Fuchsian equation

Remarks

At a finite point $a$, the condition says that each $p_k$ has at most a pole of order $k$ at $a$. At $\infty$, it says that $z^k p_k(z)$ stays bounded as $z \to \infty$, for each $k$. Since the $p_k$ are rational, there are only finitely many singular points.

Example: the hypergeometric equation $z (1 - z) y'' + \left( c - (a + b + 1) z \right) y' - a b\, y = 0$ is Fuchsian. Its normalised coefficients are $p = \frac{c - (a + b + 1) z}{z (1 - z)}$ and $q = \frac{-a b}{z (1 - z)}$. They have simple poles at $0$ and $1$, and $z p(z) \to a + b + 1$ and $z^2 q(z) \to a b$ as $z \to \infty$. So the singular points $0$, $1$ and $\infty$ are all regular.

Non-example: the Bessel equation $z^2 y'' + z y' + (z^2 - \nu^2) y = 0$ has $q(z) = 1 - \frac{\nu^2}{z^2}$. Then $z^2 q(z) = z^2 - \nu^2$ is unbounded as $z \to \infty$, so $\infty$ is an irregular singular point and the equation is not Fuchsian.