$\displaystyle {a_{n + 1}} = \sqrt {2{a_n}}
$
Is this a convergent sequence and if so, what does it converge to?
I have no idea how to beginning this...
It is easy to see that the series converges for any value of $\displaystyle a_1>0$
if $\displaystyle a_1=0$ then the series converges to zero.
Now lets assume that $\displaystyle a_n>2$ then it follows that:
$\displaystyle a_{n+1}=\sqrt{2a_n}<\sqrt{a_n\cdot a_n}=a_n$
And
$\displaystyle a_{n+1}=\sqrt{2a_n}>\sqrt{2\cdot2}=2$
So you see that if $\displaystyle a_n>2$ then: $\displaystyle 2<a_{n+1}<a_n$.
Now what happens if $\displaystyle a_n<2$ ??
Hope that helps.