The converges of $ \sqrt 2 +\sqrt { 2-\sqrt 2} +\sqrt { 2-\sqrt { 2+\sqrt 2} } + \cdots =$ 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
 A: 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:
$$ 2\sum_{n\geq 0}\sin\frac{\pi}{2^{n+2}}=2\sum_{m\geq 0}\frac{(-1)^m \pi^{2m+1}}{(2m+1)!}\sum_{n\geq 0}\frac{1}{2^{(n+2)(2m+1)}}=2\sum_{m\geq 0}\frac{(-1)^m \pi^{2m+1}}{(2m+1)!\left(2^{4m+2}-2^{2m+1}\right)}.$$
A: 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.
