# Prove that this sequence converges

1. Mar 22, 2007

### playboy

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

Let $a_{0} = a >1$ and let $a_{n+1} = a^{a_n}$.
Show that {$a_{n}$} comverges for $a < e^{e^-1} = 1.4446678$

2. Relevant equations

This is a Theorm I learned in Real Analysis and hope to apply it to this problem:

Theorm: If a sequence is montonically increasing and bounded, then it is convergent

3. The attempt at a solution

{$a_{n}$} = {$a, a^a, a^{a^a}, ...$}

Clearly, {$a_{n}$} is monotonically increasing is is bounded below by a.

How do I show that it is bounded above?

Last edited by a moderator: Mar 22, 2007
2. Mar 22, 2007

### dmoravec

try induction assuming that a(o) is less than the maximum a value...

3. Mar 22, 2007

### playboy

Sorry, I dont understand what you are trying to say...

Clearly,

$a < a^a < a^{a^a} < ...$ ... So I am having a hard time explaining how this is bounded for a>1

4. Mar 22, 2007

### Werg22

This doesn't mean anything. A sequence could be increasing and still be bounded. And $$e^{e-1} = 5.574941539$$. I don't think that's what you meant....

Last edited: Mar 22, 2007
5. Mar 22, 2007

### Dick

He meant e^(1/e). The formatting didn't come out 100% clear.

6. Mar 22, 2007

### Dick

If a<e^(1/e) and x<e, what can you tell me about a^x?

7. Mar 22, 2007

### playboy

Thank you for the correction Dick...I allways have trouble with this latex stuff

If a<e^(1/e) and x<e, we can conclude that a^x < e

so a^x < e

hence,

a < e^(1/x) ...

Can you please give me another push ?

8. Mar 22, 2007

### Dick

Ok. a=a0<e^(1/e)<e. a1=a^a0. So a1<e. a2=a^a1. a1<e. So a2<e. a3=a^a2. a2<e. So a3<e. a4=a^a3. a3<e. So a4<e. How long do you want me to keep this up?

9. Mar 23, 2007

### playboy

Okay...I see it now.

it is bounded above by e.

a
a^a
a^a^a
a^a^a^a
.
.
.
.

for 1 < a < e^(1/e)

10. Mar 23, 2007

### Dick

You've got it.

11. Mar 24, 2007

### playboy

I can follow your example, but when i practicly pund these numers in my calculator, i find that it is not bounded by e.

for example,

a = 1.3

a < a < 1^(1/e) = 1.444....

(1.3)^(1.3)^(1.3)^.....^(1.3) >>> e ... hence, not bounded by e.

12. Mar 25, 2007

### Dick

(%o1) f(x) := 1.3^x;
(%i2) f(1.3);
(%o2) 1.406456673237886
(%i3) f(%);
(%o3) 1.446293346285982
(%i4) f(%);
(%o4) 1.461488869891772
(%i5) f(%);
(%o5) 1.467327108831111
(%i6) f(%);
(%o6) 1.469576402415039
(%i7) f(%);
(%o7) 1.470443905739509
(%i8) f(%);
(%o8) 1.470778619613915
(%i9) f(%);
(%o9) 1.470907784591691
(%i10) f(%);
(%o10) 1.470957631962913
(%i11) f(%);
(%o11) 1.470976869521044
(%i12) f(%);
(%o12) 1.470984293924533
(%i13) f(%);
(%o13) 1.470987159254924
(%i14)

Are you sure the calculator isn't doing f(x):=x^1.3? That's quite different.

13. Mar 26, 2007

### playboy

>> a = 1.3

a =

1.3000

>> a^a^a^a^a^a^a^a^a^a^a^a

ans =

110.1660

>> (((((((a^a)^a)^a)^a)^a)^a)^a)

ans =

5.1877

Perhaps i am miss-understanding how you are computing it.

14. Mar 27, 2007

### NeoDevin

You are computing:
$$a_n = (a_{n-1})^a$$

You need:
$$a_n = a^{a_{n-1}}$$

It's a subtle difference in this case.

15. Mar 27, 2007

### Dick

(%i8) a^(a^(a^(a^(a^(a^(a^a))))));
(%o8) 1.470778619613915

16. Mar 27, 2007

### playboy

Why would my calculator/matlab give a different answer.

The reasoing is correct, so shouldnt the calculator give the same answer as your showing?

17. Mar 27, 2007

### Dick

Is the warranty still valid on the calculator? Seriously, did you explicitly parenthesize it like I did? Powers are not associative.