Proving irrationality of $\sum_{n = 1}^\infty 2^{−2^n}$ with basic concepts I've seen many posts about $$\sum_{n = 1}^\infty 2^{−2^n}$$ and many of them deal with irrationality of this limit. However, they all seem to try to prove its irrationality using theorems and irrationality tests like Liouville's theorem. I'm wondering if there is a way to use some basic arithmetic (or a little basic calculus if needed) to prove that this specific limit is irrational.
I tried to turn this into an infinite product, but it didn't seem to work. Any suggestions on how to approach this?
 A: There is a simple trick which I learnt from Mathologer, concerning an easy proof of the irrationality of $e$. It only works for nice series with interrelated summands, but it's probably one of the easier entirely self-contained proofs.
Put: $$\alpha:=\sum_{k=1}^\infty 2^{-2^k}$$
For any $m\in\Bbb N$, write $\alpha=N_m+T_m$ where: $$N_m:=\sum_{k=1}^m2^{-2^{k}},\,T_m:=\sum_{k=m+1}^\infty 2^{-2^k}$$
If $\alpha=p/q$ as a rational (in lowest terms), then: $$2^{2^m}qT_m=2^{2^m}p-2^{2^m}qN_m\in\Bbb Z$$
However, we can easily bound: $$0<2^{2^m}T_m<\sum_{k=1}^\infty2^{-k2^m}=2^{-2^m}\cdot\frac{1}{1-2^{-2^m}}$$Which obviously tends to zero as $m\to\infty$ (and very quickly, too). For any fixed $q$, I can choose a special (in fact, infinitely many 'special') $m\in\Bbb N$ such that: $$0<|2^{2^m}qT_m|<1$$In contradiction to it being an integer. $\blacksquare$
Loosely speaking, the proof of transcendence should be the same in spirit (I haven't checked, but $\alpha$ should be a Liouville number).
A: Lemma: Let $x\in\Bbb R.$ If for every $\delta >0$ there exist $a,b\in \Bbb Z$ with $0<|x-a/b|<\delta / |b|,$ then $x\not\in\Bbb Q.$ Proof: By contradiction. Suppose  $c,d\in\Bbb Z$ and $x=c/d.$  Take $0<\delta <1/|d|$ and take $a,b\in\Bbb Z$ such that
$0<x-a/b|<\delta / |b|.$
Then $ 0<|cb-ad|<|d|\delta $
so $ 1\le |cb-ad|<|d|\delta\;$ (...because $0<|cb-ad|\in\Bbb Z\implies 1\le |cb-ad|)$
so $1/|d|\le \delta .$ But we took $\delta <1/|d|.$ QED.
Let $x=\sum_{n=1}^{\infty}2^{2^{-n}}.$ Given $\delta > 0,$ take $m\in\Bbb N$ large enough that $2^{2^{-m}}<\delta.$ Let $b=2^{2^n}$ and let $a/b=\sum_{n=1}^m 2^{2^{-n}}.$ Then $a,b\in\Bbb N$ and
$0<|x-a/b|=\sum_{n=m+1}^{\infty}2^{2^{-n}}<1/b^2<\delta /b=\delta /|b|.$
Remarks. Note the crucial role of the "$0<$" in the statement of the Lemma. Observe how we can sharpen this to "$\le 1$" in the proof because we converted to a condition about an integer.
