# Homework Help: Limit of sequence

1. Jan 20, 2010

### HLUM

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

2. Jan 20, 2010

### vela

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

3. Jan 20, 2010

### HLUM

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

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

4. Jan 20, 2010

### vela

Staff Emeritus
Perhaps. I don't know. But you can definitely show now that $a_{n+2} > a_{n+1}+a_n$ if $a_n, a_{n+1} < 1$, which may be useful.

5. Jan 20, 2010

### HLUM

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

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

I dont 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 aN $$\geq$$ 1
Assume for all n $$\geq$$ N, an+1 = an + an - 1 > 1

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

6. Jan 20, 2010

### 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!

Last edited: Jan 20, 2010