To prove a sequence is Cauchy I have a sequence:
$ a_{n}=\sqrt{3+ \sqrt{3 + ... \sqrt { 3} } } $ , it repeats $n$-times.

and i have to prove that it is a Cauchy's sequence. 
So i did this:
As one theorem says that every convergent sequence is also Cauchy, so i proved that it's bounded between $ \sqrt{3}$  and $ 3 $ (with this one i am not sure, please check if i am right with this one.)And also i proved tat this sequence is monotonic. (with induction i proved this: $ a_{n} \leq a_{n+1} $ 
so if it's bounded and monotonic, therefore it is convergent and Cauchy.
I am just wondering if this already proved it or not? And also if the upper boundary - supremum if you wish - is chosen correctly. 
I appreciate all the help i get.
Yes, correct ideas.

For boundedness, you can use induction:


$\sqrt 3&lt;3$, good.
Suppose $a_n&lt;3$ then $a_{n+1}=\sqrt{3+a_n}&lt;\sqrt{3+3}=\sqrt6&lt;3$.
${ a }_{ n+1 }=\sqrt { a_{ n }+3 }  $ $\Rightarrow \quad { a^{ 2 } }_{ n+1 }=a_{ n }+3$ as $n\rightarrow \infty $ $\Rightarrow \quad { a^{ 2 } }_{ n+1 }=a_{ n }+3$ $\quad x^{ 2 }=x+3\quad \Rightarrow $ $x^{ 2 }-x-3=0 $ $and\quad it\quad$ convergents to the $x=\frac { 1+\sqrt { 13 }  }{ 2 } $