Prove the convergence of the following sequence:
$$x_1 = \sqrt{a}$$
$$x_{n+1} = \sqrt{a + x_n}$$
Prove the convergence of the following sequence:
$$x_1 = \sqrt{a}$$
$$x_{n+1} = \sqrt{a + x_n}$$
Hint :
Prove that the sequence $\{x_n\}$ is bounded and monotonic
(Of couse we need $a\ge 0$ as premise)
Show by induction that $0\le x_n<a+1$
Show by induction that $x_{n+1}\ge x_n$
(You can facilitate things a bit by prepending a term $x_0=0$. Not that then still $x_1=\sqrt{a+x_0}$)