As title suggests, I want to prove that $\sum_{i=1}^{\infty} 2^{-{i^{i}}}$ is not an algebraic number.
From the book I found this claim (Lectures on Discrete Geometry) it looks an immediate consequence of the fact that for every irrational an algebraic number, $\alpha$, there exists a natural number $D>0$ such that there are just finitely many pairs of integers $m,n$ such that \begin{equation}\left|\alpha - \frac{m}{n}\right|<\frac{1}{n^{D}}.\end{equation}
So I want to find infinitely many fractions satisfying the inequality for every natural number $D$ but I cannot find those fractions.
Indeed, the book is not such related with number theory is just a section about the geometry of numbers.
Let $\alpha= \sum_{i\geq 1} 2^{-i^i}$ and let $a_n$ be the sequence of partial sums (obviously $a_n$ is rational) and $a_n = m2^{-n^n}$ (for some integer $m$). By very crudely estimating the tail of the series we have that $0 < \alpha - a_n < 2\left( 2^{-(n+1)^{n+1}}\right)$. This upper bound decreases very fast, too fast for the Diophantine inequality to be satisfied for any fixed value of $D$.