# 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
$$0 < \frac{2-\frac{2}{\sqrt{2}^{b_n}}}{b_n} < 1$$