Is the Sequence x_{n+1}=\sqrt{2x_n} Converging?

[cragar]

Homework Statement

Show that \sqrt{2},\sqrt{2\sqrt{2}},\sqrt{2\sqrt{2\sqrt{2}}}
converges and find the limit.

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.
I am 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.

 

[Dick]

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

 

[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...?

 

[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.

 

[alanlu]

You got it! One nitpick though:

okay its increasing but it will eventually converge because we keep multiplying it by something smaller.

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

 

[Dick]

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.

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.

 

[cragar]

So now I need to show that the sequence A-2&lt; \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, I am 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.

 

[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, s_n -> L iff for all e > 0, there is an N such that for all n >= N, ¦L - s_n¦ < e.

 

[cragar]

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