Proving x_n < x_{n+1} for the sequence x_1 = √2, x_{n+1} = √(2+x_n)

[MrBailey]

Hi, all.
I'm working on some proof by induction problems. While I understand the concept, this one threw me for a loop.
Let $$x_1=\sqrt{2}$$ and $$x_{n+1}=\sqrt{2+x_n}$$
Show that $$x_n < x_{n+1}$$
I'd greatly appreciate help with this.
Thanks,
bailey

 

[Gale]

sure you need to use induction? i would show that the stuff under the radical for $$x_{n+1}> x_n$$ we know this because $$x_n>0$$. and 2 plus some other positive number will always be greater than two, and therefore the sq rt of that sum will be greater eh?

 

[AKG]

Uh, you certainly need induction. If x_n = 98, then x_n+1 = (2 + 98)^1/2 = 100^1/2 = 10 < 98 = x_n.

Show that x₁ < x₂
Assume that x_k < x_k+1
Use this to prove that x_k+1 < x_k+2
Write out x_k+1 and x_k+2 in terms of x_k. Then there x_k+1 < x_k+2 will follow immediately from x_k < x_k+1 as long as you know that the function f defined by f(a) = a^1/2 is an increasing function.

 

[MrBailey]

thanks...much clearer now

Bailey