Let $x_1=0 $ and for all n, $x_{n+1}$=$\sqrt{2+x_n}$
a)Prove: $x_n <2$
b)Prove: $x_{n+1} > x_n$
I made it through part a painlessly, however, b is giving me problems
Part a: If $ x_k<2 $ then $ x_{k+1}<2$
Attempt: $x_k+2<2+2$
$\sqrt{x_k+2}<\sqrt{2+2}$
$\sqrt{x_k+2}<2$
$x_{k+1}<2$ substitution from initial statement
Part b: Prove: $x_{n+1} > x_n$
Scratch work: $x_{k+1}$=$\sqrt{2+x_k}$ using algebra
$(x_{k+1})^2-2=x_k$