Any closed form for series like $$F(x)=\sum\_{p=2}^{\infty}x^p,\quad p\text{ is prime}$$ or $$F(x)=\sum\_{i=0}^{\infty}x^{i!}\quad ?$$


More generally, we can obtain a power series from decimal expansion of a number $r$ (with $0< r<1$) by replacing $$\left(\frac{1}{10}\right)^i$$ with $$x^i$$ like $$\frac{1}{3}=3\left(\frac{1}{10}\right)^1+3\left(\frac{1}{10}\right)^2+\dotsb 3\left(\frac{1}{10}\right)^i+\dotsb.$$
In this example we obtain:
$$f(x)=\sum\_{i=1}^{\infty}3x^i.$$


* When $f(x)$ is convergent, what restriction do we have to put on $r$ (if $r$ is c.e. number) to make $f(x)$ have a closed form?
* When is $f(x)$ algebraic, or transcendental?



Any closed form for series like $F(x)=\sum\limits_{p=2}^{\infty}x^p,$ where $p$ is prime?
See this blog post: <https://uniformlyatrandom.wordpress.com/tag/power-series/>


which contains a proof of the result by Fatou:


A function whose power series expansion has integer coefficients and radius of convergence 1 is either rational (in $\mathbb{Q}(x)$) or transcendental (over $\mathbb{Q}(x)$).


If $r$ is rational, then the decimal expansion will be eventually periodic. So we have rational function. (Indeed this can be done explicitly)


Otherwise, when $r$ is irrational, then the resulting function cannot be rational (plug in $1/10$, then you get irrational number). Thus, we have transcendence of $f$.


In particular, your functions $F$ in the beginning are transcendental.
However, getting closed form will be extremely hard for those examples.


