Suppose that 0 < x_1 < 2 and x_(n+1) = sqrt(2 + x_n).
Prove that 0 < x_n < x_(n+1) < 2 for all n Natural Numbers.
Suppose that $\displaystyle K > 1\,\& \,x_{K - 1} < x_K < 2$ is true.
$\displaystyle \begin{array}{rclcl}
{x_{K - 1} + 2} & < & {x_K + 2} & < & 4 \\
{\sqrt {x_{K - 1} + 2} } & < & {\sqrt {x_K + 2} } & < & {\sqrt 4 } \\
{x_K } & < & {x_{K + 1} } & < & 2 \\ \end{array} $