It's Not Infinity!

A naive approach is to let the answer be $x$ . Then, since the exponent itself is $x$ , it should satisfy $x = (\sqrt{2})^x$ . We can see that $x = 2$ is a solution, but we also find that $x = 4$ is another solution to this equation. Which one is correct? Moreover, it's even possible that the expression doesn't converge to any limit. We thus should go the hard way.

First, we need to formalize this expression. The simplest way to understand the question is to define the sequence $a_1 = \sqrt{2}, a_{n+1} = (\sqrt{2})^{a_n}$ for all $n \ge 1$ , and thus the question asks for $\lim_{n \to \infty} a_n$ .

First, we claim that the sequence $\{a_n\}$ is bounded above by 2. This will then rule out the possibility that the limit is 4; it doesn't even get anywhere close. This is simple to prove by induction. $a_1 = \sqrt{2} < 2$ is trivial, and $a_{n+1} = (\sqrt{2})^{a_n} < (\sqrt{2})^2 = 2$ is true (using the inductive hypothesis $a_n < 2$ ) because $x \mapsto (\sqrt{2})^x$ is an increasing function.

Now, we claim that the sequence $\{a_n\}$ is strictly increasing. This is harder because there doesn't seem to be any "clean" way. Instead, to prove that $a_{n+1} = (\sqrt{2})^a_n > a_n$ , we will show that the function $x \mapsto (\sqrt{2})^x - x$ is strictly positive for $x < 2$ . This comes from two results:

• It is differentiable everywhere, and its derivative is $x \mapsto \frac{\ln 2}{2} (\sqrt{2})^x - 1$ . Since $0 < \ln 2 < 1$ and $\frac{(\sqrt{2})^x}{2} < 1$ for $x < 2$ , we know that the derivative is negative for all $x < 2$ .
• At $x = 2$ , we have $(\sqrt{2})^x - x = 0$ .

What these tell is that for $x < 2$ , because the function must decrease as it approach to 2, it must be positive in this region; if it's non-positive, it cannot decrease to reach zero at $x = 2$ .

Since $x \mapsto (\sqrt{2})^x - x$ is strictly positive for $x < 2$ , plugging $x = a_n$ gives that $a_{n+1} - a_n = (\sqrt{2})^{a_n} - a_n > 0$ , so the sequence $\{a_n\}$ is strictly increasing.

Since the sequence is strictly increasing and bounded from above, it must converge by the monotonic sequence theorem , thus it has a limit. Only now that we can approach like in the naive way, finding that the limit is either 2 or 4. But our first result rules out 4, since we know the sequence doesn't even go above 2. So the limit must be $\boxed{2}$ .

Let $x=\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...}}}$ . Then, we have:

$\begin{aligned} x & =\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...}}}\\ & = \sqrt{2}^x \\ & = 2^\frac{x}{2} \\ \Rightarrow \log x & = \frac{x}{2} \log 2 \\ 2 \log x & = x \log 2 \\ \Rightarrow x & = \begin{cases} 4 & \text{but } \sqrt{2}^{4^{4^{...}}} \text{ diverges so unacceptable} \\ \boxed{2} \end{cases} \end{aligned}$

Can we simply write as following ? $sqrt{2}^{sqrt{2}^{sqrt{2}}} = lim_{n->+inf} 2^{1/2^n}$

With: $lim_{n->+inf} 1/2^n =0$

Then: $lim_{n->+inf} 2^{1/2^n} = 2^0=2$

let x=√2^√2^√2^.... Then $\sqrt{2}^x = x$ . We now raise both sides to $\frac{1}{x}$ . Then, $\sqrt{2}= x^\frac{1}{x}$ or $2^\frac{1}{2}= x^\frac{1}{x}$ Concequently $x=2$

It can be simplified as follows - $2^{\frac{1}{2}\times 2}$ It continues this way. in every level, the product is 1. so, it simplifies to - $2^{1}$ and so on( 1 at every power ). This is equal to 2!

let x=√2^√2^√2^..... Then, x=√2^x Taking log; log x = x log √2 or, log x = x/2 log 2 or, log x /x = log 2 /2 or, log x^1/x = log2^1/2 Taking antilog; x^1/x = 2^1/2 By comparison; x = 2.

i hav a more weird soln ... (i am not sure whether i did the right way though my answer is right)

as Chew-Seong Sir did below, i got

root(2)^x = x so .. root(2) = x^(1/x) taking LOGARITHM on both sides; (1/2)(log 2) = (1/x)(log x)

hence x = 2 (by observation)

AM I RIGHT???