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

  • Thread starter Thread starter kylera
  • Start date Start date
  • Tags Tags
    Limit Sequence
kylera
Messages
40
Reaction score
0

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!
 
Last edited:
Physics news on Phys.org
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.
 
Yeah I believe Borek is right.
 
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.
 
Last edited:
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

Similar threads

Replies
8
Views
2K
Replies
1
Views
1K
Replies
7
Views
984
Replies
9
Views
2K
Replies
2
Views
1K
Replies
8
Views
2K
Back
Top