Climbing powers

Well done Paul Jefferys, you got close to a complete solution here. We have to consider two different values of these climbing powers depending on the order of operations which can be shown by putting in brackets. We can define $2^{3^4}$ either as $(2^3)^4 = 2^{12}$ or as $ 2^{(3^4)} = 2^{81}$. In the same way there are two interpretations of ${\sqrt 2}^{{\sqrt 2}^{\sqrt 2}}$ The first of these is $f(f(\sqrt 2))$ where $f(x)=x^{\sqrt 2}$ which gives:



In the second case we get $g(g(\sqrt 2))$ where $g(x) = (\sqrt 2)^x$, and using a calculator to get an approximate value gives:



So



Now consider



where the powers of $\sqrt 2$ go on forever. We have seen that we have two possibilities, namely



N.B. Both iterations can be done on a calculator or computer: $X_1 = \lim x_n$ is equivalent to iterating $f(x) = x^{\sqrt 2}$ and $X_2 = \lim x_n$ is equivalent to iterating $g(x) = (\sqrt 2)^x$. If you do this experimentally, in each case starting with $x_1=\sqrt 2$, you will find that the first iteration appears to converge to infinity and the second appears to converge to $2$. We claim $X_1 = +\infty$.

Proof

We have $x_1=\sqrt{2}$ and $x_{n+1}=x_n^{\sqrt{2}}$; thus



Thus



and as $\log x_n \to +\infty$ as $n\to \infty$, we see that $x_n\to +\infty$. We now claim that $X_2=2$.

Proof

First we show that $x_n < 2$ for all $n$, and the proof is by induction. Clearly $x_1 < 2$. Now suppose that $x_n < 2$ and consider $x_{n+1}$. We have



as required. Hence (by induction) $x_n < 2$ for all $n$. Next, we show by induction that $x_n < x_{n+1}$. It is clear that



Now suppose that $x_{n-1} < x_n$. Then



Thus, as $x_n-x_{n-1}> 0$, we have $x_{n+1}/x_n > 1$ and hence $x_{n+1}> x_n$. This shows that $x_n$ is increasing with $n$, and that $x_n < 2$, and this is enough to see that $x_n$ converges to some number $X_2$, where $X_2\leq 2$. As $x_{n+1}=({\sqrt 2})^{x_n}$, if we let $n$ tend to infinity we see that $X_2$ is a solution of the equation $x=({\sqrt{2}})^x$. If we now plot the graphs of $y=x$ and $y=(\sqrt{2})^x$, we see that these two graphs meet at only two points, namely $(2,2)$ and $(4,4)$. Thus $X_2$ is either $2$ or $4$, and so it must be $2$.