When does a Fuch's type 2nd order ODE not have a singularity at infinity? I know that any second order linear ode
$$w''+p(z)w'+q(z)w=0$$
is of Fuchs type (ie, coefficients meromorphic with singular points at $z_0,...,z_n,\infty$ all regular) if
$$p(z)=\sum\frac{p_j}{z-z_j}, q(z)=\sum \frac{q_j}{(z-z_j)^2}+\frac{r_j}{z-z_j},\quad \sum r_j=0$$
where the sums are over the singular points $z_0,...,z_n$.
I wish to show that there is no singularity at $\infty$ if $$2-\sum p_j=\sum (q_j+r_jz_j)=\sum z_j(2q_j+r_jz_j)=0$$
To do this, I'm guessing that I have to make the substitution $u=1/z$ to obtain
$$w''(u)+(2/u-p(1/u)/u^2)w'(u)+1/u^4q(1/u)w(u)=0$$
but now I'm getting stuck. Anyone have any insights?
 A: It is always helpful to not reuse the same variable names for a different object. Define $\tilde w(u)=w(1/u)$, then
$$
\tilde w'(u)=-\frac1{u^2}w'(1/u),~~~ \tilde w''(u)=\frac1{u^4}w''(1/u)+\frac{2}{u^3}w'(1/u),
$$
so that the transformed equation is
$$
0=u^4\tilde w''(u)-2u^3\tilde w'(u)-p(1/u)u^2\tilde w'(u)+q(1/u)w(u),
\\~\\
0=\tilde w''(u)-\left(\frac2u+\frac{p(1/u)}{u^2}\right)\tilde w'(u)+\frac{q(1/u)}{u^4}w(u)
$$
Now the solution $w$ extends to infinity if the coefficients in the last form of the equation are at least continuous in and around $u=0$, in our context this automatically implies holomorphic around $u=0$.
For
$$
\frac1u\left(2+\sum\frac{p_j}{1-z_ju}\right)
$$
to be continuous in $u=0$ the second factor has to evaluate to zero at $u=0$, giving the first claimed condition.
The second coefficient is
$$
\frac1{u^2}\sum \frac{q_j}{(1-z_ju)^2}+\frac1{u^3}\sum \frac{r_j}{1-z_ju}
\\
=
\frac1{u^2}\sum q_j(1+2(z_ju)+3(z_ju)^2+...)+\frac1{u^3}r_j(1+z_ju+(z_ju)^2+(z_ju)^3...)
$$
using geometric or binomial series. Now collect terms of equal degree and demand that the coefficients of negative powers vanish,
$$
0=\sum r_j=\sum (q_j+r_jz_j)=\sum z_j(2q_j+r_jz_j).
$$
Was there a transmission error in the middle term?
