Homework Help: Tower of sqrt(2)

1. Jun 10, 2007

quasar987

1. The problem statement, all variables and given/known data
Let $a_1=\sqrt{2}$, $a_2=(\sqrt{2})^{a_1}$,... $a_{n+1}=(\sqrt{2})^{a_n}$. Show that a_n-->2. you may use any relevant fact from calculus.

3. The attempt at a solution
I noticed that $\ln(a_1)=\ln(\sqrt{2})=\ln(2)/2$, $$\ln(a_2)=\ln((\sqrt{2})^{\ln(2)/2})=(\ln(2)/2)^2$$, ..., $\ln(a_n)=(\ln(2)/2)^n$. But ln(2) < 2. So ln(2)/2 < 1, and thus ln(a_n)-->0 <==> a_n-->1. What's wrong?

Last edited: Jun 10, 2007
2. Jun 10, 2007

Hurkyl

Staff Emeritus
$$\log a_2 = \log (\sqrt{2})^{a_1}$$

not

$$\log a_2 = \log (\sqrt{2})^{\log a_1}$$

3. Jun 10, 2007

matt grime

we know a_n= (sqrt(2))^{a_{n-1}}

eugh.

Suppose there is a limit. Let that limit be a. Can you show from this that a=2? I.e. what equality does a satisfy, and thus what is a (by inspection).

Now we sjust need to show that a_n converges to anything? What are the only things we know converge for sure? increasing bounded above sequences, or decreasing bounded below ones.

4. Jun 10, 2007

Pseudo Statistic

Did someone say... proof by induction!?

5. Jun 10, 2007

NateTG

One way to do this is to look at the sequence:
$$b_n=2-(a_n)$$
So if $b_n \rightarrow 0$ you're good. Maybe you can look at:
$$\frac{b_n}{b_{n+1}}$$.

6. Jun 10, 2007

quasar987

Nice. Suppoes a_n-->a. Then $a_{n+1}=\sqrt{2}^{a_n}\rightarrow a$. But also, $\sqrt{2}^{a_n}\rightarrow \sqrt{2}^a$. So $\sqrt{2}^a=a$. So a=2.

Did you follow through with that idea? Cuz it did cross my mind, but I rejected it quickly because that $$\frac{b_n}{b_{n+1}}$$ fraction is not too cooperative as far as i can see.

Last edited: Jun 11, 2007
7. Jun 10, 2007

quasar987

How could we show the sequence is bounded though? :/

8. Jun 10, 2007

NateTG

$$b_{n+1}=2-a_{n+1}=2-\sqrt{2}^{2-b_n}=2-\frac{2}{\sqrt{2}^{b_n}}$$
I'm pretty sure it's possible to show that
$$0 < \frac{2-\frac{2}{\sqrt{2}^{b_n}}}{b_n} < 1$$