrt { 2 } } \right) \left( \sqrt { 2+\sqrt { 2+\sqrt { 2 } } } \right) } +\frac { 1 }{ \left( \sqrt { 2+\sqrt { 2 } } \right) \left( \sqrt { 2+\sqrt { 2+\sqrt { 2 } } } \right) \left( \sqrt { 2+\sqrt { 2+\sqrt { 2+\sqrt { 2 } } } } \right) } +... \right] $$

we can write every term as :${ a }_{ n+1 }=\frac { 1 }{ \sqrt { 2+{ a }_{ n } } } ,{ a }_{ 0 }=\sqrt { 2 } $ it is obvious that the sequences are monotone decreasing so ${ a }_{ n }>{ a }_{ n+1 }$ and I wrote sum as:$$S_{ \infty }=\sqrt { 2 } \sum _{ i=0 }^{ \infty }{ \frac { 1 }{ { a }_{ i+1 }\sqrt { 2+{ a }_{ i } } } } $$(i am not sure about it) I am stuck here,I suspect series converges but how to show,which convergence tests can i use i don't know?Any hints,help will be appriceated? P.S.I apologize for my english

Does every term in the sequence have one $2-\sqrt{\phantom{0}}$ at the front and all the rest being $2+\sqrt{\phantom{0}}$? — hmakholm left over Monica, Jul 20 '15 at 12:31
In that case I suggest writing $a_n=\sqrt{2-b_n}$ where $b_n = \sqrt{2+\sqrt{2+\sqrt{\cdots}}}$. Then the $a_n$s will converge exactly if the $b_n$s do, and you can investigate the latter by looking at the iteration $b\mapsto \sqrt{2+b}$. [Hmm, perhaps never mind that -- I had overlooked that you're trying to sum the terms rather than find the limit of the seuqence]. — hmakholm left over Monica, Jul 20 '15 at 12:34
In fact, the $n$'th term is bounded above by $\pi/2^n$, so the series converges. — David Mitra, Jul 20 '15 at 12:44

score 5 · Accepted Answer · answered Jul 20 '15 at 12:54

Let: $$ c_1=\sqrt{2},\quad c_2=\sqrt{2+\sqrt{2}},\quad c_{n+1}=\sqrt{2+c_n}$$ and $d_n=c_n/2$. Then $d_1=\cos\frac{\pi}{4}$ and $d_{n+1}=\sqrt{\frac{1+d_n}{2}}$, by recognizing the cosine duplication formula, give: $$ c_n = 2 \cos\frac{\pi}{2^{n+1}} $$ so: $$ \sqrt{2} = 2\sin\frac{\pi}{4},\quad \sqrt{2-\sqrt{2}}=2\sin\frac{\pi}{8},\qquad \sqrt{2-c_n} = 2\sin\frac{\pi}{2^{n+2}}$$ and we are asking if: $$ \sum_{n\geq 0}2\sin\frac{\pi}{2^{n+2}} $$ is convergent, but that is trivial since $0<\sin\frac{\pi}{2^{n+2}}<\frac{\pi}{2^{n+2}}$. We may also compute such a sum:

$$ 2\sum_{n\geq 0}\sin\frac{\pi}{2^{n+2}}=2\sum_{m\geq 0}\frac{(-1)^m \pi^{2m+1}}{(2m+1)!}\sum_{n\geq 0}\frac{1}{2^{(n+2)(2m+1)}}=2\sum_{m\geq 0}\frac{(-1)^m \pi^{2m+1}}{(2m+1)!\left(2^{4m+2}-2^{2m+1}\right)}.$$

another elegant solution,thanks Jack) – haqnatural Jul 20 '15 at 13:02 — haqnatural, Jul 20 '15 at 13:02

score 3 · Answer 2 · answered Jul 20 '15 at 12:53

First, the sum we want can be rewritten as $\displaystyle\;\sum_{n=0}^\infty \sqrt{\epsilon_n}\;$ where $$\epsilon_n = 2 - \overbrace{\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}^{n\; \text{terms}}$$ Notice for $n \ge 0$, $$\epsilon_{n+1} = 2 - \sqrt{4 - \epsilon_n} = \frac{\epsilon_n}{2+\sqrt{4-\epsilon_n}}$$ It is easy to see $\epsilon_n > 0 \implies \epsilon_{n+1} > 0$. Since $\epsilon_0 = 2 > 0$, we have $\epsilon_n > 0$ for all $n$. Furthermore,

$$\epsilon_{n+1} \le \frac{\epsilon_n}{2}, \forall n \ge 0 \quad\implies\quad \epsilon_n \le \frac{\epsilon_0}{2^n} = 2^{1-n}, \forall n \ge 0 $$

This leads to an upper bound for the sum at hand $$\sum_{n=0}^\infty \sqrt{\epsilon_n} \le \sum_{n=0}^\infty 2^{\frac{1-n}{2}} = \frac{\sqrt{2}}{1-\frac{1}{\sqrt{2}}} = 2(\sqrt{2}+1) < \infty$$

Since the sum is over non-negative numbers, this implies the sum converges.

The converges of $ \sqrt 2 +\sqrt { 2-\sqrt 2} +\sqrt { 2-\sqrt { 2+\sqrt 2} } + \cdots =$

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