(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

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.