# Converging sequence

1. Sep 17, 2007

### _Andreas

1. The problem statement, all variables and given/known data

Show that the sequence a1=1; an+1 = sqrt(an + 12) for all n> or =1 converges.

3. The attempt at a solution

I know that the sequence is growing, and that this means that an+1 > an for all n> or = 1, but I have no clue how to show this. Anyone got any hints?

Last edited: Sep 17, 2007
2. Sep 17, 2007

### Pseudo Statistic

Well, you should know that a bounded monotone sequence converges.
You've shown it's monotone-- hence, all you have to show is that it's bounded.
You can show using induction that it's bounded above by 4. (If you want to see why, note $$lim_{n\rightarrow\infty} a_n = lim_{n\rightarrow\infty} a_{n+1}$$)
EDIT: Oh, you haven't shown that it's monotone..
Well, $$a_{n+1}>a_{n}$$ implies $$\sqrt{a_{n} + 12}>a_{n}$$
Which is equivalent to:
$$a_{n}+12 > a_{n}^2$$, or $$a_{n}^2-a_{n}-12<0$$.
Now, complete the square. And note that:
$$y^2<x$$ is equivalent to $$-\sqrt{x}<y<\sqrt{x}$$ whenever $$x>0$$.

Last edited: Sep 17, 2007
3. Sep 17, 2007

### Dick

The limit of the sequence is likely to satisfy L=sqrt(L+12). Do you see why? Can you find the limit? To show it's increasing just means showing sqrt(x+12)>x for the region of interest. Finally you'll want to show your sequence is bounded above. I.e. if a_n<L, then a_(n+1)<L.

Last edited: Sep 17, 2007
4. Sep 17, 2007

### _Andreas

Uhm, I think I made a mistake. Sorry for that. In fact I don't know whether the sequence is monotonously increasing or decreasing, so I don't know if $$a_{n+1}>a_{n}$$; I need to show if this is so or not. So, where should I start? I really don't have any clue, since I'm fairly new to this.

Last edited: Sep 17, 2007
5. Sep 17, 2007

### Dick

If a_n=x, then a_(n+1)=sqrt(x+12). You want to show sqrt(x+12)>x. Since a_1=1, you really only need to worry about x>=1. And if you try to find the possible limit first, you don't have to worry about large values of x either.

Last edited: Sep 17, 2007
6. Sep 17, 2007

### _Andreas

Here's what I've done so far: I solved L=sqrt(L+12) as you said, and got two possible limits: -3 and 4. But since -3=/=sqrt(9), the limit has to be 4. This implies that if the sequence is converging, and its limit is 4, $$a_{n+1}>a_{n}$$. Using Pseudo Statistic's hints, this in turn yields the result -3 <an< 4. Have I now shown that the sequence is increasing monotonously, or are there more steps?

7. Sep 17, 2007

### Dick

As I said, you now want to show that sqrt(x+12)>x for x in the range [1,4). That would show it's increasing. You also want to show it's bounded above by 4. For x in that range can you show sqrt(x+12)<4?

8. Sep 17, 2007

### _Andreas

I think I can see that it is increasing, but if there's more to it than just stating that sqrt(x+12)>x if 1<x<4, then I'm lost. Pretty frustrating, to be honest.

9. Sep 17, 2007

### Dick

But there isn't any more to it than that! On what intervals is the function sqrt(x+12)-x positive?

10. Sep 17, 2007

### Dick

Oh, I see. You are worried about bounded. That's just on what intervals is 4-sqrt(x+12) positive. They are both just simple algebra problems. Don't get discouraged!

11. Sep 17, 2007

### _Andreas

Thanks for your help Dick, it's highly appreciated! I have to go to sleep now, but as I said I'm new to this, so maybe I'll have more questions tomorrow (although I really hope not ;)).