TITLE: Repeated exponentiation inequality
QUESTION [5 upvotes]: I am trying to understand a proof of the fact that the number
$$ \mu = 1 + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^{2^2}} + \dots$$ is transcendental.
(For convenience I'll write $2 \uparrow\uparrow k$ for the powers of $2$ stacked $k$ times. So $\mu = \sum_{k = 0}^\infty \frac{1}{2 \uparrow \uparrow k}$).
The method involves showing that it is rationally approximable to any integer $m$, and I mostly understand how it works... but it seems to rely on two inequalities that are stated without proof that I'm struggling to formalise.
They are: (1)
$$
\sum_{k = (n + 1)}^{\infty} \frac{1}{2 \uparrow \uparrow k} < \frac{2}{2 \uparrow \uparrow (n + 1)}
$$
and then (2)
$$
\frac{2}{2 \uparrow \uparrow (n + 1)} < \frac{2}{(2 \uparrow \uparrow n)^{n + 1}}
$$
For the first one, my guess is we would start by saying:
$$
\sum_{k = (n + 1)}^{\infty} \frac{1}{2 \uparrow \uparrow k}= \frac{1}{2 \uparrow \uparrow (n + 1)} + \underbrace{\frac{1}{2 \uparrow \uparrow (n + 2)} + \frac{1}{2 \uparrow \uparrow (n + 3)} + \dots }
$$
and then we'd want to show that the sum of all the bracketed subsequent terms is less than the first term. I'm not sure how to state that though. It's fairly intuitively true because the repeated exponentiation obviously grows extremely quickly and so the terms rapidly contribute basically nothing to the sum, but I'd like a formal argument.
For the second one, I know the problem is reduced to showing $2 \uparrow \uparrow (n + 1) > (2 \uparrow \uparrow n)^{(n + 1)}$, i.e. $2^{2 \uparrow \uparrow n} > (2 \uparrow \uparrow n)^{(n + 1)}$Again, it's fairly intuitive that having the extremely large term in the exponent rather than the base is going to make it much bigger, but again, seems tricky to prove. I suppose I could do it by induction, but that seems like overkill.
Any insights would be much appreciated.

REPLY [2 votes]: A quick sketch.
For the first one, note that $2\uparrow\uparrow (n+k)>2^k(2\uparrow\uparrow n)$ so you can just sum up and use the fact that $\sum_{k=0}^{\infty} \frac{1}{2^k}=2$.
For the second one you need to show that $2\uparrow\uparrow (n+1)=2^{2\uparrow\uparrow n}>(2\uparrow\uparrow n)^{n+1}$. If you set $x:=2\uparrow \uparrow n$, this is $2^x>x^{n+1}$. Specifically, applying $\log$, you would need that $\frac{x}{\log{x}}>\frac{n+1}{\log{2}}$ and you can find an explicit $x_0$ such that for $x>x_0$ this holds. For example, you can take $x_0=(n+1)^2$ whenever $n\geq 10$ (as a loose bound). Then you just need to take $n$ big enough to make sure that $2\uparrow\uparrow n>(n+1)^2$, which is easy as $2\uparrow\uparrow n\geq 2^n$.