Tetration: Summation of ${1 \over{}^{n}2}$

Question

I'm going to get straight to the point with this question - Can you find a closed-form solution to this sum.

$$\sum_{n=1}^\infty \displaystyle {1 \over{}^{n}2}$$

(where ${}^{n}2$ represents the nth tetration of 2)

I know the sum converges by the comparison test with the sum: ($\sum_{n=1}^\infty \displaystyle {1 \over{2}^{n}}$), but I haven't had any luck finding a solution or partial sum.

Also, let me know if there is any theorem that somehow states sums like this are impossible to find closed form solutions of.

Answer 1

I doubt there's much you can do. It converges rapidly, so a numerical computation of this is fairly easy and gives

$$S\simeq0.1101000000000000_2=0.8125152587890625$$

It's easy to observe that the digits in base $2$ are not repeating, so this is an irrational number. Since it converges rapidly, Liouville also shows this to be transcendental.