Calculate $\sum\limits_{i=0}^\infty(2^{2^{(-i)}}-1)$ 
Calculate $\sum\limits_{i=0}^\infty(2^{2^{(-i)}}-1)$

By a Python program showed below, I can calculate that it is about $1.7818$($1.7818386318393353172743493971315840477111345642771002580086952600435098253141880734976160498115058888973$ by a comment). I've worked on this problem for days, but I can't solve it. Can anyone help me?
Update: I'm a student who is interested in math. I came out with this problem about one year ago, but I think it's too difficult for me.
Also, $2^{2^{(-i)}}-1\le2^{(-i)}$, so it is a convergent serie.
Update 2: Why $2^{2^{(-i)}}-1\le2^{(-i)}$?
Let $x$ be $2^{(-i)}$, $i\ge0$, so $x\ge1$, then $2^{2^{(-i)}}-1-2^{(-i)}=2^x-1-x$.
When $x=1$, $2^x-1-x=0$, and $\frac{d(2^x-1-x)}{dx}=2^x\ln-1>0$ for $x\ge1$, so $2^{2^{(-i)}}-1-2^{(-i)}=2^x-1-x>0$, that means $2^{2^{(-i)}}-1\le2^{(-i)}$.
Update 5 on 2021-07-02: Anyone who can show whether the sum is rational can win the bounty.
from math import *
s = 0
a = 2
while a != 1:
    s += a - 1
    a = sqrt(a)
print(s)

 A: I don't think there is a closed form for this sum, but it can be rewritten to converge even more quickly
$$
\begin{align}
\sum_{k=0}^\infty\left(2^{2^{-k}}-1\right)
&=\sum_{k=0}^\infty\left(e^{\log(2)\,2^{-k}}-1\right)\tag1\\
&=\sum_{j=1}^\infty\sum_{k=0}^\infty\frac{\left(\log(2)\,2^{-k}\right)^j}{j!}\tag2\\
&=\sum_{j=1}^\infty\frac{\log(2)^j}{j!}\left(1+\frac1{2^j-1}\right)\tag3\\
&=1+\sum_{j=1}^\infty\frac{\log(2)^j}{j!}\frac1{2^j-1}\tag4\\[6pt]
&=1.7818386318393353172743493971315840477111\tag5
\end{align}
$$
Explanation:
$(1)$: $2=e^{\log(2)}$
$(2)$: use the series for $e^x-1$
$(3)$: sum the geometric series in $k$
$(4)$: recognize the series for $e^{\log(2)}-1=1$
$(5)$: evaluate to $40$ places
The number of digits for $n$ terms of the original series is
$$
n\log_{10}(2)+O(1)\sim n\log_{10}(2)\tag6
$$
whereas the number of digits for $n$ terms of the series in $(4)$ is
$$
\left(n+\frac12\right)\log_{10}\left(\frac{2n}{e\log(2)}\right)+O(1)\sim n\log_{10}(n)\tag7
$$
and it is easy to see that $(7)$ grows faster than $(6)$.
