Is this : $\sqrt{3+\sqrt{2+\sqrt{3+\sqrt{2+\sqrt{\cdots}}}}}$ irrational number? I'm confused how do i show if the bellow number could be irrational or no 
$$\sqrt{3+\sqrt{2+\sqrt{3+\sqrt{2+\sqrt{\cdots}}}}}$$  ? can I use irrationality of $\sqrt{2}+\sqrt{3}$ ?
 A: As soon as you are sure that such expression makes sense, it is a root of $(x^2-3)^2-2-x=x^4-6x^2-x+7$ which does not have any rational root.
A: As Masacroso has already pointed out in the comments, you have to find rational solutions to $c=\sqrt {3+ \sqrt {2+c}}$.
$c^2=3+\sqrt {2+c}$
$c^2-3=\sqrt {2+c}$
$(c^2-3)^2=2+c$
$c^4-6c^2+9=2+c$
$c^4-6c^2-c+7=0$
So the answer to your question is whether there are any rational solutions to $c^4-6c^2-c+7=0$ or not.  
Actually not. Answer courtesy WA.
A: As in the comment if the sequence converges to some value $c\ge 0$ then $c=\sqrt{3+\sqrt{2+c}}$.
From this it follows that we want to see that if the equation
$$c^4-6c^2-c+7=0\tag1$$
have some positive rational solution. Observe that zero cannot be a solution. Now set $c:=p/q$ such that $p$ and $q$ are coprime and positive. From $(1)$ we get
$$p^4-6p^2q^2-pq^3=-7q^4\implies p^2(p^2-6q^2)=q^3(p-7q)\tag2$$
and consequently
$$q^2\mid p^2(p^2-6q^2)\implies q^2\mid (p^2-6q^2)\implies \frac{p^2}{q^2}=c^2\in\Bbb N_{>0}\tag3$$
Now observe that the last equation of $(3)$ implies that $q=1$, thus from the RHS of $(2)$ we have
$$p^2(p^2-6)=p-7\implies\begin{cases}1\le p\le 7\implies p^2\le 6\implies p\in\{1,2\}\\p>7\end{cases}\tag4$$
Thus is easy to check that $p\in\{1,2\}$ are not solutions to the LHS of $(4)$, and if $p>7$ then $p^2(p^2-6)>p-7$, so there are no positive rational solutions to $(1)$.
