Puzzler

The quantity $\sqrt{2} ^ {\sqrt{2} ^ {\sqrt{2} ^ \ldots}}$ is defined as $\displaystyle\lim_{n \to \infty} u_n$ of the sequence $\{u_n\}$ satisfying $u_1 = \sqrt{2}$ , $u_{i+1} = \sqrt{2}^{u_i}$ .

These are the steps required:

Claim 1: $u_i < 2$ for all $i$ .

We can do induction on this. The base case, $u_1 = \sqrt{2} < 2$ , is correct. Suppose $u_i < 2$ . Since $\sqrt{2}^x$ is increasing (its derivative is always positive), by definition $\sqrt{2}^x < \sqrt{2}^y$ whenever $x < y$ . Thus $u_{i+1} = \sqrt{2}^{u_i} < \sqrt{2}^2 = 2$ , and thus the induction step is proven.

Claim 2: $u_{i+1} > u_i$ for all $i$ .

For this, we need to prove that $\sqrt{2}^x - x > 0$ for all $x < 2$ .

Observe the derivative of $\sqrt{2}^x - x$ ; that is, $\frac{\ln 2}{2} \sqrt{2}^x - 1$ . When $x < 2$ , $\sqrt{2}^x < 2$ , and so $\frac{\ln 2}{2} \sqrt{2}^x - 1 < \frac{\ln 2}{2} \cdot 2 - 1 = \ln 2 - 1 < \ln e - 1 = 0$ , thus this derivative is negative for all $x < 2$ and so the function $\sqrt{2}^x - x$ is decreasing for all $x < 2$ . Also, since $\sqrt{2}^x - x = 0$ when $x = 2$ , this establishes that $\sqrt{2}^x - x > 0$ when $x < 2$ (if for some $x$ it's negative, it contradicts the fact that it's decreasing). Thus we have proven that $\sqrt{2}^x > x$ for all $x < 2$ , and plugging $x = u_i$ and using Claim 1 gives the result.

Claim 3: $\{u_i\}$ converges.

This follows from Claim 1 and 2. It is increasing but bounded above, thus it converges. So it has a limit $L$ , which from Claim 1 cannot be larger than $2$ (otherwise the function doesn't even approach $L$ ).

Claim 4: $L = 2$ .

This is straightforward. Since $u_{i+1} = \sqrt{2}^{u_i}$ , letting $i \to \infty$ gives $u_i \to L$ , and so $L = \sqrt{2}^L$ . Taking the intermediate result for in Claim 2, we cannot have $L < 2$ as it would mean $\sqrt{2}^L - L > 0$ , and we cannot have $L > 2$ from Claim 3. And since $L = 2$ satisfies the equation, this is the sought limit.

Thus the answer is $\boxed{2}$ .

Let the desired quantity be $x$ . Then $x = \sqrt2^{x}$ . By inspection, $x=2$ satisfies this equation, so the answer is $\boxed{002}$ .

Let √2^√2^√2... be x, Or √2^x=x Or 2^x=x^2 Or x=2

This can be written as $2^{{{\frac{1}{2}}^2}^{\frac{1}{2}}}$ and it goes on

We can multiply the exponents with each other that leads us to $2^1$

Thus, $2^1 = \boxed{2}$