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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 $2$.
Climbing Powers
Age 16 to 18
Challenge Level 
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:
$({\sqrt 2}^{\sqrt 2})^{\sqrt 2} =
{\sqrt 2}^{\sqrt 2 \times \sqrt 2} = {\sqrt 2}^2 =2$
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:
${\sqrt 2}^{({\sqrt 2}^{\sqrt 2})}
= {\sqrt 2} ^{1.63...} = 1.76 $ to $2$ decimal places.
So
${\sqrt 2}^{({\sqrt 2}^{\sqrt
2})}< ({\sqrt 2}^{\sqrt 2})^{\sqrt 2}.$
Now consider
${\sqrt 2}^{{\sqrt 2}^{{\sqrt
2}^{{\sqrt 2}^{{\sqrt 2}^ {{\sqrt 2}^{..}}}}}} $
where the powers of $\sqrt 2$ go on forever. We have seen that we have two possibilities, namely
$X_1 = \lim x_n$ where $x_1 =
\sqrt 2,\ x_{n+1}= x_n^{\sqrt 2}$ or
$X_2 = \lim x_n$ where $x_1 =
\sqrt 2,\ x_{n+1} = (\sqrt 2)^{x_n}$.
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
$\log x_{n+1}=\sqrt{2}\,\log x_n,
\quad \log x_1 = \log \sqrt{2}.$
Thus
$\log x_{n+1} =
\big(\sqrt{2}\big)^n\log \sqrt{2},$
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
$x_{n+1} =
\big(\sqrt{2}\big)^{x_n} < \big(\sqrt{2}\big)^2 = 2$
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
$x_1 = \sqrt{2} <
(\sqrt{2})^{\sqrt{2}} = x_2.$
Now suppose that $x_{n-1} < x_n$. Then
${x_{n+1}\over x_n} =
{\sqrt{2}^{x_n}\over \sqrt{2}^{x_{n-1}}}
=\sqrt{2}^{x_n-x_{n-1}}.$
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 $2$.
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