How to compute the following formulas? $\sqrt{2+\sqrt{2+\sqrt{2+\dots}}}$ 
$\dots\sqrt{2+\sqrt{2+\sqrt{2}}}$
Why they are different?
 A: Define a sequence $(x_n)_{n\geq1}$ so that $x_1=\sqrt2$ and $x_{n+1}=\sqrt{2+x_n}$. Then the second formula you give can be said to colorfully describe the limit $\lim_{n\to\infty}x_n$. If we suppose the limit $L$ does exist (and it is not difficult to show it does exist!), then since for all $n$ we have $x_{n+1}^2=2+x_n$, passing to the limit we see that $L^2=2+L$. The polynomial $x^2-x-2$ has two roots, $-1$ and $2$: since all the $x_n$ are positive, the only possible value for $L$ is $2$.
Can you do the other one?
A: Suppose that the first converges to some value $x$.  Because the whole expression is identical to the first inner radical, $\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}=x=\sqrt{2+x}$ and solving for $x$ gives $x=2$.  Of course, I haven't justified that it converges to some value.
The second can be thought of as starting with $\sqrt{2}$ and repeatedly applying the function $f(x)=\sqrt{2+x}$.  Trying this numerically suggests that the values converge to 2.  Solveing $f(x)=x$ shows that $2$ is a fixed point of that function.
Looking at the second expression is actually how I'd justify (though it is perhaps not a rigorous proof) that the first expression converges.
