Any closed form for series like $F(x)=\sum_{i=p}^{\infty}x^p$, $p$ is prime? 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 $r$ ($0< r<1$) by replacing $$\left(\frac{1}{10}\right)^i$$ with $$x^i$$ like $$\frac{1}{3}=3\left(\frac{1}{10}\right)^1+3\left(\frac{1}{10}\right)^2+\dotsb 3\left(\frac{1}{10}\right)^i+\dotsb,$$
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?
 A: Also, the classical Fabry gap theorem tells you that the unit circumference is the natural boundary. Meanwhile, all "elementary" functions can be analytically continued along almost every path on the plane, so give up all hopes for a closed formula of any sort...
A: See this blog post: https://uniformlyatrandom.wordpress.com/tag/power-series/
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.
A: Functions with a natural boundary tend to satisfy functional equations, and they are sometimes the unique solutions to those functional equations. While a system of functional equations is far from being a closed form expression, systems of functional equations are probably as close as we can get to closed form expressions for these kinds of otherwise pathological functions.
The following observations apply to the function $\sum_{k=0}^{\infty}z^{k!}$ and related functions.
Observation: Suppose that $f$ is a function holomorphic in a neighborhood of $0$ with $f(z)=\sum_{k=0}^{\infty}a_k z^{k}$. Let $m/n$ be a reduced rational number. The following are equivalent:

*

*$f(z\cdot e^{2\pi im/n})-f(z)$ is a polynomial.


*For all but finitely many $k$, $a_k=0$ or $k$ is a multiple of $n$ .


*There is some $k$ where $f$ satisfies the functional differential equation
$$\frac{d^k}{dz^k}\big(f(z\cdot e^{2\pi im/n})-f(z)\big)=0.$$
Observation: Suppose that $f(z)$ is a function that is analytic in a neighborhood of $z=0$. Let $(r_n)_{n=0}^\infty$ be a strictly increasing sequence of positive integers where $r_n$ is always a factor of $r_{n+1}$. Then $f(z)$ can be put in the form $\sum_{k=0}^\infty a_kz^{r_k}$ if and only if $f^{(r)}(0)=0$ for $r<r_0$ and for all $n\geq 0$, $f(z)$ satisfies the functional differential equation $$\frac{d^{r_{n+1}}}{dz^{r_{n+1}}}\big(f(z)-f(z\cdot\exp(2\pi i/r_{n+1}))\big)=0.$$
Suppose now that $f$ satisfies these conditions. Then
$$\frac{d^{r_n}}{dz^{r_n}}\big(f(z)-f(z\cdot\exp(2\pi i/r_{n+1}))\big)=a_n\cdot (r_n)!\cdot\big(1-\exp(\frac{2\pi i r_n}{r_{n+1}})\big)$$ for each $n\geq 0$, and the function $f(z)$ is completely determined by these functional differential equations.
