Find Limit of Sequence {sqrt(2), sqrt(2sqrt(2)), ...}

[kylera]

Homework Statement

Find the limit of the sequence
{ sqrt(2), sqrt(2sqrt(2)), sqrt(2sqrt(2sqrt(2))) ... }

Homework Equations

Limit Laws?

The Attempt at a Solution

I wrote out the first five values in the sequence and came to the conclusion that this sequence could be written out as

$$A_{n} = 2^\frac{2^{n}-1}{2^{n}}$$

I then took $$\frac{2^{n}-1}{2^{n}}$$, broke it down to $$1 - \frac{1}{2^{n}}$$ which allowed me to rewrite the equation to $$2\times2^\frac{-1}{2^{n}}$$. Ignoring the 2 for now, I re-worked the fraction exponent and resulted with $$-(\frac{1}{2})^{n}$$ and made the value into a fraction $$\frac{1}{2^(\frac{1}{2})^{n}}$$.

Using the sheer power of what is known as the graphing calculator, I was able to determine that the limit of that equation is 1, and then multiplying 2 to it gave 2. Without a calculator, how can I lay out the steps?

Note to self: BUY A TABLET!

 

[Borek]

Won't it be enough to calculate limit of

$$1 - \frac 1 {2^n}$$

Seems rather obvious. But then I am mathematically challenged and could be I am missing some fine print.

 

[Defennder]

Yeah I believe Borek is right.

 

[zpconn]

I will outline a solution. The details are your job.

There's a much simpler way of writing the sequence as a recurrence relation. Use this way.

First show that each term is less than a certain constant. Next demonstrate that the sequence is increasing. Hence show the sequence is convergent with some undetermined limit L.

The continuity of a certain function (which one?) will allow you to take limits of both sides of the recurrence relation.

The exact value of the limit L should now be in sight. I'll leave the rest up to you.

I will outline a solution. The details are your job.

There's a much simpler way of writing the sequence as a recurrence relation. Use this way.

First show that each term is less than a certain constant. Next demonstrate that the sequence is increasing. Hence show the sequence is convergent with some undetermined limit L.

The continuity of a certain function (which one?) will allow you to take limits of both sides of the recurrence relation.

The exact value of the limit L should now be in sight. I'll leave the rest up to you.

If you're up for the challenge, you might also try to find the set of all x such that the sequence {x, x^x, x^(x^x), ...} is convergent.