The sum $\sum x^n$ is unbounded in $|x| \le 1$. Similarly if $p$ is prime then trivially $\sum x^p$ is also unbounded in $|x| < 1$ because all primes $> 2$ are odd so the lower bound would approach $-\infty$ as $x \to -1$ and the upper bound would approach $\infty$ as $x \to 1$.
I observed that if we take the summation over the set of composite numbers $c$ then we have a lower bound of about $-0.424$ at about $x = -0.539$. Heuristically this is because the sum over odd composite would negate the sum over even composites so we expect that at the suitable value of $x$ the total sum would reach a minima as seen the observations below.
Also observe that $\sum x^p$ has a zero at about $-0.63$ and $\sum x^c$ has a zero at about $-0.77$.
Question 1: Show that even though $\sum x^n$ does not have a non-trivial zero, both $\sum x^p$ and $\sum x^c$ must have a non-trivial zero? Trivial zero is $x = 0$.
Question 2: Show that $\sum x^c$ must have a lower bound? Is there a closed form expression of the lower bound?