Radius of Convergence and the Frobenius Method Consider the equation
$$4xy'' + 2y'+ y = 0$$
I know that $x=4$ is a regular singular point, and in the notation that my uni uses, we say that:
$$(x-x_0)^2 y'' + (x-x_0)p(x)y' + q(x)y = 0$$
where $y'' + b(x) y' +c(x) y = 0$
and $p(x) = (x-x_0)b(t)$
and $q(x) = (x-x_0)^2c(x) $
with $x_0$ being the singular point.
So in this instance, $p(x)$ would be $\frac1{2}$
and $q(x)$ is $\frac{x}{4}$.
In my notes it says that: 'The power series in $y_1$ and $y_2$ will converge
for $|x| < ∞$ since $p$ and $q$ have convergent power series in this interval. ' What does this mean? How does $p(x)=\frac1{2}$ and $q(x)=\frac{x}{4}$ relate to the convergence of the power series?
Any help would be appreciated. 
 A: $x=4$ is not a singular point. $\color{blue}{x=0}$ is the only singular point of this equation (and it is a regular singular point).
It's important to get that right since the next step is to appeal to the following theorem regarding the radius of convergence for power series solutions:

Theorem. If $x_0$ is a regular singular point of 
  $$y''+p(x)y'+q(x)y=0, \quad x>x_0,\tag{$*$}
$$
  then there exists at least one power series solution of the form 
  $$
\sum_{n=0}^\infty a_n(x-x_0)^{n+r},
$$
  where $r=r_1$ is the larger root of the indicial equation. Moreover, the series converges for all $x$ such that $0<x-x_0<R$, where $R$ is the distance from $x_0$ to the nearest other singular point (real or complex) of $(*)$.

In your problem, the singular point is at $x_0=0$ and since it is the only singular point, $R=\infty$. By the theorem, the interval of convergence is $0<x<\infty$.
The phrase about $p(x)$ and $q(x)$ is referring to the fact that we look at them to determine any and all singular points. And here the only one is at $x_0=0$.
