Prove that this sequence converges

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  • #1
playboy

Homework Statement



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



Homework 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



The Attempt at a Solution



{[itex] a_{n} [/itex]} = {[itex]a, a^a, a^{a^a}, ...[/itex]}

Clearly, {[itex] a_{n} [/itex]} is monotonically increasing is is bounded below by a.

How do I show that it is bounded above?
 
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Answers and Replies

  • #2
147
0
try induction assuming that a(o) is less than the maximum a value...
 
  • #3
playboy
try induction assuming that a(o) is less than the maximum a value...

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

Clearly,

[itex]a < a^a < a^{a^a} < ...[/itex] ... So I am having a hard time explaining how this is bounded for a>1
 
  • #4
1,425
1
This doesn't mean anything. A sequence could be increasing and still be bounded. And [tex]e^{e-1} = 5.574941539 [/tex]. I don't think that's what you meant....
 
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  • #5
Dick
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This doesn't mean anything. A sequence could be increasing and still be bounded. And [tex]e^{e-1} = 5.574941539 [/tex]. I don't think that's what you meant....
He meant e^(1/e). The formatting didn't come out 100% clear.
 
  • #6
Dick
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If a<e^(1/e) and x<e, what can you tell me about a^x?
 
  • #7
playboy
He meant e^(1/e). The formatting didn't come out 100% clear.
Thank you for the correction Dick...I allways have trouble with this latex stuff :confused:


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

so a^x < e

hence,

:confused:

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


Can you please give me another push ?
 
  • #8
Dick
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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
playboy
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?
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
Dick
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You've got it.
 
  • #11
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
Dick
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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.
(%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
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
299
1
You are computing:
[tex] a_n = (a_{n-1})^a[/tex]

You need:
[tex] a_n = a^{a_{n-1}} [/tex]

It's a subtle difference in this case.
 
  • #15
Dick
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Which do you believe, your reasoning or your calculator?

(%i8) a^(a^(a^(a^(a^(a^(a^a))))));
(%o8) 1.470778619613915
 
  • #16
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
Dick
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Is the warranty still valid on the calculator? Seriously, did you explicitly parenthesize it like I did? Powers are not associative.
 

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