(Square root of 2) power (square root of 2) power...

Question

The problem is to calculate $A$: $$ A = \sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\dots}}} $$ Here, each term (except the first and second) is a power of the previous power. I applied my usual method: $$ A = \sqrt{2}^A $$ However, this seems incorrect because $A$ could be both 2 and 4. What's wrong with my method, and what would be the correct approach?

Answer 1

When looking for $A$ such that $A=\sqrt2^A$, you're actually assuming that $A$ converges. So, when you find $A=2$ or $A=4$, you proved: "If $A$ converges, then it's either equal to $2$ or $4$".

I have no idea about proving the convergence of this, but in my opinion, the sequence diverges

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EDIT: added some more info

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Let $A_1 = \sqrt{2}$ and for $n>1$, let $$A_n=\sqrt{2}^{A_{n-1}}$$

You have shown that if this sequence of $A_n$ converges as $n\to\infty$, then it must converge to $2$ or $4$.

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It is known that this sequence converges, since $$\frac{1}{e^e}\leq \sqrt{2}\leq \sqrt[e]{e}$$

Now, what is left to show is that it converges to $2$ and not $4$.

Answer 2

First of all we need to formalize. Define the following sequence:

$$\left\{\begin{aligned}x_1 &:= \sqrt{2}\\ x_{n+1} &:= \Big(\sqrt{2}\Big)^{x_n}\end{aligned}\right.$$

$x_1 = \sqrt{2}$, $x_2 = \sqrt{2}^{\sqrt{2}}$, $x_3 = \sqrt{2}^{\sqrt{2}^{\sqrt{2}}}$, ...

The problem is equivalent to compute $\lim\limits_{n \to \infty} x_n$.

Convergence

To prove that $x_n$ converges (i.e. the limit is a real number), show monotone and bounded by induction: $$x_n \le x_{n+1}\quad\text{and}\quad x_n \le 2$$

Hint:

$$x_{n-1}\ln(\sqrt{2}) \le x_{n}\ln(\sqrt{2}) \le 2\ln(\sqrt{2})$$

Computing the limit

Let $L$ the limit. i.e. $L := \lim_{n \to \infty} x_n$.

Using the continuity of $f(x) = \sqrt{2}^x$:

$$\lim_{n \to \infty} x_{n+1} = \Big(\sqrt{2}\Big)^{\lim\limits_{n\to\infty} x_n}$$

i.e. $$L = \sqrt{2}^L$$

and $$x_n\le 2 \implies L \le 2$$

Therefore $L$ is a solution of $\sqrt{2}^x = x$ with $x \le 2$.

Finally we need to show the only solution is $L = 2$, to get $$\sqrt{2}^{\sqrt{2}^{ \sqrt{2}^{.^{.^.}} } } = 2$$

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For $x \le 2$,

$$x \le 2$$ $$x\ln(\sqrt{2}) \le 2\ln(\sqrt{2})$$ $$\sqrt{2}^x \le \sqrt{2}^2$$ $$\ln(\sqrt{2})\sqrt{2}^x - 1 \le \ln(\sqrt{2})\sqrt{2}^2 - 1$$

By monotonicity of the integral, $$\underbrace{\int_{x}^2 \ln(\sqrt{2})\sqrt{2}^t - 1 \,dt}_{0 - (\sqrt{2}^x - x)} \le \Big(\underbrace{\ln(\sqrt{2})\sqrt{2}^2 - 1}_{\ln(2)-1}\Big)(2-x) $$

Then, $$\Big(1 - \ln(2)\Big)(2-x) \le \sqrt{2}^x - x$$

For $x \lneqq 2$, $$0 < \underbrace{\Big(1 - \ln(2)\Big)}_{> 0}\underbrace{(2-x)}_{>0 }\le \sqrt{2}^x - x$$ because $e > 2 \implies 1 > \ln(2) \implies 1-\ln(2) > 0$.

In summary, the only solution is $x = 2$ because in other case $x < \sqrt{2}^x$. i.e. $x \neq \sqrt{2}^x$.

Answer 3

We can show that the sequence converges with a proof by induction.

For $n =1$,we have that $\sqrt{2} < 2$. Assume that for some $n > 1$, $\sqrt{2} \uparrow \uparrow n < 2$. Then we have that

$$\sqrt{2} \uparrow \uparrow (n + 1) = \sqrt{2}^{\sqrt{2} \uparrow \uparrow n} < \sqrt{2}^2 = 2$$

Therefore the sequence is monotone (we know $x^y > x$ for $x > 1, y > 1$), and bounded, and therefore converges.