Prove Limit of Sequence: a_n ≥ 1 for n ≥ N

[HLUM]

1. Let a₀ and a₁ be positive real numbers, and set a_n+2 = sqrt(a_n+1) + sqrt(a_n) for n \geq 0.
(a) Show that there is N such that for all n \geq N, a_n \geq 1.
(b) Let e_n = |a_n −4|. Show that e_n+2 \leq(e_n+1 +e_n)/3 for n\geq N.
(c) Prove that this sequence converges.




Can someone please give me some hints to start with a)? Thank you in advanced.

 

[vela]

If 0<x<1, is \sqrt{x} bigger or smaller than x?

 

[HLUM]

if 0 < a_n < 1 then a_n x a_n < a_n ==>
a_n < sqrt(a_n)

So you mean i should prove part a by contradiction...

 

[vela]

Perhaps. I don't know. But you can definitely show now that a_{n+2} &gt; a_{n+1}+a_n if a_n, a_{n+1} &lt; 1, which may be useful.

 

[HLUM]

So
a) Assume that for all n \geq 0, 0 < a_n < 1
then sqrt(a_n) > a_n

ie, a_n+2 > a_n+1 + a_n
a_n is increasing sequence

I don't know how to show the contradiction here, but there is no assumption of increasing sequence if you choose a₀ to start with

---> there is N st a_N \geq 1
Assume for all n \geq N, a_n+1 = a_n + a_{n - 1} > 1

therefore, by induction it is true for all n \geq N, a_n \geq 1

 

[jgens]

Edit: My previous post was so incomprehensible that I don't think that it would have been much help. I'll post again later if I can get my thoughts together, but anyway, good luck!