4
$\begingroup$

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

$(x_{k+1})^2-2=x_k$

  • 0
    Talk about "race condition" :-)2011-10-15
  • 0
    arete: Are you sure $x_{n+1}2011-10-16
  • 1
    arete: the race condition comment was about me and Arturo making the same edit at the same time. Nothing about the contents of your question.2011-10-16

1 Answers 1

0

You can get parts (a) and (b) with a single induction.

By definition $x_1=0$ and $x_2=\sqrt{2+x_1}=\sqrt2$, so $0\le x_1