- #1

Linday12

- 54

- 0

## Homework Statement

Let a_n be defined recursively by

[tex]a_{1}[/tex]=1, [tex]a_{n+1}[/tex]=sqrt(6+[tex]a_{n}[/tex]) (n=1,2,3,...).

Show that lim n->infinity [tex]a_{n}[/tex] exists and find its value

## The Attempt at a Solution

Observe that [tex]a_{2}[/tex]=[tex]\sqrt{6+1}[/tex]=[tex]\sqrt{7}[/tex] > [tex]a_{1}[/tex]. If [tex]a_{k+1}[/tex] > [tex]a_{k}[/tex], then [tex]a_{k+2}[/tex] = [tex]\sqrt{6+a_{k+1}}[/tex] > [tex]\sqrt{6+a_k}[/tex] = [tex]a_{k+1}[/tex], so {[tex]a_{n}[/tex]} is increasing by induction. I get that part, its the next part I'm a little confused about:

Now, observe that [tex]a_{1}[/tex]=1 < 3. If [tex]a_{k}[/tex] < 3, then [tex]a_{k+1}[/tex]=[tex]\sqrt{6+a_k}[/tex] < [tex]\sqrt{6+3}[/tex], so [tex]a_{n}[/tex] < 3 for every n by induction. How does this show that for all n, the value will be less than 3? Sorry, I'm not really sure how it shows that if you keep going on in the sequence it won't go past 3 eventually.