Convergence of a sequence.

1. Feb 18, 2012

cragar

1. The problem statement, all variables and given/known data
Show that $\sqrt{2},\sqrt{2\sqrt{2}},\sqrt{2\sqrt{2\sqrt{2}}}$
converges and find the limit.
3. The attempt at a solution
I can write it also like this correct
$2^{\frac{1}{2}},2^{\frac{1}{2}}2^{\frac{1}{4}},2^{\frac{1}{2}}2^{\frac{1}{4}}2^{\frac{1}{8}}$
so each time i multiply it by the new number it is getting closer to 1.
Every new number on the end of the sequence is getting closer to 1 that I am multiplying it by. so this sequence is bounded and decreasing therefore it must have a limit.
Im not sure what it converges to. but when I know I need to show that it that the sequence
A-b<ε , where A is the sequence and b is the limit.

2. Feb 18, 2012

Dick

No, it's not decreasing. Think about it again.

3. Feb 18, 2012

alanlu

I think it does converge. What happens when you multiply exponentials with a common base?

Edit: this sequence doesn't converge to 1, but...?

Last edited: Feb 18, 2012
4. Feb 18, 2012

cragar

okay its increasing but it will eventually converge because we keep multiplying it by something smaller. okay I see it now we have 2 raised to the sum of
$\sum \frac{1}{2^n}$
so it should converge to 2 .
all the exponents should add up to 1.
by a geometric series.

5. Feb 19, 2012

alanlu

You got it! One nitpick though:

Consider the sequence $$2^{-1}+2, (2^{-2}+2)(2^{-1}+2), (2^{-3}+2)(2^{-2}+2)(2^{-1}+2), \ldots$$

6. Feb 19, 2012

Dick

That's one way to do it. Another way is the call the limit $L$ and figure out what $\sqrt{ 2 L }$ must be if it converges.

7. Feb 19, 2012

cragar

So now I need to show that the sequence $A-2< \epsilon$
where A is the sequence. Do I need to write the sequence as 2 raised to a sum with a variable n and then use that to show that it converges. Your saying call the limit L and then put it in and then manipulate it to show that it converges, im mot exactly sure how that would work.
@alanlu: Are you using that other sequence as an example and I should figure out what it converges to.
thanks for all the help by the way.

Last edited: Feb 19, 2012
8. Feb 19, 2012

alanlu

The other sequence is an example of a series where the terms get multiplied by decreasing positive numbers, but the series diverges.

Further hint for the other route: can you write $\sqrt{2L}$ in terms of L?

Lastly, the definition of the limit of a sequence is a bit unwieldy for this situation: at best, you should use it to check your work. Also, it is stated, sn -> L iff for all e > 0, there is an N such that for all n >= N, ¦L - sn¦ < e.

9. Feb 23, 2012

cragar

Okay now I see what you mean by write in terms of L, write it recursively.
$x_{n+1}=\sqrt{2x_n}$

Last edited: Feb 23, 2012