Power series representation of arctangent: fails to converge everywhere

Your insistence "let's keep everything in real variables, please" is precisely the problem: the cause of the finite radius of convergence is due to the function's behavior in $\mathbf C$, not $\mathbf R$.

A much simpler example than $\arctan x$ is $1/(1+x^2)$, which is defined and infinitely differentiable on the whole real line but its power series at $0$ (a geometric series with $-x^2$ in place of $x$) has radius of convergence $1$, not $\infty$. To use your language, "there is an obvious discontinuity that you bump into as you work your way out from the center," namely at $x = \pm i$ where the function blows up. In fact, if you expand $1/(1+x^2)$ into a power series at a real number $a$, not necessarily at $0$, the radius of convergence will be $\sqrt{a^2+1} = |a-i|$ -- the distance from the center out to $i$. This phenomenon is bewildering if you refuse to use complex numbers and extremely clear if you use them. Choose wisely.

If $f(x)$ is a rational function in reduced form with a nonconstant denominator and its denominator does not vanish at $a$, its power series at $a$ has radius of convergence $|a-\rho|$ where $\rho$ is the root of the denominator in $\mathbf C$ that is closest to $a$. This simple geometric result can not be explained in terms of real variables if the roots of the denominator are not all real.

To reinforce how poorly the real numbers are compared to the complex numbers as a predictive tool for the radius of convergence, there are functions $\mathbf R \rightarrow \mathbf R$ that are infinitely differentiable on the whole real line but their power series at each real number $a$ has radius of convergence zero for all $a$ in $\mathbf R$.

Strictly speaking, the real numbers have enough information in principle to compute the radius of convergence $R$ of a power series $\sum c_n(x-a)^n$ with all real coefficients using Hadamard's formula $1/R = \varlimsup\limits_{n\to\infty} \sqrt[n]{|c_n|}$, but this formula is often not feasible to compute in practice.

Let $z$ be an arbitrary complex number. Then $\arctan(z) = \frac12 i \log(1 - i z) - \frac12 i \log(1 + i z).$

Notice that $1-iz=0$ if $z=-i,$ and $1+iz=0$ if $z=i.$ So $\arctan(z)$ has singularities at $z=-i$ and $z=i.$

A thing about the radius of convergence in complex numbers is that if you plot it on an Argand diagram, it really is a radius of a circle. Inside the circle the series always converges, outside it diverges. The series of the real-number function $\arctan(x)$ is just $\arctan(z)$ restricted to the real number line, so it has the same behavior.