Proving Convergence of a Sequence with Upper Bound of 2

[lokisapocalypse]

Hey guys,

I have a sequence, \sqrt{2}, \sqrt{2 \sqrt{2}}, \sqrt{2 \sqrt{2 \sqrt{2}}}, ...

Basically, the sequence is defined as x1 = root 2
x(n+1) = root (2 * xn).

I need to show that this sequence converges and find the limit.

I proved by induction that this sequence increases. Since it increases, its bounded below by root 2. I need to show that it is bounded above by 2. Then I can use the Monotone Convergence Theorem to show that this sequence converges.

Any ideas?

 

[Justin Lazear]

Suppose the sequence is given in terms of n by

a_n = 2^{ \frac{2^n - 1}{2^n}},

where the first term is given by a_1.

Show that it satisfies the recurrence relation.

--J

 

[lokisapocalypse]

Thanks I got it from that. But can someone tell me how to do the root thing? Is the code LaTeX code or what is it?

Also, how can I prove the limit of that sequence = 2?

Is there some theorem that says that the limit of an increasing bounded sequence is equal to the sup of that sequence?

 

[lokisapocalypse]

Oh nevermind about the limit part. It was proved in the book with the Monotone Convergence Theorem. I just didn't see it right away.

 

[Justin Lazear]

\sqrt{2 a_n} gives \sqrt{2a_n}.

--J

 

[lokisapocalypse]

Okay I guess I am just stupid then. Where exactly do you type \sqrt{2 a_n} ? I thought it was to surround it by CODE tags but that didn't do it and I tried typing it just by itself, with and without the \. What do I do then?

 

[Justin Lazear]

[ tex ] \sqrt{2a_n}[ /tex ]
\sqrt{2a_n}

and

[ itex ] \sqrt{2a_n}[ /itex ]
\sqrt{2a_n}

Remove the spaces from the tags to get them to work.

--J

 

[lokisapocalypse]

Okay got it. Thanks for all your help.