Showing that $\lim\limits_{n\to\infty}x_n$ exists, where $x_{n} = \sqrt{1 + \sqrt{2 + \sqrt{3 + ...+\sqrt{n}}}}$ Let $x_{n} = \sqrt{1 + \sqrt{2 + \sqrt{3 + ...+\sqrt{n}}}}$
a) Show that $x_{n} < x_{n+1}$
b) Show that  $x_{n+1}^{2} \leq 1+ \sqrt{2} x_{n}$
Hint : Square $x_{n+1}$ and factor a 2 out of the square root
c) Hence Show that  $x_{n}$ is bounded above by 2. Deduce that $\lim\limits_{n\to \infty} x_{n}$ exists.
Any help? I don't know where to start.

10 days old question, but .
a) Is already clear, that $ \sqrt{1 + \sqrt{2 + \sqrt{3 + ...+\sqrt{n}}}} < \sqrt{1 + \sqrt{2 + \sqrt{3 + ...+\sqrt{n+1}}}}$ , because $\sqrt{n} <\sqrt{n} + \sqrt{n+1}$ which is trivial.
My point here is to give some opinion about b) and c), for me it's better to do the c) first. We know that : $$ \sqrt{1 + \sqrt{2 + \sqrt{3 + ...+\sqrt{n}}}} < \sqrt{p+\sqrt{p+\sqrt{p+ ... }}} $$
But it is only true for $q\leq p<\infty $ for  $q \in \mathbb{Z}^{+}$. Because it is trivial that 
$$ \sqrt{1 + \sqrt{2 + \sqrt{3 + ...+\sqrt{n}}}} > \sqrt{1+\sqrt{1+\sqrt{1+ ... }}} $$
Let $x=\sqrt{2+\sqrt{2+\sqrt{2+ ... }}}$, then $x^2=2+ \sqrt{2+\sqrt{2+\sqrt{2+ ... }}} \rightarrow x^2-x-2=0 $, thus $x=2$, because $x>0$. 
Now let's probe this equation :
$$\sqrt{1 + \sqrt{2 + \sqrt{3 + ...+\sqrt{n}}}} \leq \sqrt{2+\sqrt{2+\sqrt{2+ ... }}}=2 \tag{1}$$
2 is bigger than 1 , with their difference is 1. so for $x_{n}$ to be bigger than 2, it is required for  $\sqrt{2+\sqrt{3+\sqrt{4+ ... \sqrt{n}}}} \geq 3$
but if square both sides of (1) and substract, we get that $\sqrt{2+\sqrt{3+\sqrt{4+ ... \sqrt{n}}}} \leq 3$. 
for (b) , first square both sides, the '1' is gone , square again until the '2' is gone, and we arrive to this equation :
$$\sqrt{3+\sqrt{4+...\sqrt{n}}} \leq 2.\sqrt{2+\sqrt{3+...\sqrt{n}}}$$
which is true, because from (1) we know that $\sqrt{3 +\sqrt{4 ...+\sqrt{n}}} \leq 2$ and $ \sqrt{2+\sqrt{3+...\sqrt{n}}} >0 $
 In fact, if you can prove (b) then (c) is trivial and vice versa.
